MAT 275 - MODERN DIFFERENTIAL EQUATIONS -
Conditions for exactness and methods for finding solutions
Practice Material - Set 1
1. Consider the differential equation given by
(y2+ 2x)dx + (2y)dy = 0
Determine whether the equation is exact or not, and if it is exact, find the general solution.
Ans. To determine whether the given differential equation is exact, we need to check if the
partial derivatives of the terms with respect to yare equal. Then, we need to find the integrating
factor if it is not exact. If the equation is exact, we can find the general solution by integrating
one of the terms with respect to the corresponding variable.
1. Check for Exactness:
Let M=y2+ 2xand N= 2y. Then, calculate the partial derivatives with respect to y:
∂M
∂y = 2y
∂N
∂x = 0
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Find the Integrating Factor:
To find the integrating factor, we need to calculate the integrating factor µ(x)such that
∂
∂y (µ(x)M) = ∂
∂x (µ(x)N)
Let µ(x)be the integrating factor. Then, we have:
d
dx (µ(x)N)−dµ(x)
dx N=d
dy (µ(x)M)−dµ(x)
dy M
Integrating factors generally have the form µ(x) = e∫P(x)dx, where P(x)is a function of x.
3. Solve for the Integrating Factor:
From the equation above, we get:
2µ(x)−dµ(x)
dx (2y) = µ(x)
Solving this differential equation will give us the integrating factor µ(x).
4. Multiply the Equation by the Integrating Factor:
After determining the integrating factor µ(x), multiply both sides of the equation by µ(x).
5. Find the General Solution:
Now that the equation is exact, find the general solution by integrating one of the terms with
respect to the corresponding variable. The general solution will be in the form F(x, y) = C,
where Cis the constant of integration.
2. Consider the following differential equation:
(2xy −y2)dx + (x2−2y)dy = 0
Determine whether 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 the mixed
partial derivatives of the coefficients with respect to xand yare equal. In this case, we have:
∂
∂y (2xy −y2) = 2x−2y
∂
∂x(x2−2y) = 2x
Since the mixed partial derivatives are not equal, the given differential equation is not exact.
3. Suppose we have a differential equation of the form
∂M
∂y −∂N
∂x = 0.
Determine the conditions for exactness of the equation and outline the method for finding a
solution.
Ans. To determine the conditions for exactness and find a solution, we follow the steps below:
1. The given equation is exact if and only if
∂M
∂y =∂N
∂x .
2. Once we have verified that the equation is exact, we can find a potential function Φ(x, y)
such that
∂Φ
∂x =Mand ∂Φ
∂y =N.
3. To find Φ(x, y), integrate Mwith respect to xand check if the result depends only on y.
If it does, let this result be a function of yand denote it as f(y).
4. Then, integrate Nwith respect to yand check if the result depends only on x. If it does,
let this result be a function of xand denote it as g(x).
5. Finally, let
Φ(x, y) = f(y) + g(x) + C,
where Cis a constant of integration. This function Φ(x, y)is a potential function for the
exact equation.
4. Suppose we have the following differential equation:
(x2y−2xy2)dx + (x3−3xy)dy = 0
Determine whether the given differential equation is exact. If it is exact, find the general
solution.
Ans. To determine whether the given differential equation is exact, we need to ensure that the
partial derivatives satisfy the condition for exactness: ∂M
∂y =∂N
∂x .
1. Calculate the partial derivatives:
M=x2y−2xy2,N=x3−3xy
∂M
∂y =x2−4xy and ∂N
∂x = 3x−3y
2. Check if ∂M
∂y =∂N
∂x :
x2−4xy = 3x−3y
Since the condition for exactness is not satisfied, we cannot conclude that the given differential
equation is exact. We will now proceed with finding an integrating factor to make it exact.
3. To find the integrating factor, we will solve the following differential equation:
∂M
∂y −∂N
∂x
N=−4x−3
x3−3x
4. Simplify the above expression and integrate it to find the integrating factor:
Integrating factor =e∫−4x−3
x3−3xdx
5. After finding the integrating factor, multiply it by the differential equation to make it
exact.
6. Solve the exact differential equation using standard techniques to find the general solution.
5. Consider the following differential equation:
(2x+ 3y)dx + (x+ 2y)dy = 0
Determine if the equation is exact and, if so, find its general solution.
Ans. To check if the given differential equation is exact, we need to verify if the partial derivatives
of the coefficients with respect to y and x are equal. 1. Compute the partial derivatives:
∂M
∂y = 3
∂N
∂x = 1
2. Check if the equation is exact: Since ∂M
∂y =∂N
∂x , the equation is not exact. 3. Find an
integrating factor: To make the equation exact, we need to find a function µ(x, y)such that:
µ(2x+ 3y)dx +µ(x+ 2y)dy = 0
Comparing coefficients, we get the following equations:
∂
∂y (µ(x+ 2y)) = ∂
∂x(µ(2x+ 3y))
2µ′(x+ 2y) = 3µ′(2x+ 3y)
4. Solve for µ(x, y): Solving the above equation, we find that µ(x, y) = e3x−2y
6=ex
2−y
3. 5.
