MAT 275 - MODERN DIFFERENTIAL EQUATIONS -
Conditions for exactness and methods for finding solutions
Practice Material - Set 2
1. Suppose we have a differential equation of the form M(x, y)dx +N(x, y)dy = 0. Determine
the conditions for exactness and find the solution if the conditions are met.
Ans. To determine the conditions for exactness and find the solution, we follow these steps:
1. The condition for exactness is that ∂M
∂y =∂N
∂x . 2. If the condition is met, we can find a
function f(x, y)such that df =Mdx +Ndy by integrating Mwith respect to xand Nwith
respect to y. 3. Once we find f(x, y), the general solution is given by f(x, y) = C, where Cis
a constant.
Let’s now apply these steps to a specific example.
2. Let M(x, y)=2xy2−cos(x)and N(x, y) = x2y−y+sin(y). Determine if the following
differential equation is exact, and if so, find the general solution:
∂u
∂x = 2xy2−cos(x)and ∂u
∂y =x2y−y+sin(y)
Ans. To check for exactness, we need to verify that ∂M
∂y =∂N
∂x . 1. Calculate the partial
derivatives: ∂M
∂y = 4xy and ∂N
∂x = 2xy
2. Check for exactness: Since ∂M
∂y =∂N
∂x , the differential equation is not exact. 3. Find an
integrating factor µ(x, y)where µ(x, y) = e∫My−Nx
Ndx. 4. Calculate the integrating factor:
µ(x, y) = e∫4xy−2xy
x2y−y+sin(y)dx =e∫2xy
x2y−y+sin(y)dx
5. Simplify the integrating factor:
µ(x, y) = e∫2y
x−y+sin(y)
x
dx
6. Attempt to find a solution using the integrating factor:
µ(x, y)(∂u
∂x dx +∂u
∂y dy)=µ(x, y)M(x, y)dx +µ(x, y)N(x, y)dy
7. Since the differential equation is not exact and finding a suitable integrating factor is chal-
lenging in this case, further analysis or alternative methods may be needed to find a solution.
3. Let M(x, y)dx +N(x, y)dy = 0 be a first-order ordinary differential equation. Determine
whether the equation is exact. If it is exact, find a solution f(x, y) = c, where cis a constant. If
it is not exact, find an integrating factor µ(x, y)that makes it exact, and then find the solution.
Ans. The given ODE is M(x, y)dx +N(x, y)dy = 0.
1. To check for exactness, we compute the partial derivatives ∂M
∂y and ∂N
∂x .
∂M
∂y =∂N
∂x
2. Since the equation is exact, we look for a function f(x, y)such that fx=Mand fy=N.
Integrating Mwith respect to xgives f(x, y) = ∫M dx +h(y).
3. To find the remaining term h(y), we differentiate f(x, y)with respect to yand set it equal
to N.∂f
∂y =∂
∂y (∫M dx)+h′(y) = N
So, h′(y) = N−∂
∂y (∫M dx)
4. Solving for h(y), we find f(x, y) = ∫M dx +∫(N−∂
∂y (∫M dx))dy =c, where cis
a constant.
Hence, the solution to the differential equation is f(x, y) = c.
4. Let M=y2dx + 2xydy be a differential form. Determine if Mis exact. If it is exact, find a
potential function f(x, y)such that df =M.
Ans. To determine if the given differential form M=y2dx + 2xydy is exact, we will check if
∂M
∂y =∂N
∂x where Nis the unknown function associated with M. If it is exact, we will then find
f(x, y)such that df =M. 1. Find ∂M
∂y :∂M
∂y = 2y2. Find ∂N
∂x : Comparing ∂N
∂x with ∂M
∂y , we have:
∂N
∂x = 2yIntegrating this with respect to xgives N= 2xy +h(y)where h(y)is the constant of
integration. 3. Find ∂N
∂y : Taking the partial derivative of N= 2xy +h(y)with respect to y, we
get: ∂N
∂y = 2x+h′(y)4. Comparing with M=y2dx + 2xydy, we have ∂N
∂y = 2x. Equating the
two expressions, we find h′(y)=0, which implies h(y) = Cfor some constant C. 5. Therefore,
N= 2xy +C. Since Nis the potential function associated with M, the given differential form
Mis exact and the potential function is f(x, y) = 2xy +Cwhere Cis a constant.
5. Find the value of kfor which the following differential equation is exact:
(kx +y2)dx + (2y)dy = 0
Ans. To check for exactness, we compute the partial derivatives of the terms with respect to y
and x, respectively.
1. Compute the partial derivative with respect to y:
∂(kx +y2)
∂y = 2y
2. Compute the partial derivative with respect to x:
∂(2y)
∂x = 0
Since the cross-partial derivatives are equal (∂(2y)
∂x =∂(kx+y2)
∂y ), the equation is exact.
3. Write the equation in the exact form:
(kx +y2)dx + (2y)dy = 0
M(x, y)dx +N(x, y)dy = 0
4. Apply the exactness condition: For an exact differential equation, the following must
hold: ∂M
∂y =∂N
∂x
Substitute M=kx +y2and N= 2yinto the above formula:
∂(kx +y2)
∂y =∂(2y)
∂x
2y= 0
5. Solve for k:Since this equation must hold for all xand y, we find that k= 0 to satisfy
the exactness condition. Thus, the value of kfor which the given differential equation is exact is
k= 0.
6. Question:
Consider the following differential equation:
(3x2y−2x3)dx + (2y−x3)dy = 0
Determine whether the given equation is exact. If it is exact, find a function F(x, y)such
that ∂F
∂x = 3x2y−2x3and ∂F
∂y = 2y−x3. Then, find the general solution of the differential
equation.
Ans. Step-by-step solution:
1. Check for exactness: To determine if the given differential equation is exact, we calculate
the partial derivatives of Mand Nwith respect to yand x, respectively. Let M= 3x2y−2x3
and N= 2y−x3.
2. Calculate ∂M
∂y and ∂N
∂x :
∂M
∂y = 3x2
∂N
∂x =−3x2
3. Check for exactness: Since ∂M
∂y =∂N
∂x , the equation is not exact.
4. Find integrating factor: To make the equation exact, we need to find the integrating factor
µ(x)which allows us to write the equation as (µM )dx + (µN )dy = 0 where ∂(µM)
∂y =∂(µN )
∂x .
The integrating factor is given by µ=e∫(∂N
∂x
−∂M
∂y
N)dx
.
5. Calculate the integrating factor:
µ=e∫(−3x2−3x2
2y−x3)dx
µ=e∫(−6x2
2y−x3)dx
6. Recalculate the differential equation with the integrating factor: Multiplying the given
differential equation by the integrating factor µ, we get:
e∫−6x2
2y−x3dx[(3x2y−2x3)dx + (2y−x3)dy] = 0
7. Solve the differential equation: The above equation is exact, with F(x, y) = ∫e∫−6x2
2y−x3dx dx+
C, where Cis the constant of integration. Solving for F(x, y)gives the general solution of the
differential equation.
7. Determine whether the following differential equation is exact, and if so, find the general
solution:
(2x+y2)dx + (2y−2xy)dy = 0
Ans. To determine if the given differential equation is exact, we check if the mixed partial
derivatives of the coefficients are equal. If so, we can find the general solution by integrating the
equation with respect to x and y one by one.
1. Check for exactness: Let M= 2x+y2and N= 2y−2xy. We calculate the partial
derivatives: ∂M
∂y = 2yand ∂N
∂x =−2y
Since ∂M
∂y =∂N
∂x , the equation is exact.
2. Find the general solution: Since the equation is exact, we need to find a function
ψ(x, y)such that:
∂ψ
∂x = 2x+y2and ∂ψ
∂y = 2y−2xy
Integrating the first equation with respect to x, we get:
ψ(x, y) = ∫(2x+y2)dx =x2+xy2+h(y)
where h(y)is a function of y that we will determine by differentiating with respect to y and
comparing to the second equation:
∂ψ
∂y =x(2y)+2xy +h′(y) = 2y−2xy
Equating coefficients, we get:
h′(y) = 2y
Integrating with respect to y, we find:
h(y) = y2+C
where C is the constant of integration.
3. Final solution: Thus, the general solution to the differential equation is:
x2+xy2+y2+C= 0
8. Consider the differential equation:
(3x2y+cos(y))dx + (x3−sin(x))dy = 0
Determine if the given differential equation is exact. If it is exact, find the general solution.
Ans. To determine if the given differential equation is exact, we need to check if
∂M
∂y =∂N
∂x
where the given differential equation is in the form M dx +N dy = 0.
1. Check for exactness: Let M= 3x2y+cos(y)and N=x3−sin(x). Then,
∂M
∂y = 3x2−sin(y)
∂N
∂x = 3x2−cos(x)
It is clear that ∂M
∂y =∂N
∂x . Thus, the differential equation is not exact.
2. Integrate a factor: To make the equation exact, we need to find an integrating factor
µ(x, y)such that:
µ(x, y)[∂M
∂y −∂N
∂x ]=∂µ
∂x
3. Solve for the integrating factor: Let µ(x)be a function of xonly. Then, the integrating
factor equation simplifies to:
dµ
dx =−1
N(∂M
∂y −∂N
∂x )
dµ
dx =−1
x3−sin(x)(sin(y)−cos(x))
Integrating both sides will give us the integrating factor.
4. Find the general solution: Once we have found the integrating factor, we can multiply
it to the entire differential equation to make it exact. Then, we can find the general solution
using standard methods like finding a potential function or integrating factors.
9. Given the differential equation
(3x2y−y3)dx + (x3−3xy2)dy = 0
Determine whether the given differential equation is exact. If it is exact, find a solution. If
not, explain why.
Ans. To determine whether the given differential equation is exact, we check if the partial
derivatives of the coefficients with respect to yand x, respectively, are equal. If they are equal,
the equation is exact, and we can find a potential function to obtain the general solution.
1. Check for exactness:
Let M= 3x2y−y3and N=x3−3xy2be the coefficients of dx and dy, respectively.
Calculate the partial derivatives:
∂M
∂y = 3x2−3y2
∂N
∂x = 3x2−3y2
Since ∂M
∂y =∂N
∂x , the given differential equation is exact.
2. Finding a potential function:
To find the potential function F(x, y), we integrate the coefficient of dx with respect to x:
F(x, y) = ∫(3x2y−y3)dx =x3y−y3x+h(y)
Now, we differentiate F(x, y)with respect to yand set it equal to the coefficient of dy:
∂F
∂y =x3−3xy2=x3−3xy2
This confirms that F(x, y)is the potential function, so the general solution to the given
differential equation is:
x3y−y3x=C
10. Consider the differential equation given by
(2xy +y2)dx + (x2+ 2xy)dy = 0
Determine if the equation is exact and if so, find its solution.
Ans. To determine if the given differential equation is exact, we check if
∂M
∂y =∂N
∂x
where M= 2xy +y2and N=x2+ 2xy.
1. Calculate the partial derivatives:
∂M
∂y = 2x+ 2y
∂N
∂x = 2x+ 2y
2. Since ∂M
∂y =∂N
∂x , the equation is exact.
To find the solution, we need to find a function ϕ(x, y)such that
∂ϕ
∂x = 2xy +y2
∂ϕ
∂y =x2+ 2xy
3. Integrate the first equation with respect to x:
ϕ=∫(2xy +y2)dx =x2y+xy2+h(y)
where h(y)is a function of yto be determined.
