MAT 275 - MODERN DIFFERENTIAL EQUATIONS -
Conditions for exactness and methods for finding solutions
Practice Material - Set 3
1. Consider the following differential equation:
(y−x2+ 2)dx + (2y−2x)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 that makes it exact and then find the solution.
Ans. To determine if the given differential equation is exact, we check if ∂M
∂y =∂N
∂x .
1. Calculate the partial derivatives:
∂M
∂y =−1
∂N
∂x =−2
2. Since the partial derivatives are not equal, the equation is not exact. To find an integrating
factor, we set: 1
µ(∂M
∂y −∂N
∂x )=1
µ(−1 + 2) = 1
µ= 1
3. Solving for µ, we find that µ=e∫1dx =ex.
4. Multiply the entire equation by the integrating factor µ=ex:
ex(y−x2+ 2)dx +ex(2y−2x)dy = 0
5. Now we check if the equation is exact. Calculate the partial derivatives:
∂
∂y (ex(y−x2+ 2))=ex
∂
∂x (ex(2y−2x)) = ex
6. Since the partial derivatives are equal, the equation is exact. We can now find the general
solution by integrating Mwith respect to xand Nwith respect to y:
∫(y−x2+ 2)exdx =ex(y−x2+ 2) + g(y)
∫(2y−2x)exdy =ex(2y−2x) + h(x)
7. Equating the two expressions, we find:
ex(y−x2+ 2) + g(y) = ex(2y−2x) + h(x)
8. To find g(y)and h(x), we compare coefficients and equate terms. We can then write the
general solution as ex(y−x2+ 2) = Cwhere Cis the constant of integration.
2. Question: Consider the differential equation
(2y2−x2)dx + (4xy)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 check the partial derivatives
of the coefficients with respect to yand x. 1. Compute the partial derivative of (2y2−x2)with
respect to y:
∂
∂y (2y2−x2) = 4y
2. Compute the partial derivative of (4xy)with respect to x:
∂
∂x(4xy) = 4y
Since the partial derivatives are equal, the given differential equation is exact.
To find the general solution, we proceed by finding the integrating factor µ(x). 3. Write the
differential equation in the form M(x, y)dx +N(x, y)dy = 0:
(2y2−x2)dx + (4xy)dy = 0
4. Compute ∂M
∂y and ∂N
∂x :
∂
∂y (2y2−x2) = 4yand ∂
∂x(4xy) = 4y
5. Find the integrating factor:
µ(x) = e∫(∂N
∂x −∂M
∂y
N)dx
=e∫0
4xy dx =e0= 1
6. Multiply the differential equation by the integrating factor:
(2y2−x2)dx + (4xy)dy = 0 =⇒dx(2y2−x2) + dy(4xy) = 0
7. Integrate the equation with respect to x:
∫(2y2−x2)dx +φ(y) = C
2xy2−x3
3+φ(y) = C
8. Differentiate the above equation with respect to yand compare with the given equation to
find φ(y):
∂
∂y (2xy2−x3
3) = 4xy =⇒4xy +φ′(y) = 4xy =⇒φ′(y) = 0
9. Integrate φ′(y) = 0 to find φ(y):
φ(y) = C1
10. Substitute φ(y) = C1back into the integrated equation:
2xy2−x3
3+C1=C
Therefore, the general solution is 2xy2−x3
3+C1=C.
3. Consider the following differential equation:
(2x3y−y3)dx + (3x2y2−x)dy = 0
Determine if the given equation is exact. If it is exact, find the general solution. If it is not
exact, find an integrating factor to make it exact and then find the general solution.
Ans. To determine if the given differential equation is exact, we first calculate the partial
derivatives of Mand N, where Mand Nare the coefficients of dx and dy, respectively.
1. Calculate ∂M /∂y and ∂N/∂x:
∂M
∂y = 2x3−3y2and ∂N
∂x = 6xy2−1
Since ∂M
∂y =∂N
∂x , the given equation is not exact. Now, we need to find an integrating factor
µ(x, y), such that µ(x, y)M dx +µ(x, y)Ndy = 0 is exact.
2. Find the integrating factor µ(x, y):
µ(x, y) = e∫(∂N
∂x −∂M
∂y
M)dx
µ(x, y) = e∫((6xy2−1)−(2x3−3y2)
2x3y−y3)dx
µ(x, y) = e∫(6xy2−1−2x3+3y2
2x3y−y3)dx
µ(x, y) = e∫3y2−2x3+6xy2−1
2x3y−y3dx
µ(x, y) = e∫3y2+6xy2−2x3−1
2x3y−y3dx
After solving the integral, we find the integrating factor µ(x, y).
3. Once we have found the integrating factor µ(x, y), we multiply the given differential
equation by it:
µ(x, y)[(2x3y−y3)dx + (3x2y2−x)dy] = 0
4. The new equation is exact. Find the general solution using the method for solving exact
equations.
5. Write the equation in the form F(x, y) = cand solve for cusing the initial condition, if
provided.
4. Consider the differential equation:
(2y+ 3x2)dx + (2x−2y)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 check if ∂M
∂y =∂N
∂x .
1. Calculate the partial derivatives of M= 2y+ 3x2and N= 2x−2ywith respect to yand
x:∂M
∂y = 2,∂N
∂x = 2
2. Since ∂M
∂y =∂N
∂x , the differential equation is exact.
3. To find the general solution, we need to find a function f(x, y)such that ∂f
∂x =Mand
∂f
∂y =N.
4. Integrating Mwith respect to x, we get:
f(x, y) = ∫(2y+ 3x2)dx = 2xy +x3+h(y)
where h(y)is a constant of integration that may depend on y.
5. Now, we differentiate f(x, y)with respect to y, set it equal to N, and solve for h(y):
∂f
∂y = 2x+h′(y) = 2x−2y=⇒h′(y) = −2y
6. Integrating h′(y)with respect to y, we get:
h(y) = −y2+C
where Cis an arbitrary constant.
7. Thus, the general solution to the given differential equation is:
2xy +x3−y2+C= 0
5. Consider the following differential equation:
(2x+y2)dx + (2y+ 2xy)dy = 0
Determine whether the given equation is exact or not. If it is exact, find the solution using
the exactness condition.
Ans. To determine if the given differential equation is exact, we check if ∂M
∂y =∂N
∂x . After
confirming exactness, we find the solution by integrating Mwith respect to xand then solving
for the constant of integration.
1. Exactness Condition: Let M= 2x+y2and N= 2y+ 2xy. We calculate the partial
derivatives: ∂M
∂y = 2y
∂N
∂x = 2y+ 2y= 4y
Since ∂M
∂y =∂N
∂x , the given differential equation is exact.
2. Finding the Solution: Since the equation is exact, there exists a function ψ(x, y)such
that dψ = (2x+y2)dx + (2y+ 2xy)dy.
Integrating M= 2x+y2with respect to xgives us:
ψ=∫(2x+y2)dx =x2+xy2+g(y)
Now, we differentiate the expression with respect to yand equate it to Nto find g(y):
∂ψ
∂y = 2xy +g′(y) = 2y+ 2xy
g′(y) = 2y
g(y) = y2+C
Therefore, the solution to the given differential equation is:
ψ=x2+xy2+y2+C
6. Let M(x, y)dx +N(x, y)dy be a differential equation, where Mand Nare functions of two
variables xand y. Determine whether the following differential equations are exact, and if so,
find the general solution.
1. (ey+xy)dx + (x2−2y)dy = 0
Ans. To determine if the given differential equation is exact, we check if ∂M
∂y =∂N
∂x .
1. Calculate the partial derivatives: ∂M
∂y =ey+xand ∂N
∂x = 2x
Since ∂M
∂y =∂N
∂x , the differential equation is not exact.
2. To find an integrating factor µ(x, y), calculate:
∂M
∂y −∂N
∂x
N=(ey+x)−2x
x2−2y=ey−x
x2−2y
µ(x, y) = e∫ey−x
x2−2ydy
3. Find µ(x, y)by solving the integral: µ(x, y) = e∫ey−x
x2−2ydy =e∫ey
x2−2ydy−∫x
x2−2ydy
4. Once you have found µ(x, y), multiply the entire differential equation by µ(x, y)and
check if it is exact. If it is exact, proceed to find the general solution. If not, consider alternative
methods such as integrating factors depending on the specific form of the differential equation.
7. Consider the following differential equation:
(2xy3+ex)dx + (3x2y2+ey)dy = 0
Let M= 2xy3+exand N= 3x2y2+ey.
Is the given differential equation exact? If not, find an integrating factor µ(x, y)to make it
exact and solve the differential equation.
Ans. To determine if the given differential equation is exact, we check if ∂M
∂y =∂N
∂x .
1. Calculate the partial derivatives:
∂M
∂y = 6xy2
∂N
∂x = 6xy2+ex
2. Since ∂M
∂y =∂N
∂x , the given differential equation is not exact.
3. To find an integrating factor µ(x, y), we set up the equation 1
µ(∂M
∂y −∂N
∂x ) = ∂
∂x
µ
∂y .
4. Substitute the given Mand Ninto the equation above and solve for µ:
1
µ(6xy2−(6xy2+ex)) = −∂
∂x(3x2y2+ey)
−ex
µ=−3x2y2−ey
µ=exeye3x2y2
5. Using the integrating factor µ=exeye3x2y2, multiply both sides of the differential equation
by µ.
(2xy3exeye3x2y2+ex+2y+3x2y2)dx + (3x2y2exeye3x2y2+ex+y+3x2y2)dy = 0
6. Now the differential equation is exact. We can find the solution by integrating the terms
with respect to xand yseparately.
7. Integrating the term with respect to x:
∫2xy3exeye3x2y2+ex+2y+3x2y2dx =ex+2y+3x2y2+C(y)
8. Integrating the term with respect to y:
∫3x2y2exeye3x2y2+ex+y+3x2y2dy =ex+y+3x2y2+C(x)
9. Therefore, the general solution to the differential equation is given by ex+2y+3x2y2+C(y) =
ex+y+3x2y2+C(x).
8. Let M(x, y)dx +N(x, y)dy = 0 be a first-order ordinary differential equation. Determine
conditions for exactness and explain methods for finding solutions.
Ans. To determine the conditions for exactness of the given ODE and explain methods for
finding solutions, follow these steps:
1. Check for Exactness: A first-order ODE M(x, y)dx +N(x, y)dy = 0 is exact if and only
if ∂M
∂y =∂N
∂x .
2. Determine Integrating Factor: If the ODE is not exact, find an integrating factor µ(x)
such that µ(x)M(x, y)dx +µ(x)N(x, y)dy = 0 is exact. The integrating factor is given by
µ(x) = e∫P(x)dx, where P(x)is a function of xonly that can be determined by the condition
∂
∂y (µM )
µ∂x =∂
∂x (µN )
µ∂y .
3. Solve the Exact ODE: After obtaining the exact ODE, solve it by integrating dF = 0,
where F(x, y) = Cis the general solution, and Cis the constant of integration.
4. Boundary Conditions: If initial conditions are given, use them to find a particular solution
by substituting them into the general solution.
5. Check for Correctness: Finally, check the correctness of the solution by substituting it back
into the original ODE to ensure that it satisfies the equation.