Multiply the given differential equation by the integrating factor: Multiplying the given differential
equation by µ(x, y), we get:
e3x
6−2y
6((2x+ 3y)dx + (x+ 2y)dy) = 0
2xex
2−y
3dx + 3ye x
2−y
3dx +xex
2−y
3dy + 2ye x
2−y
3dy = 0
6. Integrate the exact differential equation: By recognizing the left-hand side as the total differ-
ential of a particular function, we have:
d(ex
2−y
3(2x+ 3y)) = 0
Integrating both sides, we get:
ex
2−y
3(2x+ 3y) = C
where C is the constant of integration. 7. General solution: Therefore, the general solution of
the given differential equation is:
ex
2−y
3(2x+ 3y) = C
6. Consider the following differential equation:
(2x+y2)dx + (2xy)dy = 0
Determine whether the equation is exact. If it is exact, find a potential function ϕ(x, y)such
that dϕ = (2x+y2)dx + (2xy)dy. If it is not exact, explain why and determine an integrating
factor to make it exact.
Ans. To determine whether the equation is exact, we check the mixed partial derivatives:
∂
∂y (2x+y2) = 0 and ∂
∂x(2xy) = 2y
Since the two mixed partial derivatives are not equal, the equation is not exact. We can find
the integrating factor µ(x)by dividing the partial derivative of Mwith respect to yby the partial
derivative of Nwith respect to x:
µ(y) = e∫(∂M
∂y
−∂N
∂x
N)dy
=e∫(0−2y
2xy )dy =e−∫1
xdy =e−y
x
Multiplying the given differential equation by the integrating factor µ=e−y
x, we get:
(e−y
x(2x+y2))dx + (e−y
x(2xy))dy = 0
The equation is now exact. Let’s find ϕ.
By integrating the first term with respect to x, we get:
∫(e−y
x(2x+y2))dx =e−y
x(x2+xy2) + h(y)
where h(y)is a function of y.
Taking the partial derivative of this expression with respect to y, we find:
∂
∂y (e−y
x(x2+xy2)+h(y)) = −x
x2e−y
x(x2+xy2)+e−y
x(2xy)+h′(y) = 2xy−xy+h′(y) = h′(y)+xy
Comparing this with the original N, we see that h′(y) = 0. Thus, h(y) = Cfor some
constant C.
Therefore, the potential function ϕ(x, y)is given by:
ϕ(x, y) = e−y
x(x2+xy2) + C
7. Determine whether the following first-order ordinary differential equation is exact, and if so,
find the general solution:
(2xy + 3)dx + (x2+ 1)dy = 0
Ans. To determine whether the given differential equation is exact, we will check if the partial
derivatives of the coefficients with respect to the opposite variable are equal. If they are equal,
we will proceed to find the integrating factor and then find the general solution.
1. Calculate the partial derivatives:
∂
∂y (2xy + 3) = 2xand ∂
∂x(x2+ 1) = 2x
Since the partial derivatives are equal, the given differential equation is exact.
2. Find the integrating factor: The integrating factor, denoted by µ, is given by:
µ(x) = e∫
∂M
∂y
−∂N
∂x
Ndx
In this case:
µ(x) = e∫2x−2x
x2+1 dx =e∫0dx =eC=C(where C is a constant)
3. Multiplying the differential equation by the integrating factor:
C(2xy + 3)dx +C(x2+ 1)dy = 0
4. Integrate to find the general solution: Integrating both sides:
C(x2y+ 3x) + C1(x) = 0
where C1(x)is the constant of integration.
8. Question: Consider the differential equation
(2xy cos(x2y)−x2sin(x2y))dx + (x2cos(x2y)−4ysin(x2y))dy = 0
Determine whether the given equation is exact or not. If it is exact, find a potential function
for the equation. If it is not exact, determine an integrating factor to make it exact.
Ans. To determine if the given equation is exact, we check if the partial derivatives of the
coefficients with respect to yand xare equal. Then, we proceed to find the potential function
or integrating factor accordingly.
1. We calculate the partial derivatives of the coefficients with respect to yand x:
∂
∂y (2xy cos(x2y)−x2sin(x2y)) = 2xcos(x2y)−2x3ysin(x2y)
∂
∂x(x2cos(x2y)−4ysin(x2y)) = 2xcos(x2y)−2x3ysin(x2y)
Since the partial derivatives are equal, the given equation is exact.
2. To find the potential function, we integrate the coefficient of dx with respect to xand
then differentiate the result with respect to y.