4. Differentiate ϕwith respect to yand compare it with the second equation:
∂ϕ
∂y =x2+ 2xy +dh(y)
dy
5. Equate coefficients to get dh(y)
dy = 0 ⇒h(y) = cwhere cis an arbitrary constant.
6. Thus, the general solution to the differential equation is x2y+xy2+c= 0.
11. Consider the differential equation given by:
(2x−y)dx + (x+ 2y)dy = 0
Determine whether the equation is exact. If it is exact, find the general solution.
Ans. To determine if the given differential equation is exact, we need to check if the partial
derivatives of the terms involving xand yare equal. 1. Calculate ∂M
∂y :
∂M
∂y =−1
2. Calculate ∂N
∂x :
∂N
∂x = 1
Since ∂M
∂y =∂N
∂x , the equation is not exact. 3. To make the equation exact, we can find an
integrating factor µ(x):
µ(x) = e∫
∂N
∂x
−∂M
∂y
Ndx
µ(x) = e∫1−(−1)
x+2ydx =e∫2
x+2ydx =e2ln |x+2y|=|x+ 2y|2
4. Multiply both sides of the differential equation by the integrating factor:
|x+ 2y|2(2x−y)dx +|x+ 2y|2(x+ 2y)dy = 0
5. The left side of the equation can be rewritten as the exact differential of a function f(x, y):
f(x, y) = 1
3|x+ 2y|3=C
6. Therefore, the general solution to the differential equation is:
|x+ 2y|3= 3C
12. Question: Consider the differential equation (3ysin(x) + cos(y)) dx + (3xcos(y)−
sin(x)) dy = 0. Determine whether the given differential equation is exact. If it is exact, find the
general solution y(x).
Ans. Step-by-step solution: 1. To check if the given differential equation is exact, we calculate
the partial derivatives Myand Nx:My= 3 sin(x)−sin(y)Nx= 3 cos(y)−cos(x)
2. The differential equation is exact if My=Nx. Thus, we need to solve the following system
of equations: 3sin(x)−sin(y) = 3 cos(y)−cos(x)
3. Rearranging the terms, we get: sin(y)−3cos(y) = cos(x)−3sin(x)
4. We can rewrite the equation as: sin(y)−3cos(y) = √10 (cos(x)
√10 −3sin(x)
√10 )
5. By looking at the equation above, it is clear that both sides are linear combinations of
sin(y),cos(y),sin(x), and cos(x). Thus, the given differential equation is exact.
6. To find the general solution y(x), we integrate Mwith respect to xand compare that
to the partial integration of Nwith respect to y. The general solution will be of the form
F(x, y) = c, where Fis the potential function. Integrating M= 3ysin(x) + cos(y)with
respect to x:∫3ysin(x)dx +∫cos(y)dx = 3y(−cos(x)) + g(y), where g(y)is the constant of
integration.
7. Partially integrating N= 3xcos(y)−sin(x)with respect to y:∫3xcos(y)dy−∫sin(x)dy =
3xsin(y) + h(x), where h(x)is the constant of integration.
8. For F(x, y)to satisfy exactness, we must have g(y) = h(x). Therefore, the general
solution is given by 3ysin(x) + cos(y) = c.
13. Question 13: Consider the differential equation
(3x2y+ 2y2)dx + (x3+ 4xy)dy = 0.
Determine whether the given differential equation is exact. If it is exact, find a solution. If
not, explain why it is not exact.
Ans. To determine if the given differential equation is exact, we will check if the partial
derivatives of the terms are equal. Then, we will find a solution if the equation is exact.
1. Check for exactness: Let
M= 3x2y+ 2y2
and
N=x3+ 4xy.
Calculate the partial derivatives of Mand Nwith respect to y:
∂M
∂y = 3x2+ 4y
∂N
∂x = 3x2+ 4y
Since ∂M
∂y =∂N
∂x , the given differential equation is exact.
2. Find a solution: To find the solution, we need to find a function f(x, y)such that
∂f
∂x = 3x2y+ 2y2
∂f
∂y =x3+ 4xy
Integrating the first equation with respect to xgives:
f(x, y) = x3y+ 2xy2+g(y)
where g(y)is a constant of integration.
Now, differentiate f(x, y)with respect to y:
∂f
∂y =x3+ 4xy +dg
dy
Comparing this with ∂f
∂y =x3+ 4xy, we find that dg
dy = 0, so g(y)is a constant.
Therefore, the solution to the given differential equation is x3y+ 2xy2+C= 0, where Cis
a constant.
14. Let M(x, y)dx +N(x, y)dy be a differential 1-form. Show that the differential form is
exact if and only if the following condition is satisfied:
(∂N
∂x −∂M
∂y )dx ∧dy = 0
Moreover, if M(x, y)dx +N(x, y)dy is exact, then find the function f(x, y)such that df =
M(x, y)dx +N(x, y)dy.
Ans. To show that the differential form is exact if and only if the given condition is satisfied,
we need to show the following:
1. If the differential form M(x, y)dx+N(x, y)dy is exact, then (∂N ∂x−∂M
∂y )dx∧dy = 0:
Since the differential form is exact, there exists some function f(x, y)such that df =
M(x, y)dx +N(x, y)dy. By Clairaut’s Theorem, we have
∂2f
∂x∂y =∂2f
∂y∂x
Now, we can express df as
df =∂f
∂x dx +∂f
∂y dy
We can rewrite this as
df =(∂f
∂x ,∂f
∂y )
By comparing coefficients, we have M(x, y) = ∂f
∂x and N(x, y) = ∂f
∂y . Therefore,
∂N
∂x −∂M
∂y =∂2f
∂x∂y −∂2f
∂y∂x = 0
Hence, (∂N
∂x −∂M
∂y )dx ∧dy = 0, as required.
2. If (∂N ∂x −∂M
∂y )dx ∧dy = 0, then the differential form M(x, y)dx +N(x, y)dy is
exact:
Let f(x, y)be a scalar function defined by
∂f
∂x =M(x, y)and ∂f
∂y =N(x, y)
Then, df =M(x, y)dx +N(x, y)dy, which implies that the differential form is exact.
Therefore, the given differential form M(x, y)dx +N(x, y)dy is exact if and only if (∂N
∂x −
∂M
∂y )dx ∧dy = 0.
15. **Question:** Consider the differential equation
(2xy3−y2)dx + (3x2y2−4x)dy = 0.
Determine whether the differential equation is exact and if so, find the general solution. If
the equation is not exact, determine an integrating factor that makes it exact.
Ans. **Step-by-step solution:**
**1. Determine if the equation is exact:** To determine if the equation is exact, we check if
∂M
∂y =∂N
∂x , where M= 2xy3−y2and N= 3x2y2−4x.
Calculating the partial derivatives:
∂M
∂y = 6xy2−2y
∂N
∂x = 6xy2−4
Since ∂M
∂y =∂N
∂x , the equation is not exact.
**2. Find an integrating factor:** To make the equation exact, we need to find an integrating
factor µ(x, y)such that µM dx +µN dy is exact. We find µby the equation:
∂(µM)
∂y −∂(µN )
∂x
µ=
∂M
∂y −∂N
∂x
N
Substitute Mand Ninto the above equation to find µ.
**3. Solve the differential equation:** Once we find the integrating factor µ(x, y), we multiply
Mand Nby µto make the equation exact and solve for the general solution using the method
of exact equations.
16. Let M(x, y) = 3x2y−5y2and N(x, y) = x3+ 4xy2. Determine whether the following
differential form is exact, and if it is, find its solution.
M(x, y)dx +N(x, y)dy
Ans. To check whether the given differential form is exact, we need to verify if ∂M
∂y =∂N
∂x . If it
is exact, we can then find its solution using the method of integrating factors.
1. Check for Exactness:
Given M(x, y) = 3x2y−5y2and N(x, y) = x3+ 4xy2.
We compute the partial derivatives:
∂M
∂y = 3x2−10y
∂N
∂x = 3x2+ 4y2
Since ∂M
∂y =∂N
∂x , the given differential form is not exact.
2. Finding an Integrating Factor:
To find an integrating factor µ(x, y), we define µsuch that:
µ∂M
∂y =µ(3x2−10y) = µ∂N
∂x =µ(3x2+ 4y2)
Equating the expressions leads to:
µ(3x2−10y) = µ(3x2+ 4y2)⇒µ(−10y) = µ(4y2)
Solving this differential equation gives µ=e14y.
3. Finding the Solution:
We multiply the given differential form by the integrating factor µ(x, y) = e14y:
e14y(3x2y dx −5y2dx +x3dy + 4xy2dy)
This can be rewritten as:
d(e14y(3x2y−5y2+x3+ 4xy2))
Therefore, the solution is e14y(3x2y−5y2+x3+ 4xy2) = C, where Cis the constant of
integration.
17. Let M(x, y)dx +N(x, y)dy = 0 be a first-order ordinary differential equation. Determine
the conditions for exactness and find the general solution for the following differential equation:
(2x+y)dx + (x−y+ 1)dy = 0
Ans. To determine the conditions for exactness and find the general solution for the given
differential equation (2x+y)dx + (x−y+ 1)dy = 0:
1. Check for exactness by verifying if ∂M
∂y =∂N
∂x :∂M
∂y = 1 and ∂N
∂x = 1. Since ∂M
∂y =∂N
∂x , the
equation is exact.
2. Find the potential function Φ(x, y)by integrating Mwith respect to x:Φ(x, y) =
∫(2x+y)dx =x2+xy +g(y), where g(y)is a constant of integration that may depend on y.
3. Differentiate Φ(x, y)partially with respect to yand compare it with N:∂Φ
∂y =x+dg
dy =
x−y+ 1. Comparing terms, we get dg
dy =−y+ 1.
4. Integrate dg
dy =−y+ 1 with respect to yto find g(y):g(y) = −1
2y2+y+C, where Cis
the constant of integration.
5. Substitute g(y)back into the potential function Φ(x, y):Φ(x, y) = x2+xy −1
2y2+y+C.
This is the potential function.
6. The general solution is then given by Φ(x, y) = C, where Cis the constant of integration:
x2+xy −1
2y2+y=C.
18. Let M= (x2+ 2xy)dx + (x2−y2)dy be a differential form. Determine whether Mis
exact, and if it is, find a function f(x, y)such that df =M.
Ans. To check if Mis exact, we need to verify if ∂M
∂y =∂N
∂x , where M= (x2+ 2xy)and
N= (x2−y2).
1. Calculate ∂M
∂y :
∂M
∂y =∂
∂y (x2+ 2xy) = 2x
2. Calculate ∂N
∂x :
∂N
∂x =∂
∂x(x2−y2) = 2x
Since ∂M
∂y =∂N
∂x , the form Mis exact. To find the function f(x, y), we integrate Mwith
respect to xand set it equal to f:
f(x, y) = ∫(x2+ 2xy)dx =1
3x3+x2y+g(y)
where g(y)is the constant of integration with respect to x.
3. To find g(y), differentiate the result with respect to yand set it equal to the Nin M:
∂f
∂y =x2+ 2xy =N=x2−y2
x2+ 2xy =x2−y2
2xy =−y2
2x=−y
y=−2x
So, g(y) = −2xy. Therefore, the function f(x, y) = 1
3x3+x2y−2xy satisfies df =M.
19. Let M= 3x2y2−x3y−2yand N=x3y2+y2−2xy. Determine whether the differential
equation (3x2y2−x3y−2y)dx + (x3y2+y2−2xy)dy = 0 is exact, and if so, find the general
solution.