By following these steps, you can determine the conditions for exactness of a given ODE and
find its solution.
9. Suppose we have the following differential equation:
(2x2y+y2+exy)dx + (x2+ 2xy +exy)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 that makes it exact and find the general solution.
Ans. To determine if the given equation is exact, we check if ∂N
∂x =∂M
∂y , where the given
equation is of the form M dx +Ndy = 0.
1. Check for exactness: Let M= 2x2y+y2+exyand N=x2+ 2xy +exy. Calculating the
partial derivatives:
∂M
∂y = 2x2+ 2y+ex
∂N
∂x = 4xy +ex
The equation is not exact because ∂N
∂x =∂M
∂y .
2. Find the integrating factor: To find the integrating factor µ(x), we set up the following
equation: µ(x)(∂M
∂y −∂N
∂x ) = µ′(x)
µ(x)= 0. Therefore, µ(x) = e∫4xy+ex−2x2−2y
x2+2xy+exydx.
Calculating the integral in the exponent:
e∫4xy+ex−2x2−2y
x2+2xy+exydx =e∫2(x+y)(2x−y)+ex
(x+y)2dx =e2ln |x+y|+∫ex
x+ydx
= (x+y)2e∫ex
x+ydx
3. Find the solution with the integrating factor: Applying the integrating factor to the given
equation:
(2x2y+y2+exy)dx + (x2+ 2xy +exy)dy = 0
⇒(x2+ 2xy +exy)(x+y)2dx + (x2+ 2xy +exy)(x+y)2dy = 0
This equation is exact. Integrating both sides, we find the general solution.
The process of finding the integrating factor and solving the resulting exact differential equa-
tion can be quite complex but important in the study of differential equations.
10. Find the particular solution y(x)of the following differential equation:
(2x+y)dx + (x+ 2y)dy = 0
Ans. To find the particular solution y(x)of the given differential equation, we need to check
for exactness and then use appropriate methods to solve for y(x).
1. Check for exactness: We check the exactness of the given equation by computing the
partial derivatives of (2x+y)dx + (x+ 2y)dy with respect to yand x, respectively.
∂M
∂y = 1 =∂N
∂x = 1
Since the equation is not exact, we will find an integrating factor to make it exact.
2. Find the integrating factor: Let (x)be the integrating factor such that when multiplied
to the given equation, it becomes exact. We have the following condition for exactness:
∂(µM)
∂y =∂(µN)
∂x
Substitute M= (2x+y)and N= (x+ 2y)into the above equation and solve for (x).
3. Solve the integrating factor equation: From the condition above, we get:
∂(µ(2x+y))
∂y =∂(µ(x+ 2y))
∂x
Solve this equation to obtain (x).
4. Rewrite the equation with the integrating factor: Once we have found (x), multiply
it to the given differential equation to make it exact. The new equation will look like:
µ(2x+y)dx +µ(x+ 2y)dy = 0
5. Solve for y(x):Now that the equation is exact, we can find a function F(x, y)such that
∂F
∂x =µ(2x+y)and ∂F
∂y =µ(x+ 2y)
Solve these equations simultaneously to find F(x, y).
6. Obtain the solution: The solution of the differential equation will be given by F(x, y) =
C, where Cis a constant. Finally, solve for y(x)from the equation F(x, y) = Cto get the
particular solution.
11. Question: Determine whether the following first-order differential equation is exact, and if
it is, find the general solution:
(2x+y)dx + (x+ 2y)dy = 0
Ans. Step-by-step solution:
1. Check for exactness by verifying if ∂M
∂y =∂N
∂x . 2. Let M= 2x+yand N=x+ 2y. Then,
calculate ∂M
∂y and ∂N
∂x . 3. We have ∂M
∂y = 1 and ∂N
∂x = 1. 4. Since ∂M
∂y =∂N
∂x , the differential
equation is exact. 5. To find the general solution, integrate Mdx with respect to xand Ndy
with respect to y. 6. Integrating Mdx = (2x+y)dx with respect to xgives us ∫(2x+y)dx =
x2+xy +C(y), where C(y)is the constant of integration. 7. Integrating Ndy = (x+2y)dy with
respect to ygives us ∫(x+ 2y)dy =xy +y2+C(x), where C(x)is the constant of integration.
8. Equate these two results to find the general solution: x2+xy +C(y) = xy +y2+C(x). 9.
Rearranging terms gives y2−x2+C(x)−C(y) = xy −xy. 10. Thus, the general solution is
y2−x2+C(x)−C(y) = 0, where C(x)and C(y)are arbitrary constants.
12. Question: Consider the differential equation (x2+ 2xy)dx + (y2−x2)dy = 0.
Determine if the given equation is exact. If it is exact, find the general solution. If it is not
exact, apply an integrating factor to make it exact and then find the general solution.
Ans. To determine if the given equation is exact, we check if ∂M
∂y =∂N
∂x . If not exact, we find
an integrating factor µ(x, y)such that µ(x, y)Mdx +µ(x, y)N dy = 0 is exact.
1. Calculate ∂M
∂y and ∂N
∂x .
∂M
∂y = 2xand ∂N
∂x =−2x
2. Since ∂M
∂y =∂N
∂x , the equation is not exact.
3. To find the integrating factor µ(x, y), we solve the differential equation
∂µM
∂y −∂µN
∂x
µ=
∂N
∂x −∂M
∂y
N.
2x−(−2x)
µ=−2x
y2−x2⇒4x
µ=−2x
y2−x2⇒µ(x, y) = (y2−x2)2
4. Multiply the given equation by the integrating factor µ(x, y).
(y2−x2)2(x2+ 2xy)dx + (y2−x2)2(y2−x2)dy = 0
5. Check if the new equation is exact by calculating ∂
∂y ((y2−x2)2(x2+ 2xy)) and ∂
∂x ((y2−x2)2(y2−x2)).
6. Integrate to find the general solution for the exact differential equation.
13. Let M= 2xy3−e3x+1
yand N=x2y+1
x. Determine whether the differential form
ω=Mdx +Ndy is exact. If ωis exact, find a potential function fsuch that df =ω.
Ans. To determine if ωis exact, we check if ∂M
∂y =∂N
∂x . If it is exact, we can find fby
integrating Mwith respect to xand Nwith respect to yand adding a constant of integration.
1. Calculate the partial derivatives:
∂M
∂y = 6xy2−1
y2and ∂N
∂x = 2xy +1
x2.
2. Check if ∂M
∂y =∂N
∂x :
6xy2−1
y2= 2xy +1
x2.
Since ωis not exact, there is no potential function fthat satisfies df =ωin this case.
14. Consider the following differential equation:
(2y2+x2)dx + 2xy dy = 0
Determine if the equation is exact, and if so, find the general solution.
Ans. To check if the equation is exact, we need to verify if ∂M
∂y =∂N
∂x , where the given equation
is in the form M dx +Ndy = 0.
1. Check for exactness:
Given equation: (2y2+x2)dx + 2xy dy = 0
Here, M= 2y2+x2and N= 2xy
Calculating the partial derivatives: ∂M
∂y = 4y∂N
∂x = 2y
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 µM dx +
µNdy = 0 is exact.
Integrating factor µ(x, y)is given by: µ(x, y) = e∫
∂N
∂x −∂M
∂y
Ndx
Calculating the integrating factor:
∂N
∂x −∂M
∂y
N=2y−4y
2xy =−1
x
So, µ(x, y) = e∫−1
xdx =e−ln |x|=1
|x|
3. Multiply the equation by the integrating factor:
Multiplying the given equation by the integrating factor: 2y2+x2
|x|dx +2xy
|x|dy = 0
4. Check for exactness after multiplying:
Now, we need to check if the equation is exact after multiplying by the integrating factor.
New M=2y2+x2
|x|and N=2xy
|x|
Calculating the partial derivatives: ∂M
∂y =4y
|x|
∂N
∂x =2y
|x|
Since ∂M
∂y =∂N
∂x , the equation is now exact.
5. Find the general solution:
Now, we can find the general solution by integrating Mdx +Ndy = 0:∫2y2+x2
|x|dx +
∫2xy
|x|dy =C
Solving these integrals will give the general solution to the differential equation.
15. Let M= (2xy2−6x)dx + (x2y+ 4y)dy be a differential 1-form. Determine if Mis exact,
and if so, find a potential function for M.
Ans. To determine if Mis exact, we need to check if the partial derivatives of the coefficients
with respect to yand xare equal. If Mis exact, we can find a potential function Φ(x, y)such
that dΦ = M. Let’s start by checking for exactness:
1. Checking for Exactness: The partial derivatives of the coefficients are: ∂
∂y (2xy2−6x) =
4xy ∂
∂x (x2y+ 4y) = ySince the partial derivatives are not equal (4xy =y), Mis not exact.
2. Integrating Factor Method: To find an integrating factor µ(x)that makes Mexact,
we solve the following equation: µ(x)∂
∂y (2xy2−6x)−µ(x)∂
∂x (x2y+ 4y) = ∂µ(x)
∂x = 0
Integrating with respect to x, we get: µ(x) = Cwhere Cis a constant. Since µ(x)is
independent of x, we can say that there is no integrating factor µ(x)that makes Mexact.
Since Mis not exact and there is no integrating factor, we conclude that there does not exist
a potential function Φsuch that dΦ = M.
16. Question: Determine whether the following first-order differential equation is exact, and if
so, find the general solution.
(3y2+ 2x)dx + (6xy)dy = 0
Ans. To determine whether the given first-order differential equation is exact, we check if
the partial derivatives with respect to yof the coefficient of dx and the partial derivatives with
respect to xof the coefficient of dy are equal. If they are equal, then the equation is exact and
we can find the general solution using the method of exact equations.
1. First, let’s compute the partial derivatives:
∂M
∂y = 6y
∂N
∂x = 6y
Since ∂M
∂y =∂N
∂x , the given equation is exact.
2. Next, we find a function F(x, y)such that
∂F
∂x = 3y2+ 2x
∂F
∂y = 6xy
Integrating the first equation with respect to xgives
F(x, y) = ∫(3y2+ 2x)dx = 3y2x+x2+h(y)
where h(y)is a function of y.
3. Now, differentiate F(x, y)with respect to yand compare to ∂F
∂y to find h(y):
∂F
∂y = 6xy +dh
dy
6xy = 6xy +dh
dy
This implies that dh
dy = 0, so h(y)is a constant. Let’s denote this constant as C.
4. Therefore, the general solution to the given differential equation is
3y2x+x2+C= 0
17. Consider the differential equation given by
(2x+ 3y)dx + (3x+ 4y)dy = 0
Determine if the differential 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 .
1. Calculate ∂M
∂y :
∂M
∂y = 3
2. Calculate ∂N
∂x :
∂N
∂x = 3
Since ∂M
∂y =∂N
∂x , the given differential equation is exact.
3. To find the general solution, we integrate Mdx with respect to xto find the potential
function U(x, y)and then take the partial derivative of U(x, y)with respect to yto find the
solution.