Let f(x, y)be the potential function. Integrating the coefficient of dx with respect to x:
f(x, y) = ∫(2xy cos(x2y)−x2sin(x2y))dx =x2ysin(x2y)+g(y)
where g(y)is a function of ythat we will determine by differentiating f(x, y)with respect to
yand comparing it to the coefficient of dy:
∂f
∂y =x2cos(x2y)+g′(y) = x2cos(x2y)−4ysin(x2y)
Comparing coefficients, we find:
g′(y) = −4ysin(x2y)
Integrating both sides with respect to y:
g(y) = 2 cos(x2y)+C
Therefore, the potential function for the given equation is:
f(x, y) = x2ysin(x2y)+ 2 cos(x2y)+C
where Cis a constant.
9. Find the solution to the exact differential equation
(2x2y+y2)dx + (2x3+ 2xy)dy = 0
Ans. To determine if the given differential equation is exact, we will compute the partial
derivatives of Mand Nwith respect to yand xrespectively, and check if they are equal. If they
are equal, we can then find the integrating factor µ(x, y)and solve the differential equation.
1. Determine if the equation is exact: Given differential equation is:
M= 2x2y+y2and N= 2x3+ 2xy
Calculate the partial derivatives of Mand N:
∂M
∂y = 2x2+ 2yand ∂N
∂x = 6x+ 2y
2. Check for exactness: For the equation to be exact, we require:
∂M
∂y =∂N
∂x
Comparing the two partial derivatives, we see that they are not equal. Thus, the given differential
equation is not exact.
3. Find the integrating factor: To find the integrating factor, we calculate:
µ(x) = e∫(∂N
∂x
−∂M
∂y )
Mdx
y
4. Solve the differential equation: Multiply the differential equation by the integrating
factor found in step 3 to make it exact, then solve for the solution.
10. Consider the following differential equation:
(3x2y+ 2y2)dx + (x3+ 4xy)dy = 0
Determine if the given differential equation is exact, and if it is, find the general solution.
Ans. To check if the given differential equation is exact, we can compute the partial derivatives
of M= 3x2y+ 2y2with respect to yand N=x3+ 4xy with respect to x:
1. ∂M
∂y = 3x2+ 4y2. ∂N
∂x = 3x2+ 4y
Since ∂M
∂y =∂N
∂x , the differential equation is exact. To find the general solution, we can follow
these steps:
1. Integrate Mwith respect to xto find a potential function f(x, y).
f(x, y) = ∫(3x2y+ 2y2)dx =x3y+ 2xy2+g(y)
2. Compute ∂f
∂y .
∂f
∂y =x3+ 4xy +g′(y)
3. Set ∂f
∂y =Nto find g(y).
x3+ 4xy +g′(y) = x3+ 4xy =
⇒g′(y) = 0
4. Integrate g(y) = 0 with respect to y.
g(y) = C
5. Substitute f(x, y) = x3y+ 2xy2+Cand solve the equation f(x, y) = Cfor the general
solution.
x3y+ 2xy2+C= 0
Therefore, the general solution to the differential equation is x3y+ 2xy2=−Cwhere Cis a
constant.
11. Question 11: Consider the differential equation
(2x2y+ 3y2)dx +(x3+ 6xy)dy = 0
Determine whether the given differential equation is exact. If it is exact, find a solution to
the equation. If it is not exact, determine 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
the partial derivatives of the terms with respect to yand xare equal. Let’s start by checking
exactness.
1. We check for exactness: The given equation is (2x2y+ 3y2)dx + (x3+ 6xy)dy = 0.
Calculating the partial derivatives:
∂M
∂y = 2x2+ 6y
∂N
∂x = 6y
Since ∂M
∂y =∂N
∂x , the equation is exact.
2. Find the potential function: To find the potential function F(x, y), we integrate the term
with respect to xto get F(x, y):
F(x, y) = ∫(2x2y+ 3y2)dx =x2y2+ 3xy2+g(y)
Now, we differentiate F(x, y)with respect to yand set it equal to the Nterm to find g(y):
∂F
∂y =x2+ 6xy +g′(y) = x3+ 6xy
g′(y) = 0
Thus, g(y) = C, where Cis a constant.
3. General solution: Therefore, the general solution to the differential equation is:
x2y2+ 3xy2+C= 0
12. Find the integrating factor for the following differential equation, if possible, so that it
becomes exact:
(2xy + 4y2)dx + (x2+ 8y)dy = 0
Ans. To determine if the given differential equation is exact, we must check if the partial
derivatives of the coefficients with respect to y and x are equal:
∂M
∂y = 2x+ 8y
∂N
∂x = 2y
Since ∂M
∂y =∂N
∂x , the differential equation is not exact. To make it exact, we introduce an
integrating factor µsuch that µ(x, y)M dx +µ(x, y)Ndy = 0 is exact. The integrating factor is
given by:
µ(x, y) = 1
µ(x, y)M−∂µ
∂y N
Plugging in the values for M, N, and their partial derivatives, we have:
µ(x, y) = 1
2xy + 4y2+ 8y−2x−8y=1
2xy −2x=1
2x(y−1)
Therefore, the integrating factor for the given differential equation is µ(x, y) = 1
2x(y−1) .