Ans. To check for exactness, we calculate the partial derivatives of Mwith respect to yand
Nwith respect to x, and then verify if they are equal. If so, we find a potential function Ψ(x, y)
such that dΨ = Mdx +Ndy. Finally, we find the general solution by integrating Ψwith respect
to one variable and setting it equal to a constant.
20. Consider the following differential equation
(2x3+ 3y)dx −(x2+ 4y)dy = 0.
Ans. To determine whether the given differential equation is exact and find its solution, we will
follow these steps:
1. Check for exactness by computing the partial derivatives of the coefficients with respect
to y. 2. If the equation is exact, find a function ψsuch that dψ =Mdx +Ndy. 3. Solve for ψ
by integrating either Mor Nwith respect to the corresponding variable. 4. Use ψto find the
general solution of the differential equation.
1. By checking if ∂M
∂y =∂N
∂x , we have:
∂M
∂y = 3,∂N
∂x =−2x.
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. To make the equation exact, we need an integrating factor µsuch that:
µ(2x3+ 3y)dx −µ(x2+ 4y)dy = 0
∂
∂y (µ(2x3+ 3y)) = ∂
∂x(−µ(x2+ 4y)).
Expanding and equating the mixed partial derivatives, we get:
3µ=−µ=⇒µ=−1
4.
3. Multiply the differential equation by −1
4to make it exact:
−1
4(2x3+ 3y)dx +1
4(x2+ 4y)dy = 0.
Integrating N=1
4(x2+ 4y)with respect to y, we get the potential function ψ:
ψ=∫(1
4(x2+ 4y))dy =1
4x2y+y2+C(x),
where C(x)is the constant of integration.
4. The general solution is given by ψ(x, y) = Cwhere Cis a constant:
1
4x2y+y2+C(x) = C.
21. Suppose we have a differential equation of the form (ycos(x) + xsin(y))dx + (xcos(y)−
ysin(x))dy = 0.
1. Determine whether the given differential equation is exact. If it is exact, find the solution.
If it is not exact, determine an integrating factor to make it exact and find the solution.
Ans. To check for exactness, compute the partial derivatives of the coefficients of dx and dy:
∂
∂y (ycos(x) + xsin(y)) = −sin(y) + xcos(y),
∂
∂x(xcos(y)−ysin(x)) = cos(y)−ycos(x).
Since these partial derivatives are not equal, the given differential equation is not exact. Let’s
find the integrating factor to make it exact.
1. Compute the integrating factor µ(x)using the formula
µ(x) = e∫My−Nx
Ndx.
Here, M=ycos(x) + xsin(y),N=xcos(y)−ysin(x),My=−sin(y) + xcos(y), and
Nx=cos(y)−ycos(x).
So, we have My−Nx
N=−sin(y)+xcos(y)−(cos(y)−ycos(x))
xcos(y)−ysin(x)=−sin(y)+xcos(y)−cos(y)+ycos(x)
xcos(y)−ysin(x)=ycos(x)−sin(y)
xcos(y)−ysin(x).
Therefore, the integrating factor is µ(x) = e∫ycos(x)−sin(y)
xcos(y)−ysin(x)dx.
2. After finding the integrating factor, multiply it by the original differential equation to make
it exact.
3. Solve the exact differential equation using standard methods such as integration.
22. Let M= (3y2−6xy)dx + (2x−2x2)dy be a differential 1-form in R2. Determine whether
Mis exact. If it is exact, find a function f(x, y)such that M=df.
Ans. To determine if the differential 1-form M= (3y2−6xy)dx + (2x−2x2)dy is exact, we
need to check if ∂M1
∂y =∂M2
∂x , where M1= 3y2−6xy and M2= 2x−2x2.
1. Calculate the partial derivatives:
∂M1
∂y = 6y−6xand ∂M2
∂x = 2 −4x.
2. Check if the derivatives are equal:
6y−6x?
= 2 −4x.
3. Simplify the equation and determine if Mis exact:
6y−6x= 2 −4x=⇒6y= 8 + 2x.
Since the partial derivatives are not equal, Mis not exact.
Therefore, the given differential 1-form Mis not exact and does not have a potential function
f(x, y).
23. Question: Determine whether the following differential equation is exact and, if so, find a
potential function that satisfies it:
(2xy + 3)dx + (x2−1)dy = 0
Ans. To determine if the given equation is exact, we check if the partial derivatives of the
coefficients with respect to yand xare equal. If they are equal, we can find a potential function
that satisfies the equation.
1. Check for Exactness: Calculate the partial derivatives:
∂
∂y (2xy + 3) = 2x
∂
∂x(x2−1) = 2x
Since the partial derivatives are equal, the equation is exact.
2. Find the Potential Function: To find the potential function, integrate the coefficient
of dx with respect to x:
F(x, y) = ∫(2xy + 3)dx =x2y+ 3x+g(y)
Now, differentiate Fwith respect to yand set it equal to the coefficient of dy:
∂F
∂y =x2+dg
dy =x2−1
From this, we can see that dg
dy =−1, which implies that g(y) = −y. Thus, the potential
function is:
F(x, y) = x2y+ 3x−y+C
Therefore, the potential function that satisfies the differential equation is x2y+ 3x−y+C,
where Cis a constant.
24. Question 24: Determine whether the following differential equation is exact or not, and if
it is exact, find its general solution:
(2x2+y)dx + (2xy + 1)dy = 0
Ans. To determine if the given differential equation is exact and find its general solution, we
will follow these steps:
1. Check for exactness by verifying if ∂M
∂y =∂N
∂x . 2. If the equation is exact, find the
integrating factor µ(x, y). 3. Solve the equation by integrating to find the general solution.
1. Check for exactness: Given differential equation:
(2x2+y)dx + (2xy + 1)dy = 0
Checking exactness:
∂M
∂y =∂
∂y (2x2+y) = 1
∂N
∂x =∂
∂x(2xy + 1) = 2y
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Find the integrating factor µ(x, y):To make the equation exact, we need to find an
integrating factor µ(x, y)such that µ(x, y)·(2x2+y)dx +µ(x, y)·(2xy + 1)dy = 0 is exact.
We can find µ(x, y)by the relationship:
∂
∂y (2x2+y)−∂
∂x (2xy + 1)
2xy + 1 =1−2y
2xy + 1
3. Solve the differential equation: Note: This step can vary depending on the integrating
factor found in the previous step.
Since the given differential equation is not exact and the integrating factor is not readily
apparent, further manipulation or techniques may be required to solve the equation.
Therefore, the general solution of the given differential equation may involve additional meth-
ods beyond standard exact differential equation techniques.
25. Question 25: Consider the differential equation given by
(3x2y+ 2y3)dx + (x3+ 3xy2)dy = 0
Determine whether the equation is exact. If it is exact, find the general solution. If it is not
exact, find an integrating factor µ(x, y)to make it exact, and then solve the equation.
Ans. To determine if the given equation is exact, we check if ∂M
∂y =∂N
∂x , where M= 3x2y+ 2y3
and N=x3+ 3xy2.
1. Check for exactness: We calculate the partial derivatives:
∂M
∂y = 3x2+ 6y2
∂N
∂x = 3x2+ 3y2
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Find an integrating factor: To find an integrating factor, we look for a function µ(x, y)
such that µMdx +µNdy = 0 is exact. We can find µby solving the following PDE:
∂
∂y (µM) = ∂
∂x(µN)
Expanding, we get:
3x2∂µ
∂y + 6y2µ= 3x2∂µ
∂x + 3y2µ
By comparing coefficients:
3x2∂µ
∂y −3x2∂µ
∂x = 3y2µ−6y2µ
3x2dµ
dy −3x2dµ
dx =−3y2µ
This is a first-order linear PDE, which can be solved to find µ.
3. Solve for the integrating factor µ:Integrating the PDE gives:
dµ
µ=−3y2
3x2dx +3x2
3x2dy
ln |µ|=−y2+x2+C
µ=ex2−y2eC=Cex2−y2
4. Find the general solution: Multiplying the original equation by the integrating factor
µ=Cex2−y2, we obtain an exact differential equation:
(3x2y+ 2y3)ex2−y2dx + (x3+ 3xy2)ex2−y2dy = 0
From here, the general solution can be found by integrating both sides with respect to xand
y.
26. Let M = (2x + 3y)dx + (4x + 5y)dy and N = (3x - 4y)dx + (2x - 3y)dy.
Determine whether the differential form Mdx + Ndy is exact. If it is exact, find a function
f(x, y) such that df = Mdx + Ndy.
Ans. To check whether the form Mdx + Ndy is exact, we need to verify if �M/�y = �N/�x.
1. Calculate �M/�y: �M/�y = 3.
2. Calculate �N/�x: �N/�x = 3.
Since �M/�y = �N/�x, the form Mdx + Ndy is exact.
3. To find f(x, y), integrate M with respect to x and set it equal to the partial derivative of f
with respect to y: �f/�y = 2x + 3y.
Integrating, we get f(x, y) = x2+ 3xy +g(y), whereg(y)isafunctionofy.
4. Take the partial derivative of f with respect to y and set it equal to N to find g(y):
�(x2+ 3xy +g(y))/∂y= 2x−3y.
By comparing coefficients, we find g’(y) = -3y + constant. So, g(y) = -1.5y2+C, whereCisaconstant.
5. So, the function f(x, y) is f(x, y)= x2+ 3xy −1.5y2+C, whereCisaconstant.
27. Consider the differential equation:
(y3ex+x)dx + (3y2ex−2y)dy = 0
Ans. To determine whether the given differential equation is exact, we check if the partial
derivatives of the coefficients respect to yand respect to xare the same. If not exact, we will
show how to make it exact by multiplying by an integrating factor.
1. Exactness check: Let M=y3ex+xand N= 3y2ex−2y. Compute the partial
derivatives of Mand Nwith respect to y:
∂M
∂y = 3y2ex−2,
and with respect to x:
∂N
∂x = 3y2ex.
Since ∂M
∂y =∂N
∂x , the differential equation is not exact.
2. Find the integrating factor: To find the integrating factor, we will find a function
µ(x, y)such that
µ(x, y)M dx +µ(x, y)Ndy = 0
is exact.
We can find µ(x, y)by integrating the following equation:
∂
∂y (µM) = ∂
∂x(µN).
Expanding the derivatives gives:
µ∂M
∂y +∂µ
∂y M=µ∂N
∂x +∂µ
∂x N.
Since Mdx+N dy = 0 is our original differential equation, we can simplify the above equation
to: ∂µ
∂y
µ=
∂N
∂x −∂M
∂y
M.
Substitute M,N,∂M
∂y , and ∂N
∂x into the above equation and solve for µ(x, y).
3. Multiply by the integrating factor: Once we have found µ(x, y), multiply the original
differential equation by µ(x, y)to make it exact.
Solve the resulting exact differential equation to find the solution.
28. Let M(x, y)dx +N(x, y)dy = 0 be a first-order differential equation. Determine conditions
for exactness and find the general solution for the following equation:
2x2ydx + (x2+y)dy = 0
Ans. To determine if the given differential equation is exact, we need to check if ∂M
∂y =∂N
∂x .
If this condition is satisfied, we can find a potential function ϕ(x, y)such that ∂ϕ
∂x =Mand
∂ϕ
∂y =N.