Integrating (2x+ 3y)dx with respect to x:
U(x, y) = ∫(2x+ 3y)dx =x2+ 3xy +f(y)
4. Calculating ∂U
∂y :
∂U
∂y = 3x+df
dy = 3x+ 3
Comparing ∂U
∂y to N:
3x+ 3 = 3x+ 4y⇐⇒ df
dy = 4y⇐⇒ f(y) = 2y2+C
Therefore, the general solution to the differential equation is x2+ 3xy + 2y2=Cwhere C
is a constant.
18. Question: Determine if the following differential equation is exact, and if it is, find its
general solution:
(2y−3x)dx + (x2+ 2y)dy = 0
Ans. Step-by-step solution: 1. To check if the given differential equation is exact, we calculate
the partial derivatives of the terms with respect to yand x:
∂
∂y (2y−3x) = 2 and ∂
∂x(x2+ 2y) = 2x
Since the mixed partial derivative ∂
∂y (2y−3x) = ∂
∂x (x2+ 2y), the given differential equation is
exact.
2. To find the general solution, we need to proceed to find the function Ψ(x, y)such that
∂Ψ
∂x = 2y−3xand ∂Ψ
∂y =x2+ 2y
3. Integrating the first equation with respect to x, we get
Ψ = 2xy −3
2x2+h(y)
where h(y)is an arbitrary function of y.
4. Now, we differentiate Ψwith respect to yand equate it to the second equation:
∂Ψ
∂y = 2x+h′(y) = x2+ 2y
5. Comparing the terms, we find that h′(y) = 2y, which implies that h(y) = y2+Cfor some
constant C.
6. Substituting back into the expression for Ψ, we have
Ψ(x, y) = 2xy −3
2x2+y2+C
7. Therefore, the general solution to the given exact differential equation is
2xy −3
2x2+y2+C= 0
where Cis the constant of integration.
19. Let Mand Nbe functions of two variables xand ydefined on a simply connected region
D. Consider the following differential equation:
M(x, y)dx +N(x, y)dy = 0
where Mand Nare given functions. Determine conditions for exactness of the equation and
find the general solution if it is exact.
Ans. To determine the conditions for exactness of the given differential equation M(x, y)dx +
N(x, y)dy = 0, we need to check if the following condition holds:
∂M
∂y =∂N
∂x
If the equation is exact, we can find the general solution by following these steps:
1. Check the condition for exactness:
∂M
∂y =∂N
∂x
2. If the equation is exact, find a function ψ(x, y)such that:
dψ =Mdx +Ndy
3. Integrate dψ with respect to xand then with respect to yto find ψ(x, y).
4. The general solution is given by:
ψ(x, y) = C
where Cis a constant.
20. Let M(x, y) = x2y−2xy2+sin(x)and N(x, y) = x2−3y2. Determine if the following
differential equation is exact, and if so, find its general solution:
M(x, y)dx +N(x, y)dy = 0
Ans. To determine if the given differential equation is exact, we first check if ∂M
∂y =∂N
∂x . If it is
exact, we can find the general solution by integrating Mwith respect to xand Nwith respect
to y.
1. Calculate partial derivatives: ∂M
∂y =x2−4xy
∂N
∂x = 2x
2. Check for exactness: Since ∂M
∂y =∂N
∂x , the differential equation is not exact.
3. Find integrating factor: Let the integrating factor be µ(x)such that
µ(x)∂M
∂y =µ(x)(x2−4xy)=µ(x)∂N
∂x = 2µ(x)x
For the integrating factor to exist, we must have
∂(2µx)
∂y −∂(µ(x)(x2−4xy))
∂x
µ(x)(x2−4xy)= 0
Solving this equation will give us µ(x).
4. Solve the differential equation with the integrating factor: After finding the inte-
grating factor, multiply both sides of the differential equation by µ(x)and check if it is exact.
5. Find the general solution: If the equation is exact, integrate Mµ(x)with respect to x
and Nµ(x)with respect to yto find the general solution.
21. Question 21: Consider the differential equation given by (2x2y2−ey)dx+(4x3y−2xey)dy =
0.
Determine whether the given differential equation is exact. If it is exact, find the general
solution. If not, find an integrating factor that makes it exact and then find the solution.
Ans. To determine whether the given differential equation is exact, we need to check if
∂M
∂y =∂N
∂x . Let M= 2x2y2−eyand N= 4x3y−2xey.
1. Check for exactness: Using the given differential equation, we find:
∂M
∂y = 4x2y−ey
∂N
∂x = 12x2y−2ey
Since ∂M
∂y =∂N
∂x , the given differential equation is not exact.
2. Find the integrating factor: To find an integrating factor, we need a function µ(x, y)such
that
µ(x, y)M dx +µ(x, y)Ndy = 0
is an exact differential. We can find µ(x, y)by integrating the following equation:
∂
∂x (µN )
µ−∂
∂y (µM) = 0
After simplifying the above equation and solving for µ, we find that µ=1
2xy is an integrating
factor.
3. Find the general solution: Multiplying the given differential equation by the integrating
factor 1
2xy , we get the exact differential equation:
(ydx + 2xdy)−ey·dx
2y= 0
Integrating the above equation gives the general solution:
y2+x2=Cey/2
where Cis the constant of integration.
22. Question: Determine whether the following first-order differential equation is exact, and if
so, find the general solution:
(2y+x2)dx + (2x+ex)dy = 0
Ans. Step-by-step solution:
1. To determine if the differential equation is exact, we need to check if the partial derivatives
satisfy the condition:
∂M
∂y =∂N
∂x
2. Let M= 2y+x2and N= 2x+ex. Then, we calculate the partial derivatives:
∂M
∂y = 2
∂N
∂x = 2 + ex
3. Since ∂M
∂y =∂N
∂x , the equation is not exact.
4. To find an integrating factor µ(x, y)such that µMdx +µNdy = 0 is exact, we solve the
following equation:
∂(µM)
∂y −∂(µN )
∂x
µ= 0
5. First, calculate ∂(µM)
∂y and ∂(µN )
∂x :
∂(µM)
∂y =µ∂M
∂y +M∂µ
∂y
∂(µN)
∂x =µ∂N
∂x +N∂µ
∂x
6. Substituting the expressions above into the equation from step 4, we get:
µ∂M
∂y +M∂µ
∂y −µ∂N
∂x −N∂µ
∂x
µ= 0
7. Simplify the equation to solve for µ:
M∂µ
∂y −N∂µ
∂x = 0
8. Substitute M,N, and their partial derivatives into the equation above, and solve for µ.
9. Once you have found µ, multiply the original differential equation by µto make it exact:
µ((2y+x2)dx + (2x+ex)dy) = 0
10. After multiplying through, the equation becomes exact. Finally, solve the exact equation
to find the general solution.
23. Let M=eycos(x)−yand N=eysin(x)−x. Determine whether the differential equation
Mdx +N dy = 0 is exact, and if so, find the general solution.
Ans. To determine if the given equation is exact, we check if ∂M
∂y =∂N
∂x . 1. Calculate the
partial derivatives:
∂M
∂y =eycos(x)−1
∂N
∂x =eycos(x)
2. Since ∂M
∂y =∂N
∂x , the equation is not exact. 3. To find an integrating factor, we compute:
∂M
∂y −∂N
∂x
N=−1
x
4. An integrating factor is µ(x) = e−ln |x|=1
|x|. 5. Multiply both Mand Nby µ(x) = 1
|x|:
eycos(x)
|x|−y
|x|dx +eysin(x)
|x|dy −x
|x|dy = 0
6. Rearranging, we get:
eycos(x)
|x|−x
|x|dy =y
|x|−eysin(x)
|x|dx
7. Integrating both sides:
∫(eycos(x)
|x|−x
|x|)dy =∫(y
|x|−eysin(x)
|x|)dx
8. Simplify and solve the integrals to find the general solution.
24. Let Mdx +N dy = 0 be a differential equation, where Mand Nare functions of xand y.
Question: Determine the conditions for exactness of the given equation and find its solution
if it is exact.
Ans. To check for exactness, we need to verify if the partial derivatives satisfy the condition
∂M
∂y =∂N
∂x .
1. Calculate the partial derivatives of Mand N:
∂M
∂y and ∂N
∂x
2. Check if the equation is exact by determining if ∂M
∂y =∂N
∂x .
3. If the equation is exact, integrate Mwith respect to xto find a potential function f(x, y):
f(x, y) = ∫M dx +h(y)
4. Differentiate f(x, y)with respect to yand set it equal to Nto find the integrating factor:
∂f
∂y =N
5. Solve the differential equation by finding the partial derivative of the integrating factor
with respect to yand integrating it with respect to y:
eg(y)=∫∂N
∂x −∂M
∂y
Ndy
6. Once the integrating factor is found, the solution to the exact differential equation is given
by
u(x, y) = eg(y)·f(x, y) + c
where cis the constant of integration.
25. Question 25: Consider the differential equation
(2x−3y) + (3y−4x)dx + (4x−2y)dy = 0
Determine if the equation is exact. If it is exact, find the solution.
Ans. To determine if the given equation is exact, we check the mixed partial derivatives of the
coefficients with respect to xand y. If the mixed partial derivatives are equal, then the equation
is exact.
1. Check for exactness: Calculate the partial derivatives of the coefficients:
∂M
∂y =−3,∂N
∂x = 4
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Make the equation exact: In order 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, y) [−3−4] = −7µ(x, y)
is a function of xonly.
For an integrating factor of the form µ(x), we have dµ(x)
dx =−−7µ(x,y)
N. Integrating, we get
µ(x) = e∫−(−7)
4x−2ydx
Solving the integral, we find µ(x) = (4x−2y)−7.
3. Find the solution: Multiply the given equation by the integrating factor to make it exact:
(2x−3y)(4x−2y)−7+ (3y−4x)dx + (4x−2y)(4x−2y)−7dy = 0
Now, check if the equation is exact:
∂
∂y [(2x−3y)(4x−2y)−7]=∂
∂x [(4x−2y)(4x−2y)−7]
d
dy [2x−3y
(4x−2y)7]=d
dx [4x−2y
(4x−2y)7]
This equation is exact, hence we can now solve for the general solution.
26. 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 if the equation is exact.
Ans. To determine if the first-order ODE is exact, we need to check if ∂M
∂y =∂N
∂x . If exact,
we then find a function ψ(x, y)such that ∂ψ
∂x =Mand ∂ψ
∂y =N. Finally, the general solution is
given by ψ(x, y) = C, where Cis an arbitrary constant.
Conditions for Exactness: 1. Check if ∂M
∂y =∂N
∂x .
Solution: 1. Compute ∂M
∂y and ∂N
∂x :
∂M
∂y =∂
∂y (M)and ∂N
∂x =∂
∂x(N)
2. If ∂M
∂y =∂N
∂x , the equation is exact. Otherwise, it is not.
Finding the Solution: 1. If the equation is exact, find a function ψ(x, y)such that ∂ψ
∂x =M
and ∂ψ
∂y =N.
2. Integrate Mwith respect to xand Nwith respect to yto find ψ(x, y).
3. The general solution is then given by ψ(x, y) = C, where Cis an arbitrary constant.
27. Let M(x, y)dx +N(x, y)dy = 0 be a first-order differential equation. Determine conditions
on the functions Mand Nfor exactness and provide a method for finding a solution when these
conditions are met.