13. Consider the following differential equation:
(2x3y2+cos(y))dx + (3x2y+sin(x))dy = 0
Determine if the given differential equation is exact. If it is exact, find the general solution.
If not, determine an integrating factor to make it exact, and then find the solution.
Ans. To determine if the given differential equation is exact, we first check for exactness by
computing the partial derivatives of the terms involving xand y.
1. We check for exactness by computing ∂M
∂y and ∂N
∂x :
∂M
∂y = 4x3y, ∂N
∂x = 6x2y
Since ∂M
∂y =∂N
∂x , the differential equation is not exact. To make it exact, we need to find an
integrating factor µ(x, y)such that
µ(x, y)(2x3y2+cos(y))dx +µ(x, y)(3x2y+sin(x))dy = 0
2. We find the integrating factor µ(x, y):
∂(µM)
∂y =∂(µN)
∂x
2x3y2dµ
dy +µ(4x3y) = 3x2ydµ
dx +µ(6x2y)
3. Equating the coefficients of dx and dy, we get:
2x3y2dµ
dy = 6x2yµ and µ(4x3y) = 3x2ydµ
dx
Solving for µ(x, y)will give us our integrating factor.
4. After finding the integrating factor, the differential equation should become exact. We
can then solve it by finding the potential function, which will give the general solution.
14. Let M= 2xy2+cos(x)and N=x2y−sin(y). Determine whether the differential form
ω=Mdx +Ndy is exact. If it is exact, find a function f(x, y)such that df =ω.
Ans. The given differential form ω=M dx +Ndy is exact because ∂M
∂y = 4xy =∂N
∂x = 2xy.
1. To find f(x, y)such that df =ω, we integrate Mwith respect to xand then differentiate
the result with respect to y(or integrate Nwith respect to yand differentiate with respect to
x).
Integrating M= 2xy2+cos(x)with respect to xgives f(x, y) = x2y2+sin(x) + g(y), where
g(y)is the constant of integration.
2. Next, we differentiate f(x, y)with respect to yand set it equal to N=x2y−sin(y).
∂f
∂y = 2xy2+g′(y)
N=x2y−sin(y)
Matching the yterms on both sides, we get g′(y) = −sin(y)which implies g(y) = cos(y)+C
for some constant C.
Therefore, the function f(x, y) = x2y2+sin(x)+cos(y)+Cwhere Cis an arbitrary constant.
15. Let ω= (3x2+ 2y)dx + (2x+ 3y2)dy. Determine whether the differential form is exact,
and if so, find the potential function F(x, y)such that ω=dF .
Ans. To determine if the given differential form is exact, we check if ∂M
∂y =∂N
∂x , where
ω=Mdx +Ndy.
1. Calculate the partial derivatives: ∂M
∂y = 2 and ∂N
∂x = 2
Since ∂M
∂y =∂N
∂x , the differential form is exact.
2. Find the potential function F(x, y): We integrate Fx=Mto find Fup to a constant:
F(x, y) = ∫(3x2+ 2y)dx =x3+ 2xy +g(y)
3. Find the remaining term of Fusing Fy=N:∂F
∂y =x2+2+g′(y) = 2x+ 3y2From this,
we find g′(y) = 3y2−2Integrate g′(y)to find g(y):g(y) = y3−2y+C
Therefore, the potential function F(x, y) = x3+ 2xy +y3−2y+C
16. Consider the following differential equation:
(2xy2+ 3)dx + (x2y+ 2)dy = 0
Determine if the equation is exact. If it is exact, find its solution.
Ans. To check if the given differential equation is exact, we compute the partial derivatives of
each term with respect to yand x.
1. Partial derivatives: Let M= 2xy2+ 3 and N=x2y+ 2. Calculate ∂M
∂y and ∂N
∂x .
∂M
∂y = 4xy and ∂N
∂x = 2xy
2. Check for exactness: The equation is exact if ∂M
∂y =∂N
∂x . Since 4xy = 2xy, the equation
is not exact.
3. Integrating factor: To make the equation exact, we need to find an integrating factor
µ(x, y)such that multiplying both sides of the equation by µwill make it exact. The integrating
factor µis given by µ=e∫
∂N
∂x
−∂M
∂y
Ndx.
Calculate
∂N
∂x −∂M
∂y
Nand integrate.
∂N
∂x −∂M
∂y
N=2xy −4xy
x2y+ 2 =−2xy
x2y+ 2
4. Solution: Now, we find the integrating factor:
µ=e∫−2xy
x2y+2 dx
To complete the solution, integrate µwith respect to xand then use it to multiply the original
equation in order to find the solution.