1. Check for exactness:
Given M(x, y) = 2x2yand N(x, y) = x2+y, we calculate the partial derivatives as follows:
∂M
∂y = 2x2
∂N
∂x = 2x
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Make the equation exact:
In order to make the equation exact, we can try to find an integrating factor µ(x, y)such
that µM and µN have partial derivatives that are symmetric: ∂
∂y (µM ) = ∂
∂x (µN ).
Let’s find µby solving the following equation:
∂
∂y (2x2y)−∂
∂x (x2+y)
N=2x2−2x
x2+y
This simplifies to −2x
x2+y. Therefore, µ(x, y) = 1
x2+y.
3. Find the general solution:
Multiplying the given equation by the integrating factor µ=1
x2+y, we get:
2xydx + (1)dy = 0
This equation is now exact as it satisfies the condition ∂
∂y (2xy)−∂
∂x (1) = 0.
To find the potential function ϕ(x, y), we integrate Mwith respect to x:
ϕ(x, y) = ∫2xydx =x2y+g(y), where g(y)is an arbitrary function of y.
Now, we differentiate ϕ(x, y)with respect to yand set it equal to N(x, y)to find g(y):
∂ϕ
∂y =x2+dg
dy =x2+g′(y) = x2+y
Solving g′(y) = ygives g(y) = y2
2.
Therefore, the general solution is x2y+y2
2=C, where Cis the constant of integration.
29. Question 29: Consider the differential equation (y−x2)dx + (2xy −1)dy = 0. Determine
whether the equation is exact, and if so, find the general solution.
Ans. To determine if the given differential equation is exact, we need to check if ∂M
∂y =∂N
∂x ,
where M=y−x2and N= 2xy −1.
1. Check exactness: Compute the partial derivatives:
∂M
∂y = 1 and ∂N
∂x = 2y
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Condition for exactness: In order to make the equation exact, we need to find an
integrating factor µ(x, y)such that µ(x, y) [(y−x2)dx + (2xy −1)dy]is an exact differential.
3. Find integrating factor: Since the integrating factor µ(x, y)depends only on xin this
case, we can write µ(x). From the condition for exactness, we have ∂
∂y (µM ) = ∂
∂x (µN ). This
gives us:
µ∂M
∂y +∂µ
∂y M=µ∂N
∂x +∂µ
∂x N
µ+ 0 = 2µy +dµ
dx (2xy −1)
0 = 2µy + 2xµdµ
dx −dµ
dx
This is a first-order linear differential equation in terms of µ(x). Solve for µ(x)to find the
integrating factor.
4. Find general solution: After finding the integrating factor µ(x), multiply through the
entire equation by µ(x)to make it exact. Then, integrate to find the general solution of the
differential equation.
30. Consider the following differential equation:
(3x2+y)dx + (2y+x)dy = 0
Determine whether the differential equation is exact. If it is exact, find the general solution.
Ans. To determine whether the given differential equation is exact, we check the mixed partial
derivatives of the terms involving xand y. If the mixed partial derivatives are equal, the equation
is exact. Otherwise, it is not exact.
1. Determine if the equation is exact: Calculate the partial derivatives of the terms
involving xand y:
∂M
∂y = 1
∂N
∂x = 1
Since ∂M
∂y =∂N
∂x , the differential equation is not exact.
2. Find an integrating factor: To find an integrating factor, we use the following formula:
µ=e∫
∂M
∂y
−∂N
∂x
Ndx
In this case, we have:
µ=e∫1−1
2y+xdx =e∫0
2y+xdx =e0= 1
3. Multiply through by the integrating factor: Multiplying the original equation by the
integrating factor µ= 1 gives:
(3x2+y)dx + (2y+x)dy = 0
This equation remains the same, as the integrating factor is 1.
4. Identify the exact differential equation: After multiplying through by the integrating
factor, we find that the equation remains non-exact.
Therefore, the differential equation does not admit an integrating factor and is not exact.
Further techniques such as finding an integrating factor cannot be used to solve this differential
equation.
The general solution to the differential equation can be found by using other methods such
as separation of variables or integrating factors if applicable.
31. Consider the differential equation given by:
(3y2−2xy)dx + (x2−6y)dy = 0
Determine whether the given differential equation is exact, and if so, find the general solution.
Ans. To check for exactness, we compute the partial derivatives ∂M
∂y and ∂N
∂x , where M=
3y2−2xy and N=x2−6y.
1. Compute ∂M
∂y :
∂M
∂y = 6y−2x
2. Compute ∂N
∂x :
∂N
∂x = 2x
Since ∂M
∂y =∂N
∂x , the given differential equation is not exact. To find an integrating factor,
we first check if the integrating factor depends only on xor only on y.
3. Check if the integrating factor depends only on x: Let p(x)be the integrating factor. We
have: ∂
∂y (p(x)M) = ∂
∂x(p(x)N)
Expanding, we get:
p(x)(6y−2x) = p′(x)(x2−6y)
Since the left side depends on yand the right side depends on x, the integrating factor cannot
depend only on x.
4. Check if the integrating factor depends only on y: Let q(y)be the integrating factor. We
consider the equation:
∂
∂y (q(y)M) = ∂
∂x(q(y)N)
Expanding, we get:
q(y)(6y−2x) = q′(y)(x2−6y)
This equation is not possible to solve, so this also fails. Therefore, an integrating factor that
depends only on xor ydoesn’t exist.
5. Find a general solution: Since we couldn’t find an integrating factor, we need to use other
methods such as the method of integrating factors or an integrating factor involving both xand
y. In this case, we couldn’t find a suitable integrating factor; hence, we cannot make the given
differential equation exact.
Therefore, the general solution will involve solving the differential equation or manipulating
it to a form where it can be solved by methods other than exactness.
32. Let M(x, y)and N(x, y)be two functions defined on a domain D. Consider the first-order
ordinary differential equation
M(x, y) + N(x, y)dy
dx = 0.
Suppose that the functions Mand Nare such that the equation is exact on D, i.e., there
exists a function f(x, y)such that
∂f
∂x =M(x, y)and ∂f
∂y =N(x, y),
for all (x, y)∈D.
33. Find the general solution to the exact differential equation
(2y+ex)dx + (x+ey)dy = 0.
Ans. 1. To determine if the given equation is exact, we check if
∂M
∂y =∂N
∂x .
Calculating these partial derivatives, we have
∂M
∂y = 2 and ∂N
∂x = 1,
which are not equal. Therefore, the equation is not exact.
2. However, we can make the equation exact by multiplying through by an integrating factor,
denoted by µ(x, y), such that
µ(x, y) ((2y+ex)dx + (x+ey)dy) = 0.
In this case, our integrating factor is
µ(x, y) = 1
µ(x, y)=1
2y+ex.
3. Multiplying through by the integrating factor, we have
1
2y+ex(2y+ex)dx +1
2y+ex(x+ey)dy = 0.
Simplifying, we get
dx +(x+ey
2y+ex)dy = 0.
4. Now, we can rewrite the equation as
dx
dx +(x+ey
2y+ex)dy
dx = 0.
This allows us to look for a function f(x, y)such that
∂f
∂x = 1 and ∂f
∂y =x+ey
2y+ex.
5. Integrating the first equation with respect to x, we get f=x+h(y), where h(y)is a
function of y.
6. Substituting f=x+h(y)into the second equation and matching coefficients, we find
h′(y) = ey
2y+ex.
Integrating this expression gives us
h(y) = ln |2y+ex|+C,
where Cis an arbitrary constant.
7. Therefore, the general solution to the exact differential equation is
ln |2y+ex|+x=C,
where Cis a constant of integration.
34. Consider the differential equation:
(3x2y2−2y3)dx + (2x3y−3xy2)dy = 0
Determine whether the differential equation is exact. If it is exact, find the general solution.
Ans. To determine if the given differential equation is exact, we need to check if the following
condition is satisfied:
∂M
∂y =∂N
∂x
where Mand Nare the coefficients of dx and dy, respectively.
1. Compute ∂M
∂y :
∂
∂y (3x2y2−2y3) = 6x2y−6y2
2. Compute ∂N
∂x :
∂
∂x(2x3y−3xy2) = 6x2y−3y2
Since ∂M
∂y =∂N
∂x , the given differential equation is exact. To find the general solution, we
need to find a function ϕ(x, y)such that:
∂ϕ
∂x = 3x2y2−2y3and ∂ϕ
∂y = 2x3y−3xy2
3. Integrate the first equation with respect to x:
ϕ=∫(3x2y2−2y3)dx =x3y2−2yx2+h(y)
where h(y)is a function of yonly.
4. Differentiate ϕwith respect to yand equate it to ∂ϕ
∂y :
∂
∂y (x3y2−2yx2+h(y)) = 2x3y−3xy2
5. Solve for h(y):
2x3y−2x2+h′(y) = 2x3y−3xy2⇒h′(y) = −3xy2
Integrating h′(y)gives h(y) = −xy3+Cwhere Cis a constant.
6. Therefore, the general solution of the given exact differential equation is:
x3y2−2yx2−xy3+C= 0
where Cis a constant of integration.
35. Consider the following differential equation:
(3y2−2x)dx + (6xy −3x2)dy = 0
Determine whether the differential equation is exact or not. If it is exact, find the general
solution of the differential equation.
Ans. To determine whether the given differential equation is exact, we need to check if the partial
derivatives ∂M
∂y and ∂N
∂x are equal, where the given equation is in the form M(x, y)dx+N(x, y)dy =
0.
1. Check for exactness: Given differential equation:
(3y2−2x)dx + (6xy −3x2)dy = 0
Calculating the partial derivatives:
∂M
∂y = 6y
∂N
∂x = 6y
Since ∂M
∂y =∂N
∂x , the differential equation is exact.
2. Find the general solution: To find the general solution, we integrate Mwith respect
to xand set it equal to a function f(y)of yonly.
Integrating M= 3y2−2xwith respect to x, we get:
∫(3y2−2x)dx = 3y2x−x2+g(y)
Next, we differentiate 3y2x−x2+g(y)with respect to yand set it equal to Nto solve for
g(y).
d
dy (3y2x−x2+g(y)) = 6xy −2x+g′(y) = 6xy −3x2
Comparing coefficients, we find g′(y) = −3x2.
Integrating g′(y) = −3x2with respect to y, we get:
g(y) = −3x2y+h(x)
Thus, the general solution is given by:
3y2x−x2−3x2y+h(x) = C
where Cis the constant of integration and h(x)is an arbitrary function of x.
36. Consider the following first-order differential equation:
(3x2+ 2y)dx + (2x−3y)dy = 0
Determine if the given differential equation is exact. If it is exact, find the general solution.
If it is not exact, find an integrating factor that makes it exact, and then find the solution.
Ans. To determine if the given differential equation is exact, we check the equality of the mixed
partial derivatives of the coefficients of dx and dy.
1. Check exactness:
Given the equation (3x2+ 2y)dx + (2x−3y)dy = 0, Let M= 3x2+ 2yand N= 2x−3y.
Calculate the partial derivatives of Mand Nwith respect to yand x:
∂M
∂y = 2,∂N
∂x = 2
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Find integrating factor:
To make the equation exact, we need to find an integrating factor µ(x, y)such that
µ(x, y)((3x2+ 2y)dx + (2x−3y)dy)
is exact. The integrating factor is given by
µ=e∫
∂N
∂x
−∂M
∂y
Ndx
Calculate the integrating factor:
µ=e∫2−2
2x−3ydx =e∫0dx =e0= 1
3. Solve the equation with the integrating factor:
Multiply the given equation by the integrating factor µ= 1:
(3x2+ 2y)dx + (2x−3y)dy = 0
Now we have an exact differential equation and can write it in the form:
∂u
∂xdx +∂u
∂y dy = 0
where uis a new function.