Ans. To check for exactness and find solutions, follow these steps: 1. Check for exactness:
Calculate the partial derivatives ∂M
∂y and ∂N
∂x . If these are equal, then the equation is exact.
2. If the equation is exact: Find a function F(x, y)such that dF =M dx +Ndy. Then the
solution to the differential equation is given by F(x, y) = C, where Cis a constant.
3. Find the function F: Integrate Mwith respect to xand Nwith respect to y. Set these
two integrals equal to each other and equate the result to F(x, y).
4. General solution: Solve F(x, y) = Cfor Cto obtain the general solution to the differential
equation.
By following these steps, one can determine if a given first-order differential equation is exact
and find its general solution when the conditions for exactness are met.
28. Question 28: Consider the differential equation (2x+ 3y)dx + (x−y)dy = 0.
Determine whether the given first-order ordinary differential equation is exact or not. 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 general solution.
Ans. To determine whether the given first-order ordinary differential equation is exact, we check
if ∂M
∂y =∂N
∂x .
1. Let M= 2x+ 3yand N=x−y. Calculating the partial derivatives gives: ∂M
∂y = 3 and
∂N
∂x = 1. Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. To find an integrating factor µ(x), we use the formula:
∂M
∂y −∂N
∂x
N=3−1
x−y=µ′(x). Simplifying
this expression, we get: 2
x−y=µ′(x). Thus, the integrating factor is µ(x) = e2ln |x−y|=
|x−y|2.
3. Multiplying the given equation by the integrating factor |x−y|2gives:
2x|x−y|2dx + 3y|x−y|2dx + (x−y)|x−y|2dy = 0.
Simplifying the equation yields:
2x(x−y)2dx + 3y(x−y)2dx + (x−y)3dy = 0.
4. We can now check if the equation is exact: Let M= 2x(x−y)2and N= (x−y)3.
Calculating the partial derivatives gives: ∂M
∂y =−4x(x−y)and ∂N
∂x =−3(x−y)2. Since
∂M
∂y =∂N
∂x , the equation is exact.
5. To find the general solution, we need to solve the partial differential equation fx=Mand
fy=N. Integrating Mwith respect to xgives:
f(x, y) = ∫2x(x−y)2dx =2
5x5−4
3xy(x−y)2+h(y),
where h(y)is an arbitrary function of y.
6. Now we differentiate f(x, y)with respect to yand equate it to N:
∂f
∂y =∂N
∂y ⇒d
dy (−4
3xy(x−y)2+h(y))= (x−y)3.
Solving this differential equation will yield the general solution of the given differential
equation.
29. Question 29: Consider the differential equation given by (y−2x2)dx + (2y−x)dy = 0.
Determine whether the given differential equation is exact. If it is exact, find a potential
function F(x, y)such that ∂F
∂x =y−2x2and ∂F
∂y = 2y−x.
Ans. To determine whether the given differential equation is exact, we need to check if
∂M
∂y =∂N
∂x , where the differential equation is in the form M(x, y)dx +N(x, y)dy = 0.
1. Check for exactness: The given differential equation is (y−2x2)dx + (2y−x)dy = 0.
Let M(x, y) = y−2x2and N(x, y) = 2y−x. Calculate the partial derivatives: ∂M
∂y = 1 and
∂N
∂x =−1. Since ∂M
∂y =∂N
∂x , the given differential 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 µM dx +µNdy = 0 is exact. The integrating factor is given by µ=
e∫
∂N
∂x −∂M
∂y
Ndx.
Calculate
∂N
∂x −∂M
∂y
N:
−1−1
2y−x=−2
2y−x=−2
2y−x
So, the integrating factor is µ=e∫−2
2y−xdx.
3. Verify the exactness with the integrating factor: Multiply both sides of the differential
equation by the integrating factor µ:
e∫−2
2y−xdx(y−2x2)dx +e∫−2
2y−xdx(2y−x)dy = 0
Simplify the equation and check if it is exact. If it is, find the potential function.
Note: The solution process will continue in the next steps once the integrating factor is
determined.
30. Let M(x, y)dx +N(x, y)dy = 0 be a differential equation. Determine conditions for
exactness and provide a method for finding solutions.
Ans. To determine conditions for exactness and find solutions to a given differential equation
of the form M(x, y)dx +N(x, y)dy = 0, we check if the following condition is satisfied:
∂M
∂y =∂N
∂x .
If this condition is fulfilled, then the equation is exact. In this case, we can find the solution
by finding a function f(x, y)such that ∂f
∂x =Mand ∂f
∂y =N. The solution is then given by
f(x, y) = c, where cis an arbitrary constant.
31. Question 31: Consider the differential equation given by
(cos y+xsin y)dx + (sin x+ycos x)dy = 0
Is the equation exact? If so, find a solution. If not, determine an integrating factor and then
find the general solution.
Ans. To check for exactness, we compute the partial derivatives of the coefficients with respect
to yand x:
1. Compute ∂
∂y (cos y+xsin y) = −sin y+xcos y2. Compute ∂
∂x (sin x+ycos x) =
cos x−ysin x
Since the mixed partial derivatives are not equal, the equation is not exact. To find an
integrating factor, we can use the formula µ(x, y) = e∫(My−Nx
N)dx.
3. Compute My−Nx
N=(−sin y+xcos y)−(cos x−ysin x)
sin x+ycos x=−sin y−cos x+xcos y+ysin x
sin x+ycos x
Therefore, the integrating factor is µ(x, y) = e∫−sin y−cos x+xcos y+ysin x
sin x+ycos xdx.
After finding µ(x, y), we can multiply through the differential equation by µ(x, y)in order to
obtain an exact differential equation. Finally, we can solve the exact differential equation to find
the general solution.
32. Consider the following differential equation:
(3x2−2x)dx + (y−x2)dy = 0
Determine if the given differential equation is exact. If it is exact, find its solution. If it is not
exact, find an integrating factor that makes it exact.
Ans. To determine if the given differential equation is exact, we check if the partial derivatives
of the terms with respect to yare equal. If they are not equal, we find the integrating factor that
makes the equation exact.
1. Check for exactness: Let M= 3x2−2xand N=y−x2. Calculate the partial
derivatives: ∂M
∂y = 0
∂N
∂x =−2x
Since ∂M
∂y =∂N
∂x , the differential equation is not exact.
2. Find the integrating factor: The integrating factor µ(x)can be found from the relation:
1
µ(x)(∂
∂x(µ(x)M)−∂
∂y (µ(x)N))= 0
Substitute M,N, and their partial derivatives into the above equation and solve for µ(x).
We have:
µ(x)∂M
∂x −µ(x)∂N
∂y −dµ(x)
dx N= 0
µ(x)(6x−2) −dµ(x)
dx (y−x2) = 0
Solving for µ(x):
6xµ(x)−2µ(x)−ydµ(x)
dx +x2dµ(x)
dx = 0
(6x−2)µ(x) + x2dµ(x)
dx −ydµ(x)
dx = 0
Separate variables and integrate to find µ(x).
3. Find the solution: Once we have the integrating factor µ(x), multiply the given differ-
ential equation by µ(x)and proceed to solve the equation as an exact differential equation.
Thus, for question number 32, we would proceed to find the integrating factor and solve the
differential equation using the above steps.
33. What are the conditions for a differential equation to be exact, and what are the methods
for finding solutions to exact differential equations?
Ans. To determine if a first-order differential equation of the form M(x, y)dx +N(x, y)dy = 0
is exact, we need to check if ∂M
∂y =∂N
∂x . If the equation is exact, then we can find a solution by
integrating M(x, y)dx with respect to xand N(x, y)dy with respect to y, and setting the sum
equal to a constant.
1. Check for exactness by verifying if ∂M
∂y =∂N
∂x .2. If the equation is exact, then find a
function ψ(x, y)such that ψx=Mand ψy=N.3. Integrate M(x, y)dx with respect to xto
obtain ψ(x, y) = ∫M(x, y)dx+h(y), where h(y)is a function of y.4. Take the partial derivative
of ψ(x, y)with respect to yand set it equal to N(x, y)to solve for h′(y).5. Substitute the
value of h(y)back into ψ(x, y)to get the general solution ψ(x, y) = C, where Cis a constant.
34. Let ω=ey−xcos y
x2+ydx +xsin y+ey
x2+ydy be a differential 1-form. Determine whether ωis exact,
and if it is, find a function f(x, y)such that df =ω.
Ans. To determine whether the differential form ωis exact, we calculate its partial derivatives
and check if they satisfy the condition for exactness. If so, we integrate these partial derivatives
to find the function f(x, y).
1. Calculate the partial derivatives of ω:We have ∂
∂y (ey−xcos y
x2+y)=ey+xsin y
x2+y−2xy(ey−xcos y)
(x2+y)2
and ∂
∂x (xsin y+ey
x2+y)=sin y(x2+y)−(xsin y+ey)2x
(x2+y)2.
2. Check if the partial derivatives are equal: We need to verify if ∂
∂y (ey−xcos y
x2+y)=
∂
∂x (xsin y+ey
x2+y)holds.
3. If the partial derivatives are equal, the form is exact: If ωis exact, we proceed to
find f(x, y)such that ∂f
∂x =ey−xcos y
x2+yand ∂f
∂y =xsin y+ey
x2+y.
4. Integrate the partial derivatives to find f(x, y):By integrating the expressions from
step 1 with respect to xand yseparately, we can find the function f(x, y).
Therefore, the challenging task here is to calculate the partial derivatives of the given differ-
ential form and verify if it is exact, then find the potential function f(x, y)if it is exact.
35. Consider the following first-order ordinary differential equation (ODE):
(2xy2+ex)dx + (x2y+ 2ex)dy = 0
Question 35: Determine whether the ODE above is exact. If it is exact, find a function
f(x, y)such that ∂f
∂x = 2xy2+exand ∂f
∂y =x2y+ 2ex.
Ans. To determine whether the ODE is exact, we need to check if ∂M
∂y =∂N
∂x , where the ODE
is of the form M dx +Ndy = 0.
1. Check for exactness: Let M= 2xy2+exand N=x2y+ 2ex. Calculate the partial
derivatives: ∂M
∂y = 4xy and ∂N
∂x = 2xy
Since ∂M
∂y =∂N
∂x , the ODE is not exact.
2. Find integrating factor: To find an integrating factor, we can use the formula µ=
e∫Nx−My
Mdx.
Calculate Nx−My:
Nx−My= (2xy)−(2xy) = 0
Thus, the integrating factor µ=e∫0dx =e0= 1.
3. Multiply the ODE by the integrating factor: Multiply both sides of the ODE by
µ= 1:
(2xy2+ex)dx + (x2y+ 2ex)dy = 0
4. Check for exactness with the integrating factor: The ODE is exact if ∂(µM)
∂y =∂(µN )
∂x .
Calculate: ∂
∂y (M) = 2x(2y) = 4xy
∂
∂x(N) = y(2) = 2y
Since ∂(µM)
∂y =∂(µN )
∂x , the ODE is exact with the integrating factor.