17. Consider the following differential equation:
(2x2y2+e2xy)dx + (4xy3+e2xy −1)dy = 0
Determine if the differential equation is exact, and if so, find the general solution.
Ans. To determine if the given equation is exact, we check if the partial derivatives of the
coefficients of dx and dy are equal. If so, we can find a potential function whose total derivative
yields the given equation.
1. Check for exactness: We have the differential equation:
(2x2y2+e2xy)dx + (4xy3+e2xy −1)dy = 0
Calculating the partial derivatives:
∂
∂y (2x2y2+e2xy ) = 4xy + 2xe2xy
∂
∂x(4xy3+e2xy −1) = 4y3+ (2y)e2xy
Since the mixed partial derivatives are equal (2xe2xy = 2ye2xy), the equation is exact.
2. Find potential function: We need to find a function fsuch that:
∂f
∂x = 2x2y2+e2xy
∂f
∂y = 4xy3+e2xy −1
Integrating the first equation with respect to xgives us:
f=∫(2x2y2+e2xy)dx =x2y2+1
2e2xy +h(y)
where h(y)is a constant of integration depending on y.
Now we differentiate fwith respect to y:
∂f
∂y = 4xy3+e2xy +h′(y)
Comparing with the given ∂f
∂y , we get:
h′(y) = −1
h(y) = −y+C
Therefore, the potential function is:
f=x2y2+1
2e2xy −y+C
3. General solution: Since the equation is exact, the general solution is obtained by setting
the potential function fequal to a constant:
x2y2+1
2e2xy −y+C= 0
Thus, the general solution to the given differential equation is:
x2y2+1
2e2xy −y=−C
where Cis a constant.
18. Suppose we have a differential equation given by (y2−x3) + (2xy −3y2)dy = 0.
Determine whether the equation is exact, and if so, find its general solution.
Ans. To check if the given differential equation is exact, we compute the partial derivatives of
the expression F(x, y) = y2−x3with respect to yand ∂(2xy −3y2) /∂x.
1. Compute ∂F /∂y:
∂F
∂y = 2y
2. Compute ∂(2xy −3y2)/∂x:
∂
∂x (2xy −3y2)= 2y
Since ∂F
∂y =∂
∂x (2xy −3y2), the given differential equation is exact.
3. To find the general solution, solve the equation Mdx +Ndy = 0 where M=y2−x3and
N= 2xy −3y2.
4. Find potential function U(x, y)such that dU =Mdx +Ndy: Integrating M=∂U/∂x:
U=∫(y2−x3)dx =−x3y+1
3y3+h(y)
where h(y)is the constant of integration with respect to x.
5. Partial derivative of Uwith respect to y:
∂U
∂y =−x3+y2+h′(y)
6. Equate ∂U
∂y to N= 2xy −3y2to find h′(y):
−x3+y2+h′(y) = 2xy −3y2
h′(y) = 2xy −3y2+x3−y2
h′(y) = 2xy −4y2+x3
7. Integrate h′(y)to find h(y):
h(y) = ∫(2xy −4y2+x3)dy
h(y) = x3y−4
3y3+x3y+c
where cis the constant of integration.
8. Thus, the general solution is given by:
−x3y+1
3y3+x3y−4
3y3+x3y+c= 0
2x3y−y3+c= 0
19. Consider the following differential equation:
(3y2+ 2x)dx + (6xy + 4)dy = 0
Determine whether the given differential equation is exact or not. If it is exact, find a function
u(x, y)such that the given differential equation can be written in the form ∂u
∂x dx +∂u
∂y dy = 0.
Ans. To check if the given differential equation is exact, we need to calculate the mixed partial
derivatives. If ∂
∂y (3y2+ 2x) = ∂
∂x (6xy + 4), then the equation is exact.
1. Check for exactness:
∂
∂y (3y2+ 2x) = 6y, ∂
∂x(6xy + 4) = 6y
Since the mixed partial derivatives are equal, the given differential equation is exact.
2. Find the function u(x, y): The function u(x, y)can be found by integrating the
expression with respect to xand setting it equal to a function of y.
u(x, y) = ∫(3y2+ 2x)dx = 3y2x+x2+h(y)
3. Find ∂u ∂y and compare coefficients to find h(y):
∂u
∂y = 6yx +dh
dy
Comparing coefficients with (6xy + 4), we get:
6x= 6xand dh
dy = 4
Integrating 4with respect to ygives h(y) = 4y+C, where C is a constant.
Therefore, the function u(x, y) = 3y2x+x2+ 4y+Csatisfies the condition for the exact
differential equation.
20. Consider the following differential equation:
(3x2y4+ 4y3)dx +(12x3y3+ 3xy2)dy = 0
Determine whether the given differential equation is exact or not. If it is exact, find the
general solution.