4. Find the general solution:
Integrating the coefficients of dx and dy with respect to the respective variables:
u(x, y) = ∫(3x2+ 2y)dx +g(y)
u(x, y) = x3+ 2xy +g(y)
Taking the partial derivative of uwith respect to ygives:
∂u
∂y = 2x+g′(y)
Comparing with the coefficient of dy in the original equation, we have:
2x+g′(y) = 2x−3y
6. Attempt to find a solution using the integrating factor:
µ(x, y)(∂u
∂x dx +∂u
∂y dy)=µ(x, y)M(x, y)dx +µ(x, y)N(x, y)dy
7. Since the differential equation is not exact and finding a suitable integrating factor is chal-
lenging in this case, further analysis or alternative methods may be needed to find a solution.
3. Let M(x, y)dx +N(x, y)dy = 0 be a first-order ordinary differential equation. Determine
whether the equation is exact. If it is exact, find a solution f(x, y) = c, where cis a constant. If
it is not exact, find an integrating factor µ(x, y)that makes it exact, and then find the solution.
Ans. The given ODE is M(x, y)dx +N(x, y)dy = 0.
1. To check for exactness, we compute the partial derivatives ∂M
∂y and ∂N
∂x .
∂M
∂y =∂N
∂x
2. Since the equation is exact, we look for a function f(x, y)such that fx=Mand fy=N.
Integrating Mwith respect to xgives f(x, y) = ∫M dx +h(y).
3. To find the remaining term h(y), we differentiate f(x, y)with respect to yand set it equal
to N.∂f
∂y =∂
∂y (∫M dx)+h′(y) = N
So, h′(y) = N−∂
∂y (∫M dx)
4. Solving for h(y), we find f(x, y) = ∫M dx +∫(N−∂
∂y (∫M dx))dy =c, where cis
a constant.
Hence, the solution to the differential equation is f(x, y) = c.
4. Let M=y2dx + 2xydy be a differential form. Determine if Mis exact. If it is exact, find a
potential function f(x, y)such that df =M.
Ans. To determine if the given differential form M=y2dx + 2xydy is exact, we will check if
∂M
∂y =∂N
∂x where Nis the unknown function associated with M. If it is exact, we will then find
f(x, y)such that df =M. 1. Find ∂M
∂y :∂M
∂y = 2y2. Find ∂N
∂x : Comparing ∂N
∂x with ∂M
∂y , we have:
∂N
∂x = 2yIntegrating this with respect to xgives N= 2xy +h(y)where h(y)is the constant of
integration. 3. Find ∂N
∂y : Taking the partial derivative of N= 2xy +h(y)with respect to y, we
get: ∂N
∂y = 2x+h′(y)4. Comparing with M=y2dx + 2xydy, we have ∂N
∂y = 2x. Equating the
two expressions, we find h′(y)=0, which implies h(y) = Cfor some constant C. 5. Therefore,
N= 2xy +C. Since Nis the potential function associated with M, the given differential form
Mis exact and the potential function is f(x, y) = 2xy +Cwhere Cis a constant.
5. Find the value of kfor which the following differential equation is exact:
(kx +y2)dx + (2y)dy = 0
Ans. To check for exactness, we compute the partial derivatives of the terms with respect to y
and x, respectively.
1. Compute the partial derivative with respect to y:
∂(kx +y2)
∂y = 2y
2. Compute the partial derivative with respect to x:
∂(2y)
∂x = 0
Since the cross-partial derivatives are equal (∂(2y)
∂x =∂(kx+y2)
∂y ), the equation is exact.
3. Write the equation in the exact form:
(kx +y2)dx + (2y)dy = 0
M(x, y)dx +N(x, y)dy = 0
4. Apply the exactness condition: For an exact differential equation, the following must
hold: ∂M
∂y =∂N
∂x
Substitute M=kx +y2and N= 2yinto the above formula:
∂(kx +y2)
∂y =∂(2y)
∂x
2y= 0
5. Solve for k:Since this equation must hold for all xand y, we find that k= 0 to satisfy
the exactness condition. Thus, the value of kfor which the given differential equation is exact is
k= 0.
6. Question:
Consider the following differential equation:
(3x2y−2x3)dx + (2y−x3)dy = 0
Determine whether the given equation is exact. If it is exact, find a function F(x, y)such
that ∂F
∂x = 3x2y−2x3and ∂F
∂y = 2y−x3. Then, find the general solution of the differential
equation.
Ans. Step-by-step solution:
1. Check for exactness: To determine if the given differential equation is exact, we calculate
the partial derivatives of Mand Nwith respect to yand x, respectively. Let M= 3x2y−2x3
and N= 2y−x3.
2. Calculate ∂M
∂y and ∂N
∂x :
∂M
∂y = 3x2
∂N
∂x =−3x2
3. Check for exactness: Since ∂M
∂y =∂N
∂x , the equation is not exact.
4. Find integrating factor: To make the equation exact, we need to find the integrating factor
µ(x)which allows us to write the equation as (µM )dx + (µN )dy = 0 where ∂(µM )
∂y =∂(µN )
∂x .
The integrating factor is given by µ=e∫(∂N
∂x
−∂M
∂y
N)dx
.
5. Calculate the integrating factor:
µ=e∫(−3x2−3x2
2y−x3)dx
µ=e∫(−6x2
2y−x3)dx
6. Recalculate the differential equation with the integrating factor: Multiplying the given
differential equation by the integrating factor µ, we get:
e∫−6x2
2y−x3dx[(3x2y−2x3)dx + (2y−x3)dy] = 0
7. Solve the differential equation: The above equation is exact, with F(x, y) = ∫e∫−6x2
2y−x3dx dx+
C, where Cis the constant of integration. Solving for F(x, y)gives the general solution of the
differential equation.
7. Determine whether the following differential equation is exact, and if so, find the general
solution:
(2x+y2)dx + (2y−2xy)dy = 0
Ans. To determine if the given differential equation is exact, we check if the mixed partial
derivatives of the coefficients are equal. If so, we can find the general solution by integrating the
equation with respect to x and y one by one.
1. Check for exactness: Let M= 2x+y2and N= 2y−2xy. We calculate the partial
derivatives: ∂M
∂y = 2yand ∂N
∂x =−2y
Since ∂M
∂y =∂N
∂x , the equation is exact.
2. Find the general solution: Since the equation is exact, we need to find a function
ψ(x, y)such that:
∂ψ
∂x = 2x+y2and ∂ψ
∂y = 2y−2xy
Integrating the first equation with respect to x, we get:
ψ(x, y) = ∫(2x+y2)dx =x2+xy2+h(y)
where h(y)is a function of y that we will determine by differentiating with respect to y and
comparing to the second equation:
∂ψ
∂y =x(2y)+2xy +h′(y) = 2y−2xy
Equating coefficients, we get:
h′(y) = 2y
Integrating with respect to y, we find:
h(y) = y2+C
where C is the constant of integration.
3. Final solution: Thus, the general solution to the differential equation is:
x2+xy2+y2+C= 0
8. Consider the differential equation:
(3x2y+cos(y))dx + (x3−sin(x))dy = 0
Determine if the given differential equation is exact. If it is exact, find the general solution.
Ans. To determine if the given differential equation is exact, we need to check if
∂M
∂y =∂N
∂x
where the given differential equation is in the form M dx +N dy = 0.
1. Check for exactness: Let M= 3x2y+cos(y)and N=x3−sin(x). Then,
∂M
∂y = 3x2−sin(y)
∂N
∂x = 3x2−cos(x)
It is clear that ∂M
∂y =∂N
∂x . Thus, the differential equation is not exact.
2. Integrate a factor: To make the equation exact, we need to find an integrating factor
µ(x, y)such that:
µ(x, y)[∂M
∂y −∂N
∂x ]=∂µ
∂x
3. Solve for the integrating factor: Let µ(x)be a function of xonly. Then, the integrating
factor equation simplifies to:
dµ
dx =−1
N(∂M
∂y −∂N
∂x )
dµ
dx =−1
x3−sin(x)(sin(y)−cos(x))
Integrating both sides will give us the integrating factor.
4. Find the general solution: Once we have found the integrating factor, we can multiply
it to the entire differential equation to make it exact. Then, we can find the general solution
using standard methods like finding a potential function or integrating factors.
9. Given the differential equation
(3x2y−y3)dx + (x3−3xy2)dy = 0
Determine whether the given differential equation is exact. If it is exact, find a solution. If
not, explain why.
Ans. To determine whether the given differential equation is exact, we check if the partial
derivatives of the coefficients with respect to yand x, respectively, are equal. If they are equal,
the equation is exact, and we can find a potential function to obtain the general solution.
1. Check for exactness:
Let M= 3x2y−y3and N=x3−3xy2be the coefficients of dx and dy, respectively.
Calculate the partial derivatives:
∂M
∂y = 3x2−3y2
∂N
∂x = 3x2−3y2
Since ∂M
∂y =∂N
∂x , the given differential equation is exact.
2. Finding a potential function:
To find the potential function F(x, y), we integrate the coefficient of dx with respect to x:
F(x, y) = ∫(3x2y−y3)dx =x3y−y3x+h(y)
Now, we differentiate F(x, y)with respect to yand set it equal to the coefficient of dy:
∂F
∂y =x3−3xy2=x3−3xy2
This confirms that F(x, y)is the potential function, so the general solution to the given
differential equation is:
x3y−y3x=C
10. Consider the differential equation given by
(2xy +y2)dx + (x2+ 2xy)dy = 0
Determine if the equation is exact and if so, find its solution.
Ans. To determine if the given differential equation is exact, we check if
∂M
∂y =∂N
∂x
where M= 2xy +y2and N=x2+ 2xy.
1. Calculate the partial derivatives:
∂M
∂y = 2x+ 2y
∂N
∂x = 2x+ 2y
2. Since ∂M
∂y =∂N
∂x , the equation is exact.
To find the solution, we need to find a function ϕ(x, y)such that
∂ϕ
∂x = 2xy +y2
∂ϕ
∂y =x2+ 2xy
3. Integrate the first equation with respect to x:
ϕ=∫(2xy +y2)dx =x2y+xy2+h(y)
where h(y)is a function of yto be determined.
4. Differentiate ϕwith respect to yand compare it with the second equation:
∂ϕ
∂y =x2+ 2xy +dh(y)
dy
5. Equate coefficients to get dh(y)
dy = 0 ⇒h(y) = cwhere cis an arbitrary constant.
6. Thus, the general solution to the differential equation is x2y+xy2+c= 0.
11. Consider the differential equation given by:
(2x−y)dx + (x+ 2y)dy = 0
Determine whether the equation is exact. If it is exact, find the general solution.