5. Find the function f(x, y):Since the ODE is exact with the integrating factor, the
function f(x, y)exists. We integrate the terms Mand Nwith respect to xand yrespectively
to find f(x, y).
f(x, y) = ∫(2xy2+ex)dx +g(y) = x2y2+ex+g(y)
f(x, y) = ∫(x2y+ 2ex)dy +h(x) = x2y2+ 2ex+h(x)
Equating the two expressions for f(x, y), we conclude that g(y) = 2ex+Cand h(x) = ex+C
where Cis an arbitrary constant.
Therefore, the function f(x, y) = x2y2+ex+ 2ex+C=x2y2+ 3ex+Csatisfies the
conditions, making it a solution to the exact ODE.
7. Equating the two expressions, we find:
ex(y−x2+ 2) + g(y) = ex(2y−2x) + h(x)
8. To find g(y)and h(x), we compare coefficients and equate terms. We can then write the
general solution as ex(y−x2+ 2) = Cwhere Cis the constant of integration.
2. Question: Consider the differential equation
(2y2−x2)dx + (4xy)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 check the partial derivatives
of the coefficients with respect to yand x. 1. Compute the partial derivative of (2y2−x2)with
respect to y:
∂
∂y (2y2−x2) = 4y
2. Compute the partial derivative of (4xy)with respect to x:
∂
∂x(4xy) = 4y
Since the partial derivatives are equal, the given differential equation is exact.
To find the general solution, we proceed by finding the integrating factor µ(x). 3. Write the
differential equation in the form M(x, y)dx +N(x, y)dy = 0:
(2y2−x2)dx + (4xy)dy = 0
4. Compute ∂M
∂y and ∂N
∂x :
∂
∂y (2y2−x2) = 4yand ∂
∂x(4xy) = 4y
5. Find the integrating factor:
µ(x) = e∫(∂N
∂x −∂M
∂y
N)dx
=e∫0
4xy dx =e0= 1
6. Multiply the differential equation by the integrating factor:
(2y2−x2)dx + (4xy)dy = 0 =⇒dx(2y2−x2) + dy(4xy) = 0
7. Integrate the equation with respect to x:
∫(2y2−x2)dx +φ(y) = C
2xy2−x3
3+φ(y) = C
8. Differentiate the above equation with respect to yand compare with the given equation to
find φ(y):
∂
∂y (2xy2−x3
3) = 4xy =⇒4xy +φ′(y) = 4xy =⇒φ′(y) = 0
9. Integrate φ′(y) = 0 to find φ(y):
φ(y) = C1
10. Substitute φ(y) = C1back into the integrated equation:
2xy2−x3
3+C1=C
Therefore, the general solution is 2xy2−x3
3+C1=C.
3. Consider the following differential equation:
(2x3y−y3)dx + (3x2y2−x)dy = 0
Determine if the given equation is exact. If it is exact, find the general solution. If it is not
exact, find an integrating factor to make it exact and then find the general solution.
Ans. To determine if the given differential equation is exact, we first calculate the partial
derivatives of Mand N, where Mand Nare the coefficients of dx and dy, respectively.
1. Calculate ∂M /∂y and ∂N/∂x:
∂M
∂y = 2x3−3y2and ∂N
∂x = 6xy2−1
Since ∂M
∂y =∂N
∂x , the given equation is not exact. Now, we need to find an integrating factor
µ(x, y), such that µ(x, y)M dx +µ(x, y)Ndy = 0 is exact.
2. Find the integrating factor µ(x, y):
µ(x, y) = e∫(∂N
∂x −∂M
∂y
M)dx
µ(x, y) = e∫((6xy2−1)−(2x3−3y2)
2x3y−y3)dx
µ(x, y) = e∫(6xy2−1−2x3+3y2
2x3y−y3)dx
µ(x, y) = e∫3y2−2x3+6xy2−1
2x3y−y3dx
µ(x, y) = e∫3y2+6xy2−2x3−1
2x3y−y3dx
After solving the integral, we find the integrating factor µ(x, y).
3. Once we have found the integrating factor µ(x, y), we multiply the given differential
equation by it:
µ(x, y)[(2x3y−y3)dx + (3x2y2−x)dy] = 0
4. The new equation is exact. Find the general solution using the method for solving exact
equations.
5. Write the equation in the form F(x, y) = cand solve for cusing the initial condition, if
provided.
4. Consider the differential equation:
(2y+ 3x2)dx + (2x−2y)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 check if ∂M
∂y =∂N
∂x .
1. Calculate the partial derivatives of M= 2y+ 3x2and N= 2x−2ywith respect to yand
x:∂M
∂y = 2,∂N
∂x = 2
2. Since ∂M
∂y =∂N
∂x , the differential equation is exact.
3. To find the general solution, we need to find a function f(x, y)such that ∂f
∂x =Mand
∂f
∂y =N.
4. Integrating Mwith respect to x, we get:
f(x, y) = ∫(2y+ 3x2)dx = 2xy +x3+h(y)
where h(y)is a constant of integration that may depend on y.
5. Now, we differentiate f(x, y)with respect to y, set it equal to N, and solve for h(y):
∂f
∂y = 2x+h′(y) = 2x−2y=⇒h′(y) = −2y
6. Integrating h′(y)with respect to y, we get:
h(y) = −y2+C
where Cis an arbitrary constant.
7. Thus, the general solution to the given differential equation is:
2xy +x3−y2+C= 0
5. Consider the following differential equation:
(2x+y2)dx + (2y+ 2xy)dy = 0
Determine whether the given equation is exact or not. If it is exact, find the solution using
the exactness condition.
Ans. To determine if the given differential equation is exact, we check if ∂M
∂y =∂N
∂x . After
confirming exactness, we find the solution by integrating Mwith respect to xand then solving
for the constant of integration.
1. Exactness Condition: Let M= 2x+y2and N= 2y+ 2xy. We calculate the partial
derivatives: ∂M
∂y = 2y
∂N
∂x = 2y+ 2y= 4y
Since ∂M
∂y =∂N
∂x , the given differential equation is exact.
2. Finding the Solution: Since the equation is exact, there exists a function ψ(x, y)such
that dψ = (2x+y2)dx + (2y+ 2xy)dy.
Integrating M= 2x+y2with respect to xgives us:
ψ=∫(2x+y2)dx =x2+xy2+g(y)
Now, we differentiate the expression with respect to yand equate it to Nto find g(y):
∂ψ
∂y = 2xy +g′(y) = 2y+ 2xy
g′(y) = 2y
g(y) = y2+C
Therefore, the solution to the given differential equation is:
ψ=x2+xy2+y2+C
6. Let M(x, y)dx +N(x, y)dy be a differential equation, where Mand Nare functions of two
variables xand y. Determine whether the following differential equations are exact, and if so,
find the general solution.
1. (ey+xy)dx + (x2−2y)dy = 0
Ans. To determine if the given differential equation is exact, we check if ∂M
∂y =∂N
∂x .
1. Calculate the partial derivatives: ∂M
∂y =ey+xand ∂N
∂x = 2x
Since ∂M
∂y =∂N
∂x , the differential equation is not exact.
2. To find an integrating factor µ(x, y), calculate:
∂M
∂y −∂N
∂x
N=(ey+x)−2x
x2−2y=ey−x
x2−2y
µ(x, y) = e∫ey−x
x2−2ydy
3. Find µ(x, y)by solving the integral: µ(x, y) = e∫ey−x
x2−2ydy =e∫ey
x2−2ydy−∫x
x2−2ydy
4. Once you have found µ(x, y), multiply the entire differential equation by µ(x, y)and
check if it is exact. If it is exact, proceed to find the general solution. If not, consider alternative
methods such as integrating factors depending on the specific form of the differential equation.
7. Consider the following differential equation:
(2xy3+ex)dx + (3x2y2+ey)dy = 0
Let M= 2xy3+exand N= 3x2y2+ey.
Is the given differential equation exact? If not, find an integrating factor µ(x, y)to make it
exact and solve the differential equation.
Ans. To determine if the given differential equation is exact, we check if ∂M
∂y =∂N
∂x .
1. Calculate the partial derivatives:
∂M
∂y = 6xy2
∂N
∂x = 6xy2+ex
2. Since ∂M
∂y =∂N
∂x , the given differential equation is not exact.
3. To find an integrating factor µ(x, y), we set up the equation 1
µ(∂M
∂y −∂N
∂x ) = ∂
∂x
µ
∂y .
4. Substitute the given Mand Ninto the equation above and solve for µ:
1
µ(6xy2−(6xy2+ex)) = −∂
∂x(3x2y2+ey)
−ex
µ=−3x2y2−ey
µ=exeye3x2y2
5. Using the integrating factor µ=exeye3x2y2, multiply both sides of the differential equation
by µ.
(2xy3exeye3x2y2+ex+2y+3x2y2)dx + (3x2y2exeye3x2y2+ex+y+3x2y2)dy = 0
6. Now the differential equation is exact. We can find the solution by integrating the terms
with respect to xand yseparately.
7. Integrating the term with respect to x:
∫2xy3exeye3x2y2+ex+2y+3x2y2dx =ex+2y+3x2y2+C(y)
8. Integrating the term with respect to y:
∫3x2y2exeye3x2y2+ex+y+3x2y2dy =ex+y+3x2y2+C(x)
9. Therefore, the general solution to the differential equation is given by ex+2y+3x2y2+C(y) =
ex+y+3x2y2+C(x).
8. Let M(x, y)dx +N(x, y)dy = 0 be a first-order ordinary differential equation. Determine
conditions for exactness and explain methods for finding solutions.
Ans. To determine the conditions for exactness of the given ODE and explain methods for
finding solutions, follow these steps:
1. Check for Exactness: A first-order ODE M(x, y)dx +N(x, y)dy = 0 is exact if and only
if ∂M
∂y =∂N
∂x .
2. Determine Integrating Factor: If the ODE is not exact, find an integrating factor µ(x)
such that µ(x)M(x, y)dx +µ(x)N(x, y)dy = 0 is exact. The integrating factor is given by
µ(x) = e∫P(x)dx, where P(x)is a function of xonly that can be determined by the condition
∂
∂y (µM )
µ∂x =∂
∂x (µN )
µ∂y .
3. Solve the Exact ODE: After obtaining the exact ODE, solve it by integrating dF = 0,
where F(x, y) = Cis the general solution, and Cis the constant of integration.
4. Boundary Conditions: If initial conditions are given, use them to find a particular solution
by substituting them into the general solution.
5. Check for Correctness: Finally, check the correctness of the solution by substituting it back
into the original ODE to ensure that it satisfies the equation.
By following these steps, you can determine the conditions for exactness of a given ODE and
find its solution.
9. Suppose we have the following differential equation:
(2x2y+y2+exy)dx + (x2+ 2xy +exy)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 that makes it exact and find the general solution.
Ans. To determine if the given equation is exact, we check if ∂N
∂x =∂M
∂y , where the given
equation is of the form M dx +Ndy = 0.
1. Check for exactness: Let M= 2x2y+y2+exyand N=x2+ 2xy +exy. Calculating the
partial derivatives:
∂M
∂y = 2x2+ 2y+ex
∂N
∂x = 4xy +ex
The equation is not exact because ∂N
∂x =∂M
∂y .