Ans. To check for exactness, we verify if ∂P
∂y =∂Q
∂x . 1. Calculate the partial derivatives:
∂P
∂y = 12x2y3+ 12y2and ∂Q
∂x = 36x2y3+ 3y2
Since ∂P
∂y =∂Q
∂x , the differential equation is not exact. We proceed to find an integrating factor
µ(x, y).
2. To find µ(x, y)such that µP dx +µQdy = 0 becomes exact, we solve ∂(µQ)
∂x =∂(µP )
∂y .
Expanding, we have:
µ(12x3y3+ 3xy2)=∂(µP )
∂y = 12x2y3+ 12y2
µ(12x3y3+ 3xy2)=∂(µQ)
∂x = 36x2y3+ 3y2
Equating the two expressions, we have:
dµ
µ=dx
x
3. Integrating both sides, we find µ(x, y) = xα, where αis a constant. Substituting this into
the differential equation, we have:
3x2+αy4+ 4xαy3dx + 12x3+αy3+ 3xα+1y2dy = 0
4. Matching coefficients of dx and dy, we find that 2 + α= 3 + αand 1 = 0, which is not
possible. Hence, an integrating factor does not exist, and the differential equation is not exact.
One possible way to solve this equation is using Exact Differential Equation methods.
21. Consider the following first-order ordinary differential equation (ODE):
(x3y−3xy + 5)dx + (x2−3y)dy = 0
Determine whether the ODE is exact. If so, find the general solution.
Ans. To determine whether the ODE is exact, we check the partial derivatives of the coefficients
with respect to yin the order Pyand with respect to xin the order Qx, where P(x, y)and Q(x, y)
are the coefficients of dx and dy respectively.
1. Determine exactness:
Let P(x, y) = x3y−3xy + 5 and Q(x, y) = x2−3y. Calculate Py=x3−3xand Qx= 2x.
Since Pydoes not equal Qx, the ODE is not exact.
2. Make exact:
To make the ODE exact, we need to find a function u(x, y)such that multiplying the entire
equation by umakes it exact. We use the integrating factor method. Our goal is to find a
function u(x, y)such that when we multiply each term by u, the equation becomes exact, that
is, it should satisfy the condition ∂
∂y (uP ) = ∂
∂x (uQ).
Thus, we need to solve for uin the following equation:
∂
∂y (uP ) = ∂
∂x(uQ)
3. Find integrating factor:
We have ∂
∂y [(x3y−3xy + 5)u] = ∂
∂x[(x2−3y)u]
This becomes:
x3u+ (x3−3x)uy = 2xu
Letting M(x, y) = x3and N(x, y) = (x3−3x), we get the nonlinear first-order partial
differential equation (PDE):
dy
M=dx
N=du
Q−Mu
4. Solve the PDE:
Solving the PDE, we equate the first two expressions and solve for y(x)in terms of xand a
constant. We then integrate the third expression to find u(x, y).
Finally, we substitute uback into the ODE and solve for yto obtain the general solution.
22. Let M=x2y2and N= 2xy3−1. Determine if the differential form Mdx +N dy is exact,
and if so, find the general solution f(x, y) = Cwhere Cis a constant.
Ans. To determine if the given form is exact, we need to check if ∂M
∂y =∂N
∂x . 1. Calculate the
partial derivatives:
∂M
∂y = 2x2y, ∂N
∂x = 2y3.
2. Check if the derivatives are equal: Since ∂M
∂y =∂N
∂x , the form is not exact. 3. To find an
integrating factor, calculate I(x) = eP(x), where P(x) = ∫∂N
∂x −∂M
∂y
Mdx.
∂N
∂x −∂M
∂y
M=2y3−2x2y
x2y2=2y(y2−x2)
x2y2=2(y2−x2)
x2y.
4. Solve for P(x):
P(x) = ∫2(y2−x2)
x2ydx = 2 ∫(y2
x−x
y)dx = 2y2ln |x| − 2x2
y+h(y).
5. Now, multiply the original form by the integrating factor
I(x)Mdx +I(x)Ndy =(e2y2ln |x|−2x2/y)(x2y2dx + (2xy3−1)dy).
6. Simplify and integrate the equation
e2y2ln |x|−2x2/y(x2y2dx + (2xy3−1)dy) = 0.
This is the general solution.
23. Question 23: Consider the differential equation
(2xy +1
y)dx + (x2−1
y2)dy = 0
1. Determine the conditions for exactness for the given differential equation. 2. If the
equation is exact, find the solution by finding a potential function.
Ans. 1. To check for exactness, we need to verify that
∂M
∂y =∂N
∂x
where M= 2xy +1
yand N=x2−1
y2.
Calculating the partial derivatives:
∂M
∂y = 2x−1
y2
∂N
∂x = 2x
Thus, the equation is exact if and only if
2x−1
y2= 2x
−1
y2= 0
This is false, so the equation is not exact.
2. To find an integrating factor, we can examine the function µ(x, y), which is a function of
xand ythat makes the equation exact when multiplied by both terms. We assume that µ(x, y)
is a function of xonly, i.e., µ(x).