Ans. To determine if the given differential equation is exact, we need to check if the partial
derivatives of the terms involving xand yare equal. 1. Calculate ∂M
∂y :
∂M
∂y =−1
2. Calculate ∂N
∂x :
∂N
∂x = 1
Since ∂M
∂y =∂N
∂x , the equation is not exact. 3. To make the equation exact, we can find an
integrating factor µ(x):
µ(x) = e∫
∂N
∂x
−∂M
∂y
Ndx
µ(x) = e∫1−(−1)
x+2ydx =e∫2
x+2ydx =e2ln |x+2y|=|x+ 2y|2
4. Multiply both sides of the differential equation by the integrating factor:
|x+ 2y|2(2x−y)dx +|x+ 2y|2(x+ 2y)dy = 0
5. The left side of the equation can be rewritten as the exact differential of a function f(x, y):
f(x, y) = 1
3|x+ 2y|3=C
6. Therefore, the general solution to the differential equation is:
|x+ 2y|3= 3C
12. Question: Consider the differential equation (3ysin(x) + cos(y)) dx + (3xcos(y)−
sin(x)) dy = 0. Determine whether the given differential equation is exact. If it is exact, find the
general solution y(x).
Ans. Step-by-step solution: 1. To check if the given differential equation is exact, we calculate
the partial derivatives Myand Nx:My= 3 sin(x)−sin(y)Nx= 3 cos(y)−cos(x)
2. The differential equation is exact if My=Nx. Thus, we need to solve the following system
of equations: 3sin(x)−sin(y) = 3 cos(y)−cos(x)
3. Rearranging the terms, we get: sin(y)−3cos(y) = cos(x)−3sin(x)
4. We can rewrite the equation as: sin(y)−3cos(y) = √10 (cos(x)
√10 −3sin(x)
√10 )
5. By looking at the equation above, it is clear that both sides are linear combinations of
sin(y),cos(y),sin(x), and cos(x). Thus, the given differential equation is exact.
6. To find the general solution y(x), we integrate Mwith respect to xand compare that
to the partial integration of Nwith respect to y. The general solution will be of the form
F(x, y) = c, where Fis the potential function. Integrating M= 3ysin(x) + cos(y)with
respect to x:∫3ysin(x)dx +∫cos(y)dx = 3y(−cos(x)) + g(y), where g(y)is the constant of
integration.
7. Partially integrating N= 3xcos(y)−sin(x)with respect to y:∫3xcos(y)dy−∫sin(x)dy =
3xsin(y) + h(x), where h(x)is the constant of integration.
8. For F(x, y)to satisfy exactness, we must have g(y) = h(x). Therefore, the general
solution is given by 3ysin(x) + cos(y) = c.
13. Question 13: Consider the differential equation
(3x2y+ 2y2)dx + (x3+ 4xy)dy = 0.
Determine whether the given differential equation is exact. If it is exact, find a solution. If
not, explain why it is not exact.
Ans. To determine if the given differential equation is exact, we will check if the partial
derivatives of the terms are equal. Then, we will find a solution if the equation is exact.
1. Check for exactness: Let
M= 3x2y+ 2y2
and
N=x3+ 4xy.
Calculate the partial derivatives of Mand Nwith respect to y:
∂M
∂y = 3x2+ 4y
∂N
∂x = 3x2+ 4y
Since ∂M
∂y =∂N
∂x , the given differential equation is exact.
2. Find a solution: To find the solution, we need to find a function f(x, y)such that
∂f
∂x = 3x2y+ 2y2
∂f
∂y =x3+ 4xy
Integrating the first equation with respect to xgives:
f(x, y) = x3y+ 2xy2+g(y)
where g(y)is a constant of integration.
Now, differentiate f(x, y)with respect to y:
∂f
∂y =x3+ 4xy +dg
dy
Comparing this with ∂f
∂y =x3+ 4xy, we find that dg
dy = 0, so g(y)is a constant.
Therefore, the solution to the given differential equation is x3y+ 2xy2+C= 0, where Cis
a constant.
14. Let M(x, y)dx +N(x, y)dy be a differential 1-form. Show that the differential form is
exact if and only if the following condition is satisfied:
(∂N
∂x −∂M
∂y )dx ∧dy = 0
Moreover, if M(x, y)dx +N(x, y)dy is exact, then find the function f(x, y)such that df =
M(x, y)dx +N(x, y)dy.
Ans. To show that the differential form is exact if and only if the given condition is satisfied,
we need to show the following:
1. If the differential form M(x, y)dx+N(x, y)dy is exact, then (∂N ∂x−∂M
∂y )dx∧dy = 0:
Since the differential form is exact, there exists some function f(x, y)such that df =
M(x, y)dx +N(x, y)dy. By Clairaut’s Theorem, we have
∂2f
∂x∂y =∂2f
∂y∂x
Now, we can express df as
df =∂f
∂x dx +∂f
∂y dy
We can rewrite this as
df =(∂f
∂x ,∂f
∂y )
By comparing coefficients, we have M(x, y) = ∂f
∂x and N(x, y) = ∂f
∂y . Therefore,
∂N
∂x −∂M
∂y =∂2f
∂x∂y −∂2f
∂y∂x = 0
Hence, (∂N
∂x −∂M
∂y )dx ∧dy = 0, as required.
2. If (∂N ∂x −∂M
∂y )dx ∧dy = 0, then the differential form M(x, y)dx +N(x, y)dy is
exact:
Let f(x, y)be a scalar function defined by
∂f
∂x =M(x, y)and ∂f
∂y =N(x, y)
Then, df =M(x, y)dx +N(x, y)dy, which implies that the differential form is exact.
Therefore, the given differential form M(x, y)dx +N(x, y)dy is exact if and only if (∂N
∂x −
∂M
∂y )dx ∧dy = 0.
15. **Question:** Consider the differential equation
(2xy3−y2)dx + (3x2y2−4x)dy = 0.
Determine whether the differential equation is exact and if so, find the general solution. If
the equation is not exact, determine an integrating factor that makes it exact.
Ans. **Step-by-step solution:**
**1. Determine if the equation is exact:** To determine if the equation is exact, we check if
∂M
∂y =∂N
∂x , where M= 2xy3−y2and N= 3x2y2−4x.
Calculating the partial derivatives:
∂M
∂y = 6xy2−2y
∂N
∂x = 6xy2−4
Since ∂M
∂y =∂N
∂x , the equation is not exact.
**2. Find an integrating factor:** To make the equation exact, we need to find an integrating
factor µ(x, y)such that µM dx +µN dy is exact. We find µby the equation:
∂(µM)
∂y −∂(µN )
∂x
µ=
∂M
∂y −∂N
∂x
N
Substitute Mand Ninto the above equation to find µ.
**3. Solve the differential equation:** Once we find the integrating factor µ(x, y), we multiply
Mand Nby µto make the equation exact and solve for the general solution using the method
of exact equations.
16. Let M(x, y) = 3x2y−5y2and N(x, y) = x3+ 4xy2. Determine whether the following
differential form is exact, and if it is, find its solution.
M(x, y)dx +N(x, y)dy
Ans. To check whether the given differential form is exact, we need to verify if ∂M
∂y =∂N
∂x . If it
is exact, we can then find its solution using the method of integrating factors.
1. Check for Exactness:
Given M(x, y) = 3x2y−5y2and N(x, y) = x3+ 4xy2.
We compute the partial derivatives:
∂M
∂y = 3x2−10y
∂N
∂x = 3x2+ 4y2
Since ∂M
∂y =∂N
∂x , the given differential form is not exact.
2. Finding an Integrating Factor:
To find an integrating factor µ(x, y), we define µsuch that:
µ∂M
∂y =µ(3x2−10y) = µ∂N
∂x =µ(3x2+ 4y2)
Equating the expressions leads to:
µ(3x2−10y) = µ(3x2+ 4y2)⇒µ(−10y) = µ(4y2)
Solving this differential equation gives µ=e14y.
3. Finding the Solution:
We multiply the given differential form by the integrating factor µ(x, y) = e14y:
e14y(3x2y dx −5y2dx +x3dy + 4xy2dy)
This can be rewritten as:
d(e14y(3x2y−5y2+x3+ 4xy2))
Therefore, the solution is e14y(3x2y−5y2+x3+ 4xy2) = C, where Cis the constant of
integration.
17. Let M(x, y)dx +N(x, y)dy = 0 be a first-order ordinary differential equation. Determine
the conditions for exactness and find the general solution for the following differential equation:
(2x+y)dx + (x−y+ 1)dy = 0
Ans. To determine the conditions for exactness and find the general solution for the given
differential equation (2x+y)dx + (x−y+ 1)dy = 0:
1. Check for exactness by verifying if ∂M
∂y =∂N
∂x :∂M
∂y = 1 and ∂N
∂x = 1. Since ∂M
∂y =∂N
∂x , the
equation is exact.
2. Find the potential function Φ(x, y)by integrating Mwith respect to x:Φ(x, y) =
∫(2x+y)dx =x2+xy +g(y), where g(y)is a constant of integration that may depend on y.
3. Differentiate Φ(x, y)partially with respect to yand compare it with N:∂Φ
∂y =x+dg
dy =
x−y+ 1. Comparing terms, we get dg
dy =−y+ 1.
4. Integrate dg
dy =−y+ 1 with respect to yto find g(y):g(y) = −1
2y2+y+C, where Cis
the constant of integration.
5. Substitute g(y)back into the potential function Φ(x, y):Φ(x, y) = x2+xy −1
2y2+y+C.
This is the potential function.
6. The general solution is then given by Φ(x, y) = C, where Cis the constant of integration:
x2+xy −1
2y2+y=C.
18. Let M= (x2+ 2xy)dx + (x2−y2)dy be a differential form. Determine whether Mis
exact, and if it is, find a function f(x, y)such that df =M.
Ans. To check if Mis exact, we need to verify if ∂M
∂y =∂N
∂x , where M= (x2+ 2xy)and
N= (x2−y2).
1. Calculate ∂M
∂y :
∂M
∂y =∂
∂y (x2+ 2xy) = 2x
2. Calculate ∂N
∂x :
∂N
∂x =∂
∂x(x2−y2) = 2x
Since ∂M
∂y =∂N
∂x , the form Mis exact. To find the function f(x, y), we integrate Mwith
respect to xand set it equal to f:
f(x, y) = ∫(x2+ 2xy)dx =1
3x3+x2y+g(y)
where g(y)is the constant of integration with respect to x.
3. To find g(y), differentiate the result with respect to yand set it equal to the Nin M:
∂f
∂y =x2+ 2xy =N=x2−y2
x2+ 2xy =x2−y2
2xy =−y2
2x=−y
y=−2x
So, g(y) = −2xy. Therefore, the function f(x, y) = 1
3x3+x2y−2xy satisfies df =M.
19. Let M= 3x2y2−x3y−2yand N=x3y2+y2−2xy. Determine whether the differential
equation (3x2y2−x3y−2y)dx + (x3y2+y2−2xy)dy = 0 is exact, and if so, find the general
solution.
Ans. To check for exactness, we calculate the partial derivatives of Mwith respect to yand
Nwith respect to x, and then verify if they are equal. If so, we find a potential function Ψ(x, y)
such that dΨ = Mdx +Ndy. Finally, we find the general solution by integrating Ψwith respect
to one variable and setting it equal to a constant.
20. Consider the following differential equation
(2x3+ 3y)dx −(x2+ 4y)dy = 0.
Ans. To determine whether the given differential equation is exact and find its solution, we will
follow these steps:
1. Check for exactness by computing the partial derivatives of the coefficients with respect
to y. 2. If the equation is exact, find a function ψsuch that dψ =Mdx +Ndy. 3. Solve for ψ
by integrating either Mor Nwith respect to the corresponding variable. 4. Use ψto find the
general solution of the differential equation.