2. Find the integrating factor: To find the integrating factor µ(x), we set up the following
equation: µ(x)(∂M
∂y −∂N
∂x ) = µ′(x)
µ(x)= 0. Therefore, µ(x) = e∫4xy+ex−2x2−2y
x2+2xy+exydx.
Calculating the integral in the exponent:
e∫4xy+ex−2x2−2y
x2+2xy+exydx =e∫2(x+y)(2x−y)+ex
(x+y)2dx =e2ln |x+y|+∫ex
x+ydx
= (x+y)2e∫ex
x+ydx
3. Find the solution with the integrating factor: Applying the integrating factor to the given
equation:
(2x2y+y2+exy)dx + (x2+ 2xy +exy)dy = 0
⇒(x2+ 2xy +exy)(x+y)2dx + (x2+ 2xy +exy)(x+y)2dy = 0
This equation is exact. Integrating both sides, we find the general solution.
The process of finding the integrating factor and solving the resulting exact differential equa-
tion can be quite complex but important in the study of differential equations.
10. Find the particular solution y(x)of the following differential equation:
(2x+y)dx + (x+ 2y)dy = 0
Ans. To find the particular solution y(x)of the given differential equation, we need to check
for exactness and then use appropriate methods to solve for y(x).
1. Check for exactness: We check the exactness of the given equation by computing the
partial derivatives of (2x+y)dx + (x+ 2y)dy with respect to yand x, respectively.
∂M
∂y = 1 =∂N
∂x = 1
Since the equation is not exact, we will find an integrating factor to make it exact.
2. Find the integrating factor: Let (x)be the integrating factor such that when multiplied
to the given equation, it becomes exact. We have the following condition for exactness:
∂(µM)
∂y =∂(µN)
∂x
Substitute M= (2x+y)and N= (x+ 2y)into the above equation and solve for (x).
3. Solve the integrating factor equation: From the condition above, we get:
∂(µ(2x+y))
∂y =∂(µ(x+ 2y))
∂x
Solve this equation to obtain (x).
4. Rewrite the equation with the integrating factor: Once we have found (x), multiply
it to the given differential equation to make it exact. The new equation will look like:
µ(2x+y)dx +µ(x+ 2y)dy = 0
5. Solve for y(x):Now that the equation is exact, we can find a function F(x, y)such that
∂F
∂x =µ(2x+y)and ∂F
∂y =µ(x+ 2y)
Solve these equations simultaneously to find F(x, y).
6. Obtain the solution: The solution of the differential equation will be given by F(x, y) =
C, where Cis a constant. Finally, solve for y(x)from the equation F(x, y) = Cto get the
particular solution.
11. Question: Determine whether the following first-order differential equation is exact, and if
it is, find the general solution:
(2x+y)dx + (x+ 2y)dy = 0
Ans. Step-by-step solution:
1. Check for exactness by verifying if ∂M
∂y =∂N
∂x . 2. Let M= 2x+yand N=x+ 2y. Then,
calculate ∂M
∂y and ∂N
∂x . 3. We have ∂M
∂y = 1 and ∂N
∂x = 1. 4. Since ∂M
∂y =∂N
∂x , the differential
equation is exact. 5. To find the general solution, integrate Mdx with respect to xand Ndy
with respect to y. 6. Integrating Mdx = (2x+y)dx with respect to xgives us ∫(2x+y)dx =
x2+xy +C(y), where C(y)is the constant of integration. 7. Integrating Ndy = (x+2y)dy with
respect to ygives us ∫(x+ 2y)dy =xy +y2+C(x), where C(x)is the constant of integration.
8. Equate these two results to find the general solution: x2+xy +C(y) = xy +y2+C(x). 9.
Rearranging terms gives y2−x2+C(x)−C(y) = xy −xy. 10. Thus, the general solution is
y2−x2+C(x)−C(y) = 0, where C(x)and C(y)are arbitrary constants.
12. Question: Consider the differential equation (x2+ 2xy)dx + (y2−x2)dy = 0.
Determine if the given equation is exact. If it is exact, find the general solution. If it is not
exact, apply an integrating factor to make it exact and then find the general solution.
Ans. To determine if the given equation is exact, we check if ∂M
∂y =∂N
∂x . If not exact, we find
an integrating factor µ(x, y)such that µ(x, y)Mdx +µ(x, y)N dy = 0 is exact.
1. Calculate ∂M
∂y and ∂N
∂x .
∂M
∂y = 2xand ∂N
∂x =−2x
2. Since ∂M
∂y =∂N
∂x , the equation is not exact.
3. To find the integrating factor µ(x, y), we solve the differential equation
∂µM
∂y −∂µN
∂x
µ=
∂N
∂x −∂M
∂y
N.
2x−(−2x)
µ=−2x
y2−x2⇒4x
µ=−2x
y2−x2⇒µ(x, y) = (y2−x2)2
4. Multiply the given equation by the integrating factor µ(x, y).
(y2−x2)2(x2+ 2xy)dx + (y2−x2)2(y2−x2)dy = 0
5. Check if the new equation is exact by calculating ∂
∂y ((y2−x2)2(x2+ 2xy)) and ∂
∂x ((y2−x2)2(y2−x2)).
6. Integrate to find the general solution for the exact differential equation.
13. Let M= 2xy3−e3x+1
yand N=x2y+1
x. Determine whether the differential form
ω=Mdx +N dy is exact. If ωis exact, find a potential function fsuch that df =ω.
Ans. To determine if ωis exact, we check if ∂M
∂y =∂N
∂x . If it is exact, we can find fby
integrating Mwith respect to xand Nwith respect to yand adding a constant of integration.
1. Calculate the partial derivatives:
∂M
∂y = 6xy2−1
y2and ∂N
∂x = 2xy +1
x2.
2. Check if ∂M
∂y =∂N
∂x :
6xy2−1
y2= 2xy +1
x2.
Since ωis not exact, there is no potential function fthat satisfies df =ωin this case.
14. Consider the following differential equation:
(2y2+x2)dx + 2xy dy = 0
Determine if the equation is exact, and if so, find the general solution.
Ans. To check if the equation is exact, we need to verify if ∂M
∂y =∂N
∂x , where the given equation
is in the form M dx +Ndy = 0.
1. Check for exactness:
Given equation: (2y2+x2)dx + 2xy dy = 0
Here, M= 2y2+x2and N= 2xy
Calculating the partial derivatives: ∂M
∂y = 4y∂N
∂x = 2y
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 µM dx +
µNdy = 0 is exact.
Integrating factor µ(x, y)is given by: µ(x, y) = e∫
∂N
∂x −∂M
∂y
Ndx
Calculating the integrating factor:
∂N
∂x −∂M
∂y
N=2y−4y
2xy =−1
x
So, µ(x, y) = e∫−1
xdx =e−ln |x|=1
|x|
3. Multiply the equation by the integrating factor:
Multiplying the given equation by the integrating factor: 2y2+x2
|x|dx +2xy
|x|dy = 0
4. Check for exactness after multiplying:
Now, we need to check if the equation is exact after multiplying by the integrating factor.
New M=2y2+x2
|x|and N=2xy
|x|
Calculating the partial derivatives: ∂M
∂y =4y
|x|
∂N
∂x =2y
|x|
Since ∂M
∂y =∂N
∂x , the equation is now exact.
5. Find the general solution:
Now, we can find the general solution by integrating Mdx +N dy = 0:∫2y2+x2
|x|dx +
∫2xy
|x|dy =C
Solving these integrals will give the general solution to the differential equation.
15. Let M= (2xy2−6x)dx + (x2y+ 4y)dy be a differential 1-form. Determine if Mis exact,
and if so, find a potential function for M.
Ans. To determine if Mis exact, we need to check if the partial derivatives of the coefficients
with respect to yand xare equal. If Mis exact, we can find a potential function Φ(x, y)such
that dΦ = M. Let’s start by checking for exactness:
1. Checking for Exactness: The partial derivatives of the coefficients are: ∂
∂y (2xy2−6x) =
4xy ∂
∂x (x2y+ 4y) = ySince the partial derivatives are not equal (4xy =y), Mis not exact.
2. Integrating Factor Method: To find an integrating factor µ(x)that makes Mexact,
we solve the following equation: µ(x)∂
∂y (2xy2−6x)−µ(x)∂
∂x (x2y+ 4y) = ∂µ(x)
∂x = 0
Integrating with respect to x, we get: µ(x) = Cwhere Cis a constant. Since µ(x)is
independent of x, we can say that there is no integrating factor µ(x)that makes Mexact.
Since Mis not exact and there is no integrating factor, we conclude that there does not exist
a potential function Φsuch that dΦ = M.
16. Question: Determine whether the following first-order differential equation is exact, and if
so, find the general solution.
(3y2+ 2x)dx + (6xy)dy = 0
Ans. To determine whether the given first-order differential equation is exact, we check if
the partial derivatives with respect to yof the coefficient of dx and the partial derivatives with
respect to xof the coefficient of dy are equal. If they are equal, then the equation is exact and
we can find the general solution using the method of exact equations.
1. First, let’s compute the partial derivatives:
∂M
∂y = 6y
∂N
∂x = 6y
Since ∂M
∂y =∂N
∂x , the given equation is exact.
2. Next, we find a function F(x, y)such that
∂F
∂x = 3y2+ 2x
∂F
∂y = 6xy
Integrating the first equation with respect to xgives
F(x, y) = ∫(3y2+ 2x)dx = 3y2x+x2+h(y)
where h(y)is a function of y.
3. Now, differentiate F(x, y)with respect to yand compare to ∂F
∂y to find h(y):
∂F
∂y = 6xy +dh
dy
6xy = 6xy +dh
dy
This implies that dh
dy = 0, so h(y)is a constant. Let’s denote this constant as C.
4. Therefore, the general solution to the given differential equation is
3y2x+x2+C= 0
17. Consider the differential equation given by
(2x+ 3y)dx + (3x+ 4y)dy = 0
Determine if the differential 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 .
1. Calculate ∂M
∂y :
∂M
∂y = 3
2. Calculate ∂N
∂x :
∂N
∂x = 3
Since ∂M
∂y =∂N
∂x , the given differential equation is exact.
3. To find the general solution, we integrate Mdx with respect to xto find the potential
function U(x, y)and then take the partial derivative of U(x, y)with respect to yto find the
solution.
Integrating (2x+ 3y)dx with respect to x:
U(x, y) = ∫(2x+ 3y)dx =x2+ 3xy +f(y)
4. Calculating ∂U
∂y :
∂U
∂y = 3x+df
dy = 3x+ 3
Comparing ∂U
∂y to N:
3x+ 3 = 3x+ 4y⇐⇒ df
dy = 4y⇐⇒ f(y) = 2y2+C
Therefore, the general solution to the differential equation is x2+ 3xy + 2y2=Cwhere C
is a constant.
18. Question: Determine if the following differential equation is exact, and if it is, find its
general solution:
(2y−3x)dx + (x2+ 2y)dy = 0
Ans. Step-by-step solution: 1. To check if the given differential equation is exact, we calculate
the partial derivatives of the terms with respect to yand x:
∂
∂y (2y−3x) = 2 and ∂
∂x(x2+ 2y) = 2x
Since the mixed partial derivative ∂
∂y (2y−3x) = ∂
∂x (x2+ 2y), the given differential equation is
exact.