The integrating factor is given by
µ(x) = e∫
∂M
∂y
−∂N
∂x
Ndx
Plugging in the values,
µ(x) = e∫(−1
y2−2x)
x2−1
y2
dx
Solving the integral and simplifying gives
µ(x) = e−2ln |x|
µ(x) = eln(1
x2)
µ(x) = 1
x2
Now, by multiplying the given equation by the integrating factor µ(x) = 1
x2, we can obtain
an exact equation to solve.
24. Question: Consider the differential equation
(1 + 2xy)dx + (x2+y2)dy = 0
Determine whether the given differential equation is exact or not. If it is exact, find the
general solution.
Ans. Step-by-step solution: 1. To determine if the given differential equation is exact, we need
to check if
∂M
∂y =∂N
∂x
where M= 1 + 2xy and N=x2+y2.
2. Calculate ∂M
∂y and ∂N
∂x :
∂M
∂y = 2x
∂N
∂x = 2x
Since ∂M
∂y =∂N
∂x , the given differential equation is exact.
3. To find the potential function f(x, y), we integrate Mwith respect to xand set it equal
to N:
∫(1 + 2xy)dx =x+x2y=f(x, y) + C(y)
Since f(x, y)is a function of xand C(y)is a function of y, the partial derivative with respect
to yof f(x, y)will give the missing term:
∂f
∂y =x2
Integrating this with respect to ygives us the potential function f(x, y):
f(x, y) = 1
3x3+g(y)
4. The general solution of the exact differential equation is given by f(x, y) = C, where C
is the constant of integration. Combining the expressions for f(x, y):
1
3x3+g(y) = C
where g(y)is a constant of integration. Thus, the general solution of the given differential
equation is
1
3x3+g(y) = C
25. Question: Determine whether the following first-order differential equation is exact. If it
is exact, find its solution.
(2xy +y)dx + (x+ 2x2)dy = 0
Ans. Solution: 1. To check for exactness, we need to calculate the partial derivatives of the
terms with respect to yin the equation.
2. Taking the partial derivative of the term with dx with respect to y, we get:
∂
∂y (2xy +y) = 2x+ 1
3. Taking the partial derivative of the term with dy with respect to x, we get:
∂
∂x(x+ 2x2) = 1 + 4x
4. Since the mixed partial derivatives are not equal (2x+ 1 = 1 + 4x), the given differential
equation is not exact.
5. To make it exact, we need to find an integrating factor µ(x, y), such that
µ((2xy +y)dx + (x+ 2x2)dy)= 0
6. We find the integrating factor µ(x, y)by solving the following equation:
∂
∂y (µ(2xy +y)) = ∂
∂x (µ(x+ 2x2))
7. Simplifying the above equation, we have:
2xµy+µ=µ+ 4xµ
8. Solving the above equation, we find that µ(x, y) = e3yis the integrating factor.
9. Multiplying the differential equation by the integrating factor, we get:
e3y(2xy +y)dx +e3y(x+ 2x2)dy = 0
10. Now, we can check whether this new equation is exact. After verifying that it is exact,
we can find the solution by integrating with respect to xand y.
(Alternatively, if the problem requires showing the step-by-step solution, it can be continued
from here.)
26. Let M(x, y) = ey−xand N(x, y) = x2−2xy. Determine if the following differential form
is exact, and if so, find its solution:
(x2−2xy)dx + (ey−x)dy = 0
Ans. To check if the given differential form is exact, we need to verify if the partial derivatives
satisfy the condition:
1. Compute the partial derivative of Mw.r.t. y:∂M
∂y =ey.
2. Compute the partial derivative of Nw.r.t. x:∂N
∂x = 2x−2y.
3. Check if ∂M
∂y =∂N
∂x .
Since ey= 2x−2y, the given differential form is not exact.
To find an integrating factor µ, we divide
∂N
∂x −∂M
∂y
Nand set it equal to a function of xand y:
1
N(∂N
∂x −∂M
∂y ) = 2x−2y−ey
x2−2xy .
This expression can be written as a function of xand ysolely, say f(x, y). In this case,
f(x, y) = 2x−2y−ey
x2−2xy . Now, we can multiply the given differential form by the integrating factor
and then check for exactness again.
27. Question 27: Consider the differential equation
(2xy +cos(x))dx + (x2−sin(y))dy = 0
Determine whether the equation is exact. If it is exact, find the general solution.
Ans. To check for exactness, we will verify if
∂M
∂y =∂N
∂x
where M= 2xy +cos(x)and N=x2−sin(y).
1. Calculate the partial derivatives:
∂M
∂y = 2x
∂N
∂x = 2x
Since ∂M
∂y =∂N
∂x , the equation is exact.