1. By checking if ∂M
∂y =∂N
∂x , we have:
∂M
∂y = 3,∂N
∂x =−2x.
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. To make the equation exact, we need an integrating factor µsuch that:
µ(2x3+ 3y)dx −µ(x2+ 4y)dy = 0
∂
∂y (µ(2x3+ 3y)) = ∂
∂x(−µ(x2+ 4y)).
Expanding and equating the mixed partial derivatives, we get:
3µ=−µ=⇒µ=−1
4.
3. Multiply the differential equation by −1
4to make it exact:
−1
4(2x3+ 3y)dx +1
4(x2+ 4y)dy = 0.
Integrating N=1
4(x2+ 4y)with respect to y, we get the potential function ψ:
ψ=∫(1
4(x2+ 4y))dy =1
4x2y+y2+C(x),
where C(x)is the constant of integration.
4. The general solution is given by ψ(x, y) = Cwhere Cis a constant:
1
4x2y+y2+C(x) = C.
21. Suppose we have a differential equation of the form (ycos(x) + xsin(y))dx + (xcos(y)−
ysin(x))dy = 0.
1. Determine whether the given differential equation is exact. If it is exact, find the solution.
If it is not exact, determine an integrating factor to make it exact and find the solution.
Ans. To check for exactness, compute the partial derivatives of the coefficients of dx and dy:
∂
∂y (ycos(x) + xsin(y)) = −sin(y) + xcos(y),
∂
∂x(xcos(y)−ysin(x)) = cos(y)−ycos(x).
Since these partial derivatives are not equal, the given differential equation is not exact. Let’s
find the integrating factor to make it exact.
1. Compute the integrating factor µ(x)using the formula
µ(x) = e∫My−Nx
Ndx.
Here, M=ycos(x) + xsin(y),N=xcos(y)−ysin(x),My=−sin(y) + xcos(y), and
Nx=cos(y)−ycos(x).
So, we have My−Nx
N=−sin(y)+xcos(y)−(cos(y)−ycos(x))
xcos(y)−ysin(x)=−sin(y)+xcos(y)−cos(y)+ycos(x)
xcos(y)−ysin(x)=ycos(x)−sin(y)
xcos(y)−ysin(x).
Therefore, the integrating factor is µ(x) = e∫ycos(x)−sin(y)
xcos(y)−ysin(x)dx.
2. After finding the integrating factor, multiply it by the original differential equation to make
it exact.
3. Solve the exact differential equation using standard methods such as integration.
22. Let M= (3y2−6xy)dx + (2x−2x2)dy be a differential 1-form in R2. Determine whether
Mis exact. If it is exact, find a function f(x, y)such that M=df.
Ans. To determine if the differential 1-form M= (3y2−6xy)dx + (2x−2x2)dy is exact, we
need to check if ∂M1
∂y =∂M2
∂x , where M1= 3y2−6xy and M2= 2x−2x2.
1. Calculate the partial derivatives:
∂M1
∂y = 6y−6xand ∂M2
∂x = 2 −4x.
2. Check if the derivatives are equal:
6y−6x?
= 2 −4x.
3. Simplify the equation and determine if Mis exact:
6y−6x= 2 −4x=⇒6y= 8 + 2x.
Since the partial derivatives are not equal, Mis not exact.
Therefore, the given differential 1-form Mis not exact and does not have a potential function
f(x, y).
23. Question: Determine whether the following differential equation is exact and, if so, find a
potential function that satisfies it:
(2xy + 3)dx + (x2−1)dy = 0
Ans. To determine if the given equation is exact, we check if the partial derivatives of the
coefficients with respect to yand xare equal. If they are equal, we can find a potential function
that satisfies the equation.
1. Check for Exactness: Calculate the partial derivatives:
∂
∂y (2xy + 3) = 2x
∂
∂x(x2−1) = 2x
Since the partial derivatives are equal, the equation is exact.
2. Find the Potential Function: To find the potential function, integrate the coefficient
of dx with respect to x:
F(x, y) = ∫(2xy + 3)dx =x2y+ 3x+g(y)
Now, differentiate Fwith respect to yand set it equal to the coefficient of dy:
∂F
∂y =x2+dg
dy =x2−1
From this, we can see that dg
dy =−1, which implies that g(y) = −y. Thus, the potential
function is:
F(x, y) = x2y+ 3x−y+C
Therefore, the potential function that satisfies the differential equation is x2y+ 3x−y+C,
where Cis a constant.
24. Question 24: Determine whether the following differential equation is exact or not, and if
it is exact, find its general solution:
(2x2+y)dx + (2xy + 1)dy = 0
Ans. To determine if the given differential equation is exact and find its general solution, we
will follow these steps:
1. Check for exactness by verifying if ∂M
∂y =∂N
∂x . 2. If the equation is exact, find the
integrating factor µ(x, y). 3. Solve the equation by integrating to find the general solution.
1. Check for exactness: Given differential equation:
(2x2+y)dx + (2xy + 1)dy = 0
Checking exactness:
∂M
∂y =∂
∂y (2x2+y) = 1
∂N
∂x =∂
∂x(2xy + 1) = 2y
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Find the integrating factor µ(x, y):To make the equation exact, we need to find an
integrating factor µ(x, y)such that µ(x, y)·(2x2+y)dx +µ(x, y)·(2xy + 1)dy = 0 is exact.
We can find µ(x, y)by the relationship:
∂
∂y (2x2+y)−∂
∂x (2xy + 1)
2xy + 1 =1−2y
2xy + 1
3. Solve the differential equation: Note: This step can vary depending on the integrating
factor found in the previous step.
Since the given differential equation is not exact and the integrating factor is not readily
apparent, further manipulation or techniques may be required to solve the equation.
Therefore, the general solution of the given differential equation may involve additional meth-
ods beyond standard exact differential equation techniques.
25. Question 25: Consider the differential equation given by
(3x2y+ 2y3)dx + (x3+ 3xy2)dy = 0
Determine whether the equation is exact. If it is exact, find the general solution. If it is not
exact, find an integrating factor µ(x, y)to make it exact, and then solve the equation.
Ans. To determine if the given equation is exact, we check if ∂M
∂y =∂N
∂x , where M= 3x2y+ 2y3
and N=x3+ 3xy2.
1. Check for exactness: We calculate the partial derivatives:
∂M
∂y = 3x2+ 6y2
∂N
∂x = 3x2+ 3y2
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Find an integrating factor: To find an integrating factor, we look for a function µ(x, y)
such that µMdx +µNdy = 0 is exact. We can find µby solving the following PDE:
∂
∂y (µM) = ∂
∂x(µN)
Expanding, we get:
3x2∂µ
∂y + 6y2µ= 3x2∂µ
∂x + 3y2µ
By comparing coefficients:
3x2∂µ
∂y −3x2∂µ
∂x = 3y2µ−6y2µ
3x2dµ
dy −3x2dµ
dx =−3y2µ
This is a first-order linear PDE, which can be solved to find µ.
3. Solve for the integrating factor µ:Integrating the PDE gives:
dµ
µ=−3y2
3x2dx +3x2
3x2dy
ln |µ|=−y2+x2+C
µ=ex2−y2eC=Cex2−y2
4. Find the general solution: Multiplying the original equation by the integrating factor
µ=Cex2−y2, we obtain an exact differential equation:
(3x2y+ 2y3)ex2−y2dx + (x3+ 3xy2)ex2−y2dy = 0
From here, the general solution can be found by integrating both sides with respect to xand
y.
26. Let M = (2x + 3y)dx + (4x + 5y)dy and N = (3x - 4y)dx + (2x - 3y)dy.
Determine whether the differential form Mdx + Ndy is exact. If it is exact, find a function
f(x, y) such that df = Mdx + Ndy.
Ans. To check whether the form Mdx + Ndy is exact, we need to verify if �M/�y = �N/�x.
1. Calculate �M/�y: �M/�y = 3.
2. Calculate �N/�x: �N/�x = 3.
Since �M/�y = �N/�x, the form Mdx + Ndy is exact.
3. To find f(x, y), integrate M with respect to x and set it equal to the partial derivative of f
with respect to y: �f/�y = 2x + 3y.
Integrating, we get f(x, y) = x2+ 3xy +g(y), whereg(y)isafunctionofy.
4. Take the partial derivative of f with respect to y and set it equal to N to find g(y):
�(x2+ 3xy +g(y))/∂y= 2x−3y.
By comparing coefficients, we find g’(y) = -3y + constant. So, g(y) = -1.5y2+C, whereCisaconstant.
5. So, the function f(x, y) is f(x, y)= x2+ 3xy −1.5y2+C, whereCisaconstant.
27. Consider the differential equation:
(y3ex+x)dx + (3y2ex−2y)dy = 0
Ans. To determine whether the given differential equation is exact, we check if the partial
derivatives of the coefficients respect to yand respect to xare the same. If not exact, we will
show how to make it exact by multiplying by an integrating factor.
1. Exactness check: Let M=y3ex+xand N= 3y2ex−2y. Compute the partial
derivatives of Mand Nwith respect to y:
∂M
∂y = 3y2ex−2,
and with respect to x:
∂N
∂x = 3y2ex.
Since ∂M
∂y =∂N
∂x , the differential equation is not exact.
2. Find the integrating factor: To find the integrating factor, we will find a function
µ(x, y)such that
µ(x, y)M dx +µ(x, y)Ndy = 0
is exact.
We can find µ(x, y)by integrating the following equation:
∂
∂y (µM) = ∂
∂x(µN).
Expanding the derivatives gives:
µ∂M
∂y +∂µ
∂y M=µ∂N
∂x +∂µ
∂x N.
Since Mdx+N dy = 0 is our original differential equation, we can simplify the above equation
to: ∂µ
∂y
µ=
∂N
∂x −∂M
∂y
M.
Substitute M,N,∂M
∂y , and ∂N
∂x into the above equation and solve for µ(x, y).
3. Multiply by the integrating factor: Once we have found µ(x, y), multiply the original
differential equation by µ(x, y)to make it exact.
Solve the resulting exact differential equation to find the solution.
28. Let M(x, y)dx +N(x, y)dy = 0 be a first-order differential equation. Determine conditions
for exactness and find the general solution for the following equation:
2x2ydx + (x2+y)dy = 0
Ans. To determine if the given differential equation is exact, we need to check if ∂M
∂y =∂N
∂x .
If this condition is satisfied, we can find a potential function ϕ(x, y)such that ∂ϕ
∂x =Mand
∂ϕ
∂y =N.
1. Check for exactness:
Given M(x, y) = 2x2yand N(x, y) = x2+y, we calculate the partial derivatives as follows:
∂M
∂y = 2x2
∂N
∂x = 2x
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Make the equation exact:
In order to make the equation exact, we can try to find an integrating factor µ(x, y)such
that µM and µN have partial derivatives that are symmetric: ∂
∂y (µM ) = ∂
∂x (µN ).
Let’s find µby solving the following equation:
∂
∂y (2x2y)−∂
∂x (x2+y)
N=2x2−2x
x2+y
This simplifies to −2x
x2+y. Therefore, µ(x, y) = 1
x2+y.
3. Find the general solution:
Multiplying the given equation by the integrating factor µ=1
x2+y, we get:
2xydx + (1)dy = 0
This equation is now exact as it satisfies the condition ∂
∂y (2xy)−∂
∂x (1) = 0.
To find the potential function ϕ(x, y), we integrate Mwith respect to x:
ϕ(x, y) = ∫2xydx =x2y+g(y), where g(y)is an arbitrary function of y.