2. To find the general solution, we need to proceed to find the function Ψ(x, y)such that
∂Ψ
∂x = 2y−3xand ∂Ψ
∂y =x2+ 2y
3. Integrating the first equation with respect to x, we get
Ψ = 2xy −3
2x2+h(y)
where h(y)is an arbitrary function of y.
4. Now, we differentiate Ψwith respect to yand equate it to the second equation:
∂Ψ
∂y = 2x+h′(y) = x2+ 2y
5. Comparing the terms, we find that h′(y) = 2y, which implies that h(y) = y2+Cfor some
constant C.
6. Substituting back into the expression for Ψ, we have
Ψ(x, y) = 2xy −3
2x2+y2+C
7. Therefore, the general solution to the given exact differential equation is
2xy −3
2x2+y2+C= 0
where Cis the constant of integration.
19. Let Mand Nbe functions of two variables xand ydefined on a simply connected region
D. Consider the following differential equation:
M(x, y)dx +N(x, y)dy = 0
where Mand Nare given functions. Determine conditions for exactness of the equation and
find the general solution if it is exact.
Ans. To determine the conditions for exactness of the given differential equation M(x, y)dx +
N(x, y)dy = 0, we need to check if the following condition holds:
∂M
∂y =∂N
∂x
If the equation is exact, we can find the general solution by following these steps:
1. Check the condition for exactness:
∂M
∂y =∂N
∂x
2. If the equation is exact, find a function ψ(x, y)such that:
dψ =Mdx +N dy
3. Integrate dψ with respect to xand then with respect to yto find ψ(x, y).
4. The general solution is given by:
ψ(x, y) = C
where Cis a constant.
20. Let M(x, y) = x2y−2xy2+sin(x)and N(x, y) = x2−3y2. Determine if the following
differential equation is exact, and if so, find its general solution:
M(x, y)dx +N(x, y)dy = 0
Ans. To determine if the given differential equation is exact, we first check if ∂M
∂y =∂N
∂x . If it is
exact, we can find the general solution by integrating Mwith respect to xand Nwith respect
to y.
1. Calculate partial derivatives: ∂M
∂y =x2−4xy
∂N
∂x = 2x
2. Check for exactness: Since ∂M
∂y =∂N
∂x , the differential equation is not exact.
3. Find integrating factor: Let the integrating factor be µ(x)such that
µ(x)∂M
∂y =µ(x)(x2−4xy)=µ(x)∂N
∂x = 2µ(x)x
For the integrating factor to exist, we must have
∂(2µx)
∂y −∂(µ(x)(x2−4xy))
∂x
µ(x)(x2−4xy)= 0
Solving this equation will give us µ(x).
4. Solve the differential equation with the integrating factor: After finding the inte-
grating factor, multiply both sides of the differential equation by µ(x)and check if it is exact.
5. Find the general solution: If the equation is exact, integrate Mµ(x)with respect to x
and Nµ(x)with respect to yto find the general solution.
21. Question 21: Consider the differential equation given by (2x2y2−ey)dx+(4x3y−2xey)dy =
0.
Determine whether the given differential equation is exact. If it is exact, find the general
solution. If not, find an integrating factor that makes it exact and then find the solution.
Ans. To determine whether the given differential equation is exact, we need to check if
∂M
∂y =∂N
∂x . Let M= 2x2y2−eyand N= 4x3y−2xey.
1. Check for exactness: Using the given differential equation, we find:
∂M
∂y = 4x2y−ey
∂N
∂x = 12x2y−2ey
Since ∂M
∂y =∂N
∂x , the given differential equation is not exact.
2. Find the integrating factor: To find an integrating factor, we need a function µ(x, y)such
that
µ(x, y)M dx +µ(x, y)Ndy = 0
is an exact differential. We can find µ(x, y)by integrating the following equation:
∂
∂x (µN )
µ−∂
∂y (µM) = 0
After simplifying the above equation and solving for µ, we find that µ=1
2xy is an integrating
factor.
3. Find the general solution: Multiplying the given differential equation by the integrating
factor 1
2xy , we get the exact differential equation:
(ydx + 2xdy)−ey·dx
2y= 0
Integrating the above equation gives the general solution:
y2+x2=Cey/2
where Cis the constant of integration.
22. Question: Determine whether the following first-order differential equation is exact, and if
so, find the general solution:
(2y+x2)dx + (2x+ex)dy = 0
Ans. Step-by-step solution:
1. To determine if the differential equation is exact, we need to check if the partial derivatives
satisfy the condition:
∂M
∂y =∂N
∂x
2. Let M= 2y+x2and N= 2x+ex. Then, we calculate the partial derivatives:
∂M
∂y = 2
∂N
∂x = 2 + ex
3. Since ∂M
∂y =∂N
∂x , the equation is not exact.
4. To find an integrating factor µ(x, y)such that µMdx +µNdy = 0 is exact, we solve the
following equation:
∂(µM)
∂y −∂(µN )
∂x
µ= 0
5. First, calculate ∂(µM)
∂y and ∂(µN )
∂x :
∂(µM)
∂y =µ∂M
∂y +M∂µ
∂y
∂(µN)
∂x =µ∂N
∂x +N∂µ
∂x
6. Substituting the expressions above into the equation from step 4, we get:
µ∂M
∂y +M∂µ
∂y −µ∂N
∂x −N∂µ
∂x
µ= 0
7. Simplify the equation to solve for µ:
M∂µ
∂y −N∂µ
∂x = 0
8. Substitute M,N, and their partial derivatives into the equation above, and solve for µ.
9. Once you have found µ, multiply the original differential equation by µto make it exact:
µ((2y+x2)dx + (2x+ex)dy) = 0
10. After multiplying through, the equation becomes exact. Finally, solve the exact equation
to find the general solution.
23. Let M=eycos(x)−yand N=eysin(x)−x. Determine whether the differential equation
Mdx +N dy = 0 is exact, and if so, find the general solution.
Ans. To determine if the given equation is exact, we check if ∂M
∂y =∂N
∂x . 1. Calculate the
partial derivatives:
∂M
∂y =eycos(x)−1
∂N
∂x =eycos(x)
2. Since ∂M
∂y =∂N
∂x , the equation is not exact. 3. To find an integrating factor, we compute:
∂M
∂y −∂N
∂x
N=−1
x
4. An integrating factor is µ(x) = e−ln |x|=1
|x|. 5. Multiply both Mand Nby µ(x) = 1
|x|:
eycos(x)
|x|−y
|x|dx +eysin(x)
|x|dy −x
|x|dy = 0
6. Rearranging, we get:
eycos(x)
|x|−x
|x|dy =y
|x|−eysin(x)
|x|dx
7. Integrating both sides:
∫(eycos(x)
|x|−x
|x|)dy =∫(y
|x|−eysin(x)
|x|)dx
8. Simplify and solve the integrals to find the general solution.
24. Let Mdx +N dy = 0 be a differential equation, where Mand Nare functions of xand y.
Question: Determine the conditions for exactness of the given equation and find its solution
if it is exact.
Ans. To check for exactness, we need to verify if the partial derivatives satisfy the condition
∂M
∂y =∂N
∂x .
1. Calculate the partial derivatives of Mand N:
∂M
∂y and ∂N
∂x
2. Check if the equation is exact by determining if ∂M
∂y =∂N
∂x .
3. If the equation is exact, integrate Mwith respect to xto find a potential function f(x, y):
f(x, y) = ∫M dx +h(y)
4. Differentiate f(x, y)with respect to yand set it equal to Nto find the integrating factor:
∂f
∂y =N
5. Solve the differential equation by finding the partial derivative of the integrating factor
with respect to yand integrating it with respect to y:
eg(y)=∫∂N
∂x −∂M
∂y
Ndy
6. Once the integrating factor is found, the solution to the exact differential equation is given
by
u(x, y) = eg(y)·f(x, y) + c
where cis the constant of integration.
25. Question 25: Consider the differential equation
(2x−3y) + (3y−4x)dx + (4x−2y)dy = 0
Determine if the equation is exact. If it is exact, find the solution.
Ans. To determine if the given equation is exact, we check the mixed partial derivatives of the
coefficients with respect to xand y. If the mixed partial derivatives are equal, then the equation
is exact.
1. Check for exactness: Calculate the partial derivatives of the coefficients:
∂M
∂y =−3,∂N
∂x = 4
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Make the equation exact: In order 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, y) [−3−4] = −7µ(x, y)
is a function of xonly.
For an integrating factor of the form µ(x), we have dµ(x)
dx =−−7µ(x,y)
N. Integrating, we get
µ(x) = e∫−(−7)
4x−2ydx
Solving the integral, we find µ(x) = (4x−2y)−7.
3. Find the solution: Multiply the given equation by the integrating factor to make it exact:
(2x−3y)(4x−2y)−7+ (3y−4x)dx + (4x−2y)(4x−2y)−7dy = 0
Now, check if the equation is exact:
∂
∂y [(2x−3y)(4x−2y)−7]=∂
∂x [(4x−2y)(4x−2y)−7]
d
dy [2x−3y
(4x−2y)7]=d
dx [4x−2y
(4x−2y)7]
This equation is exact, hence we can now solve for the general solution.
26. 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 if the equation is exact.
Ans. To determine if the first-order ODE is exact, we need to check if ∂M
∂y =∂N
∂x . If exact,
we then find a function ψ(x, y)such that ∂ψ
∂x =Mand ∂ψ
∂y =N. Finally, the general solution is
given by ψ(x, y) = C, where Cis an arbitrary constant.
Conditions for Exactness: 1. Check if ∂M
∂y =∂N
∂x .
Solution: 1. Compute ∂M
∂y and ∂N
∂x :
∂M
∂y =∂
∂y (M)and ∂N
∂x =∂
∂x(N)
2. If ∂M
∂y =∂N
∂x , the equation is exact. Otherwise, it is not.
Finding the Solution: 1. If the equation is exact, find a function ψ(x, y)such that ∂ψ
∂x =M
and ∂ψ
∂y =N.
2. Integrate Mwith respect to xand Nwith respect to yto find ψ(x, y).
3. The general solution is then given by ψ(x, y) = C, where Cis an arbitrary constant.
27. Let M(x, y)dx +N(x, y)dy = 0 be a first-order differential equation. Determine conditions
on the functions Mand Nfor exactness and provide a method for finding a solution when these
conditions are met.
Ans. To check for exactness and find solutions, follow these steps: 1. Check for exactness:
Calculate the partial derivatives ∂M
∂y and ∂N
∂x . If these are equal, then the equation is exact.
2. If the equation is exact: Find a function F(x, y)such that dF =M dx +Ndy. Then the
solution to the differential equation is given by F(x, y) = C, where Cis a constant.
3. Find the function F: Integrate Mwith respect to xand Nwith respect to y. Set these
two integrals equal to each other and equate the result to F(x, y).
4. General solution: Solve F(x, y) = Cfor Cto obtain the general solution to the differential
equation.