Now, we differentiate ϕ(x, y)with respect to yand set it equal to N(x, y)to find g(y):
∂ϕ
∂y =x2+dg
dy =x2+g′(y) = x2+y
Solving g′(y) = ygives g(y) = y2
2.
Therefore, the general solution is x2y+y2
2=C, where Cis the constant of integration.
29. Question 29: Consider the differential equation (y−x2)dx + (2xy −1)dy = 0. Determine
whether the equation is exact, and if so, find the general solution.
Ans. To determine if the given differential equation is exact, we need to check if ∂M
∂y =∂N
∂x ,
where M=y−x2and N= 2xy −1.
1. Check exactness: Compute the partial derivatives:
∂M
∂y = 1 and ∂N
∂x = 2y
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Condition for exactness: In order to make the equation exact, we need to find an
integrating factor µ(x, y)such that µ(x, y) [(y−x2)dx + (2xy −1)dy]is an exact differential.
3. Find integrating factor: Since the integrating factor µ(x, y)depends only on xin this
case, we can write µ(x). From the condition for exactness, we have ∂
∂y (µM ) = ∂
∂x (µN ). This
gives us:
µ∂M
∂y +∂µ
∂y M=µ∂N
∂x +∂µ
∂x N
µ+ 0 = 2µy +dµ
dx (2xy −1)
0 = 2µy + 2xµdµ
dx −dµ
dx
This is a first-order linear differential equation in terms of µ(x). Solve for µ(x)to find the
integrating factor.
4. Find general solution: After finding the integrating factor µ(x), multiply through the
entire equation by µ(x)to make it exact. Then, integrate to find the general solution of the
differential equation.
30. Consider the following differential equation:
(3x2+y)dx + (2y+x)dy = 0
Determine whether the differential equation is exact. If it is exact, find the general solution.
Ans. To determine whether the given differential equation is exact, we check the mixed partial
derivatives of the terms involving xand y. If the mixed partial derivatives are equal, the equation
is exact. Otherwise, it is not exact.
1. Determine if the equation is exact: Calculate the partial derivatives of the terms
involving xand y:
∂M
∂y = 1
∂N
∂x = 1
Since ∂M
∂y =∂N
∂x , the differential equation is not exact.
2. Find an integrating factor: To find an integrating factor, we use the following formula:
µ=e∫
∂M
∂y
−∂N
∂x
Ndx
In this case, we have:
µ=e∫1−1
2y+xdx =e∫0
2y+xdx =e0= 1
3. Multiply through by the integrating factor: Multiplying the original equation by the
integrating factor µ= 1 gives:
(3x2+y)dx + (2y+x)dy = 0
This equation remains the same, as the integrating factor is 1.
4. Identify the exact differential equation: After multiplying through by the integrating
factor, we find that the equation remains non-exact.
Therefore, the differential equation does not admit an integrating factor and is not exact.
Further techniques such as finding an integrating factor cannot be used to solve this differential
equation.
The general solution to the differential equation can be found by using other methods such
as separation of variables or integrating factors if applicable.
31. Consider the differential equation given by:
(3y2−2xy)dx + (x2−6y)dy = 0
Determine whether the given differential equation is exact, and if so, find the general solution.
Ans. To check for exactness, we compute the partial derivatives ∂M
∂y and ∂N
∂x , where M=
3y2−2xy and N=x2−6y.
1. Compute ∂M
∂y :
∂M
∂y = 6y−2x
2. Compute ∂N
∂x :
∂N
∂x = 2x
Since ∂M
∂y =∂N
∂x , the given differential equation is not exact. To find an integrating factor,
we first check if the integrating factor depends only on xor only on y.
3. Check if the integrating factor depends only on x: Let p(x)be the integrating factor. We
have: ∂
∂y (p(x)M) = ∂
∂x(p(x)N)
Expanding, we get:
p(x)(6y−2x) = p′(x)(x2−6y)
Since the left side depends on yand the right side depends on x, the integrating factor cannot
depend only on x.
4. Check if the integrating factor depends only on y: Let q(y)be the integrating factor. We
consider the equation:
∂
∂y (q(y)M) = ∂
∂x(q(y)N)
Expanding, we get:
q(y)(6y−2x) = q′(y)(x2−6y)
This equation is not possible to solve, so this also fails. Therefore, an integrating factor that
depends only on xor ydoesn’t exist.
5. Find a general solution: Since we couldn’t find an integrating factor, we need to use other
methods such as the method of integrating factors or an integrating factor involving both xand
y. In this case, we couldn’t find a suitable integrating factor; hence, we cannot make the given
differential equation exact.
Therefore, the general solution will involve solving the differential equation or manipulating
it to a form where it can be solved by methods other than exactness.
32. Let M(x, y)and N(x, y)be two functions defined on a domain D. Consider the first-order
ordinary differential equation
M(x, y) + N(x, y)dy
dx = 0.
Suppose that the functions Mand Nare such that the equation is exact on D, i.e., there
exists a function f(x, y)such that
∂f
∂x =M(x, y)and ∂f
∂y =N(x, y),
for all (x, y)∈D.
33. Find the general solution to the exact differential equation
(2y+ex)dx + (x+ey)dy = 0.
Ans. 1. To determine if the given equation is exact, we check if
∂M
∂y =∂N
∂x .
Calculating these partial derivatives, we have
∂M
∂y = 2 and ∂N
∂x = 1,
which are not equal. Therefore, the equation is not exact.
2. However, we can make the equation exact by multiplying through by an integrating factor,
denoted by µ(x, y), such that
µ(x, y) ((2y+ex)dx + (x+ey)dy) = 0.
In this case, our integrating factor is
µ(x, y) = 1
µ(x, y)=1
2y+ex.
3. Multiplying through by the integrating factor, we have
1
2y+ex(2y+ex)dx +1
2y+ex(x+ey)dy = 0.
Simplifying, we get
dx +(x+ey
2y+ex)dy = 0.
4. Now, we can rewrite the equation as
dx
dx +(x+ey
2y+ex)dy
dx = 0.
This allows us to look for a function f(x, y)such that
∂f
∂x = 1 and ∂f
∂y =x+ey
2y+ex.
5. Integrating the first equation with respect to x, we get f=x+h(y), where h(y)is a
function of y.
6. Substituting f=x+h(y)into the second equation and matching coefficients, we find
h′(y) = ey
2y+ex.
Integrating this expression gives us
h(y) = ln |2y+ex|+C,
where Cis an arbitrary constant.
7. Therefore, the general solution to the exact differential equation is
ln |2y+ex|+x=C,
where Cis a constant of integration.
34. Consider the differential equation:
(3x2y2−2y3)dx + (2x3y−3xy2)dy = 0
Determine whether the differential equation is exact. If it is exact, find the general solution.
Ans. To determine if the given differential equation is exact, we need to check if the following
condition is satisfied:
∂M
∂y =∂N
∂x
where Mand Nare the coefficients of dx and dy, respectively.
1. Compute ∂M
∂y :
∂
∂y (3x2y2−2y3) = 6x2y−6y2
2. Compute ∂N
∂x :
∂
∂x(2x3y−3xy2) = 6x2y−3y2
Since ∂M
∂y =∂N
∂x , the given differential equation is exact. To find the general solution, we
need to find a function ϕ(x, y)such that:
∂ϕ
∂x = 3x2y2−2y3and ∂ϕ
∂y = 2x3y−3xy2
3. Integrate the first equation with respect to x:
ϕ=∫(3x2y2−2y3)dx =x3y2−2yx2+h(y)
where h(y)is a function of yonly.
4. Differentiate ϕwith respect to yand equate it to ∂ϕ
∂y :
∂
∂y (x3y2−2yx2+h(y)) = 2x3y−3xy2
5. Solve for h(y):
2x3y−2x2+h′(y) = 2x3y−3xy2⇒h′(y) = −3xy2
Integrating h′(y)gives h(y) = −xy3+Cwhere Cis a constant.
6. Therefore, the general solution of the given exact differential equation is:
x3y2−2yx2−xy3+C= 0
where Cis a constant of integration.
35. Consider the following differential equation:
(3y2−2x)dx + (6xy −3x2)dy = 0
Determine whether the differential equation is exact or not. If it is exact, find the general
solution of the differential equation.
Ans. To determine whether the given differential equation is exact, we need to check if the partial
derivatives ∂M
∂y and ∂N
∂x are equal, where the given equation is in the form M(x, y)dx+N(x, y)dy =
0.
1. Check for exactness: Given differential equation:
(3y2−2x)dx + (6xy −3x2)dy = 0
Calculating the partial derivatives:
∂M
∂y = 6y
∂N
∂x = 6y
Since ∂M
∂y =∂N
∂x , the differential equation is exact.
2. Find the general solution: To find the general solution, we integrate Mwith respect
to xand set it equal to a function f(y)of yonly.
Integrating M= 3y2−2xwith respect to x, we get:
∫(3y2−2x)dx = 3y2x−x2+g(y)
Next, we differentiate 3y2x−x2+g(y)with respect to yand set it equal to Nto solve for
g(y).
d
dy (3y2x−x2+g(y)) = 6xy −2x+g′(y) = 6xy −3x2
Comparing coefficients, we find g′(y) = −3x2.
Integrating g′(y) = −3x2with respect to y, we get:
g(y) = −3x2y+h(x)
Thus, the general solution is given by:
3y2x−x2−3x2y+h(x) = C
where Cis the constant of integration and h(x)is an arbitrary function of x.
36. Consider the following first-order differential equation:
(3x2+ 2y)dx + (2x−3y)dy = 0
Determine if the given differential equation is exact. If it is exact, find the general solution.
If it is not exact, find an integrating factor that makes it exact, and then find the solution.
Ans. To determine if the given differential equation is exact, we check the equality of the mixed
partial derivatives of the coefficients of dx and dy.
1. Check exactness:
Given the equation (3x2+ 2y)dx + (2x−3y)dy = 0, Let M= 3x2+ 2yand N= 2x−3y.
Calculate the partial derivatives of Mand Nwith respect to yand x:
∂M
∂y = 2,∂N
∂x = 2
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Find integrating factor:
To make the equation exact, we need to find an integrating factor µ(x, y)such that
µ(x, y)((3x2+ 2y)dx + (2x−3y)dy)
is exact. The integrating factor is given by
µ=e∫
∂N
∂x
−∂M
∂y
Ndx
Calculate the integrating factor:
µ=e∫2−2
2x−3ydx =e∫0dx =e0= 1
3. Solve the equation with the integrating factor:
Multiply the given equation by the integrating factor µ= 1:
(3x2+ 2y)dx + (2x−3y)dy = 0
Now we have an exact differential equation and can write it in the form:
∂u
∂xdx +∂u
∂y dy = 0
where uis a new function.
4. Find the general solution:
Integrating the coefficients of dx and dy with respect to the respective variables:
u(x, y) = ∫(3x2+ 2y)dx +g(y)
u(x, y) = x3+ 2xy +g(y)
Taking the partial derivative of uwith respect to ygives:
∂u
∂y = 2x+g′(y)
Comparing with the coefficient of dy in the original equation, we have:
2x+g′(y) = 2x−3y
Therefore, g′(y) = −3y⇒g(y) = −3
2y2+Cfor some constant C.
Thus the general solution of the differential equation is:
x3+ 2xy −3
2y2=C
where Cis an arbitrary constant.