By following these steps, one can determine if a given first-order differential equation is exact
and find its general solution when the conditions for exactness are met.
28. Question 28: Consider the differential equation (2x+ 3y)dx + (x−y)dy = 0.
Determine whether the given first-order ordinary differential equation is exact or not. 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 general solution.
Ans. To determine whether the given first-order ordinary differential equation is exact, we check
if ∂M
∂y =∂N
∂x .
1. Let M= 2x+ 3yand N=x−y. Calculating the partial derivatives gives: ∂M
∂y = 3 and
∂N
∂x = 1. Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. To find an integrating factor µ(x), we use the formula:
∂M
∂y −∂N
∂x
N=3−1
x−y=µ′(x). Simplifying
this expression, we get: 2
x−y=µ′(x). Thus, the integrating factor is µ(x) = e2ln |x−y|=
|x−y|2.
3. Multiplying the given equation by the integrating factor |x−y|2gives:
2x|x−y|2dx + 3y|x−y|2dx + (x−y)|x−y|2dy = 0.
Simplifying the equation yields:
2x(x−y)2dx + 3y(x−y)2dx + (x−y)3dy = 0.
4. We can now check if the equation is exact: Let M= 2x(x−y)2and N= (x−y)3.
Calculating the partial derivatives gives: ∂M
∂y =−4x(x−y)and ∂N
∂x =−3(x−y)2. Since
∂M
∂y =∂N
∂x , the equation is exact.
5. To find the general solution, we need to solve the partial differential equation fx=Mand
fy=N. Integrating Mwith respect to xgives:
f(x, y) = ∫2x(x−y)2dx =2
5x5−4
3xy(x−y)2+h(y),
where h(y)is an arbitrary function of y.
6. Now we differentiate f(x, y)with respect to yand equate it to N:
∂f
∂y =∂N
∂y ⇒d
dy (−4
3xy(x−y)2+h(y))= (x−y)3.
Solving this differential equation will yield the general solution of the given differential
equation.
29. Question 29: Consider the differential equation given by (y−2x2)dx + (2y−x)dy = 0.
Determine whether the given differential equation is exact. If it is exact, find a potential
function F(x, y)such that ∂F
∂x =y−2x2and ∂F
∂y = 2y−x.
Ans. To determine whether the given differential equation is exact, we need to check if
∂M
∂y =∂N
∂x , where the differential equation is in the form M(x, y)dx +N(x, y)dy = 0.
1. Check for exactness: The given differential equation is (y−2x2)dx + (2y−x)dy = 0.
Let M(x, y) = y−2x2and N(x, y) = 2y−x. Calculate the partial derivatives: ∂M
∂y = 1 and
∂N
∂x =−1. Since ∂M
∂y =∂N
∂x , the given differential 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 µM dx +µNdy = 0 is exact. The integrating factor is given by µ=
e∫
∂N
∂x −∂M
∂y
Ndx.
Calculate
∂N
∂x −∂M
∂y
N:
−1−1
2y−x=−2
2y−x=−2
2y−x
So, the integrating factor is µ=e∫−2
2y−xdx.
3. Verify the exactness with the integrating factor: Multiply both sides of the differential
equation by the integrating factor µ:
e∫−2
2y−xdx(y−2x2)dx +e∫−2
2y−xdx(2y−x)dy = 0
Simplify the equation and check if it is exact. If it is, find the potential function.
Note: The solution process will continue in the next steps once the integrating factor is
determined.
30. Let M(x, y)dx +N(x, y)dy = 0 be a differential equation. Determine conditions for
exactness and provide a method for finding solutions.
Ans. To determine conditions for exactness and find solutions to a given differential equation
of the form M(x, y)dx +N(x, y)dy = 0, we check if the following condition is satisfied:
∂M
∂y =∂N
∂x .
If this condition is fulfilled, then the equation is exact. In this case, we can find the solution
by finding a function f(x, y)such that ∂f
∂x =Mand ∂f
∂y =N. The solution is then given by
f(x, y) = c, where cis an arbitrary constant.
31. Question 31: Consider the differential equation given by
(cos y+xsin y)dx + (sin x+ycos x)dy = 0
Is the equation exact? If so, find a solution. If not, determine an integrating factor and then
find the general solution.
Ans. To check for exactness, we compute the partial derivatives of the coefficients with respect
to yand x:
1. Compute ∂
∂y (cos y+xsin y) = −sin y+xcos y2. Compute ∂
∂x (sin x+ycos x) =
cos x−ysin x
Since the mixed partial derivatives are not equal, the equation is not exact. To find an
integrating factor, we can use the formula µ(x, y) = e∫(My−Nx
N)dx.
3. Compute My−Nx
N=(−sin y+xcos y)−(cos x−ysin x)
sin x+ycos x=−sin y−cos x+xcos y+ysin x
sin x+ycos x
Therefore, the integrating factor is µ(x, y) = e∫−sin y−cos x+xcos y+ysin x
sin x+ycos xdx.
After finding µ(x, y), we can multiply through the differential equation by µ(x, y)in order to
obtain an exact differential equation. Finally, we can solve the exact differential equation to find
the general solution.
32. Consider the following differential equation:
(3x2−2x)dx + (y−x2)dy = 0
Determine if the given differential equation is exact. If it is exact, find its solution. If it is not
exact, find an integrating factor that makes it exact.
Ans. To determine if the given differential equation is exact, we check if the partial derivatives
of the terms with respect to yare equal. If they are not equal, we find the integrating factor that
makes the equation exact.
1. Check for exactness: Let M= 3x2−2xand N=y−x2. Calculate the partial
derivatives: ∂M
∂y = 0
∂N
∂x =−2x
Since ∂M
∂y =∂N
∂x , the differential equation is not exact.
2. Find the integrating factor: The integrating factor µ(x)can be found from the relation:
1
µ(x)(∂
∂x(µ(x)M)−∂
∂y (µ(x)N))= 0
Substitute M,N, and their partial derivatives into the above equation and solve for µ(x).
We have:
µ(x)∂M
∂x −µ(x)∂N
∂y −dµ(x)
dx N= 0
µ(x)(6x−2) −dµ(x)
dx (y−x2) = 0
Solving for µ(x):
6xµ(x)−2µ(x)−ydµ(x)
dx +x2dµ(x)
dx = 0
(6x−2)µ(x) + x2dµ(x)
dx −ydµ(x)
dx = 0
Separate variables and integrate to find µ(x).
3. Find the solution: Once we have the integrating factor µ(x), multiply the given differ-
ential equation by µ(x)and proceed to solve the equation as an exact differential equation.
Thus, for question number 32, we would proceed to find the integrating factor and solve the
differential equation using the above steps.
33. What are the conditions for a differential equation to be exact, and what are the methods
for finding solutions to exact differential equations?
Ans. To determine if a first-order differential equation of the form M(x, y)dx +N(x, y)dy = 0
is exact, we need to check if ∂M
∂y =∂N
∂x . If the equation is exact, then we can find a solution by
integrating M(x, y)dx with respect to xand N(x, y)dy with respect to y, and setting the sum
equal to a constant.
1. Check for exactness by verifying if ∂M
∂y =∂N
∂x .2. If the equation is exact, then find a
function ψ(x, y)such that ψx=Mand ψy=N.3. Integrate M(x, y)dx with respect to xto
obtain ψ(x, y) = ∫M(x, y)dx+h(y), where h(y)is a function of y.4. Take the partial derivative
of ψ(x, y)with respect to yand set it equal to N(x, y)to solve for h′(y).5. Substitute the
value of h(y)back into ψ(x, y)to get the general solution ψ(x, y) = C, where Cis a constant.
34. Let ω=ey−xcos y
x2+ydx +xsin y+ey
x2+ydy be a differential 1-form. Determine whether ωis exact,
and if it is, find a function f(x, y)such that df =ω.
Ans. To determine whether the differential form ωis exact, we calculate its partial derivatives
and check if they satisfy the condition for exactness. If so, we integrate these partial derivatives
to find the function f(x, y).
1. Calculate the partial derivatives of ω:We have ∂
∂y (ey−xcos y
x2+y)=ey+xsin y
x2+y−2xy(ey−xcos y)
(x2+y)2
and ∂
∂x (xsin y+ey
x2+y)=sin y(x2+y)−(xsin y+ey)2x
(x2+y)2.
2. Check if the partial derivatives are equal: We need to verify if ∂
∂y (ey−xcos y
x2+y)=
∂
∂x (xsin y+ey
x2+y)holds.
3. If the partial derivatives are equal, the form is exact: If ωis exact, we proceed to
find f(x, y)such that ∂f
∂x =ey−xcos y
x2+yand ∂f
∂y =xsin y+ey
x2+y.
4. Integrate the partial derivatives to find f(x, y):By integrating the expressions from
step 1 with respect to xand yseparately, we can find the function f(x, y).
Therefore, the challenging task here is to calculate the partial derivatives of the given differ-
ential form and verify if it is exact, then find the potential function f(x, y)if it is exact.
35. Consider the following first-order ordinary differential equation (ODE):
(2xy2+ex)dx + (x2y+ 2ex)dy = 0
Question 35: Determine whether the ODE above is exact. If it is exact, find a function
f(x, y)such that ∂f
∂x = 2xy2+exand ∂f
∂y =x2y+ 2ex.
Ans. To determine whether the ODE is exact, we need to check if ∂M
∂y =∂N
∂x , where the ODE
is of the form M dx +Ndy = 0.
1. Check for exactness: Let M= 2xy2+exand N=x2y+ 2ex. Calculate the partial
derivatives: ∂M
∂y = 4xy and ∂N
∂x = 2xy
Since ∂M
∂y =∂N
∂x , the ODE is not exact.
2. Find integrating factor: To find an integrating factor, we can use the formula µ=
e∫Nx−My
Mdx.
Calculate Nx−My:
Nx−My= (2xy)−(2xy) = 0
Thus, the integrating factor µ=e∫0dx =e0= 1.
3. Multiply the ODE by the integrating factor: Multiply both sides of the ODE by
µ= 1:
(2xy2+ex)dx + (x2y+ 2ex)dy = 0
4. Check for exactness with the integrating factor: The ODE is exact if ∂(µM)
∂y =∂(µN )
∂x .
Calculate: ∂
∂y (M) = 2x(2y) = 4xy
∂
∂x(N) = y(2) = 2y
Since ∂(µM)
∂y =∂(µN )
∂x , the ODE is exact with the integrating factor.
5. Find the function f(x, y):Since the ODE is exact with the integrating factor, the
function f(x, y)exists. We integrate the terms Mand Nwith respect to xand yrespectively
to find f(x, y).
f(x, y) = ∫(2xy2+ex)dx +g(y) = x2y2+ex+g(y)
f(x, y) = ∫(x2y+ 2ex)dy +h(x) = x2y2+ 2ex+h(x)
Equating the two expressions for f(x, y), we conclude that g(y) = 2ex+Cand h(x) = ex+C
where Cis an arbitrary constant.
Therefore, the function f(x, y) = x2y2+ex+ 2ex+C=x2y2+ 3ex+Csatisfies the
conditions, making it a solution to the exact ODE.