2. Find the potential function F: Integrating Mwith respect to xand Nwith respect to y,
we get
F=∫(2xy +cos(x))dx =x2y+sin(x) + h(y)
F=∫(x2−sin(y))dy =x2y+cos(y) + g(x)
3. Equate the two expressions for Fand solve for h(y)and g(x):
x2y+sin(x) + h(y) = x2y+cos(y) + g(x)
Since the two sides are equal to each other, we have
h(y) = cos(y)
g(x) = sin(x)
4. The general solution is given by F=x2y+sin(x) + cos(y) = C, where Cis a constant.
28. Consider the differential equation given by:
(3x2y−y3)dx + (x3−3xy2)dy = 0
Determine whether the equation is exact or not, and 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 :
M= 3x2y−y3, N =x3−3xy2
∂M
∂y = 3x2−3y2,∂N
∂x = 3x2−3y2
Since ∂M
∂y =∂N
∂x , the equation is exact.
1. To find the potential function F(x, y), we integrate Mwith respect to dx:
F(x, y) = ∫(3x2y−y3)dx =x3y−xy3+g(y)
2. We differentiate F(x, y)partially with respect to yand set it equal to N:
∂F
∂y =x3−3xy2+g′(y) = x3−3xy2
g′(y) = 0 ⇒g(y) = C
3. Therefore, the potential function F(x, y)is:
F(x, y) = x3y−xy3+C
4. The general solution to the exact differential equation is then:
x3y−xy3+C= 0
where Cis an arbitrary constant.
29. Let P(x, y)and Q(x, y)be continuous functions such that the partial derivatives Py=∂P
∂y
and Qx=∂Q
∂x exist.
Consider the first-order ordinary differential equation
P(x, y)dx +Q(x, y)dy = 0.
If the equation is exact, find the general solution.
Ans. To check if the given equation is exact, we must verify if
∂P
∂y =∂Q
∂x .
1. Check exactness: ∂P
∂y =Pyand ∂Q
∂x =Qx.
2. If ∂P
∂y =∂Q
∂x , then the equation is exact. In this case, we can find a function F(x, y)such
that ∂F
∂x =Pand ∂F
∂y =Q.
3. Integrate ∂F
∂x =Pwith respect to x:
F=∫P dx +g(y),
where g(y)is a constant of integration that may depend on y.
4. Differentiate Fwith respect to yand set it equal to Qto solve for g(y):
∂F
∂y =∂
∂y (∫P dx)+dg
dy =Q.
5. Solve for dg
dy :
dg
dy =Q−∂
∂y (∫P dx).
6. Integrate dg
dy with respect to yto find g(y):
g(y) = ∫(Q−∂
∂y (∫P dx)) dy.
7. Substitute g(y)back into Fto find the general solution:
F(x, y) = ∫P dx +∫(Q−∂
∂y (∫P dx)) dy.
Therefore, the general solution to the exact ODE is given by F(x, y) = C, where Cis a
constant.
30. Question 30: Determine whether the following first-order differential equation is exact, and
if so, find its general solution:
(2x+ 4y)dx + (x+ 4)dy = 0
Ans. To determine whether the given differential equation is exact, we will check if the partial
derivatives of the coefficients with respect to yare equal. If the equation is exact, we will then
find its general solution by integrating the terms.
1. Check exactness: Let M= 2x+ 4yand N=x+ 4. Compute the partial derivatives:
∂M
∂y = 4
∂N
∂x = 1
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Make the equation exact: To make the equation exact, we need to find an integrating
factor µ(x)such that:
µ(x)Mdx +µ(x)Ndy = 0
is an exact differential equation.
We can find µ(x)by the formula:
µ(x) = e∫
∂N
∂x
−∂M
∂y
Ndx
Calculating the values, we have:
∂N
∂x −∂M
∂y
N=1−4
x+ 4 =−3
x+ 4
Therefore, the integrating factor µ(x)is µ(x) = e−3ln(x+4) =1
(x+4)3.
3. Multiply the given equation by the integrating factor: Multiply both sides of the given
equation by µ(x):2x+ 4y
(x+ 4)3dx +x+ 4
(x+ 4)3dy = 0
4. Check for exactness and find the general solution: Now that the equation is exact, we can
find the general solution by integrating. The solution will be of the form F(x, y) = C, where C
is a constant.
Integrating the terms with respect to xand yand setting the result equal to F(x, y), we find
the general solution.
General Solution: 1
2(x+ 4)2+y
(x+ 4)3=C
31. Determine whether the following first-order ordinary differential equation is exact:
(2x−y)dx + (x−2y)dy = 0
Ans. To determine if the given 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. Calculate ∂M
∂y and ∂N
∂x :
∂M
∂y =−1and ∂N
∂x = 1
2. Since ∂M
∂y =∂N
∂x , the given differential equation is not exact.
To find an integrating factor µ(x, y), we need to use the formula:
1
µ(∂(µM)
∂y −∂(µN)
∂x )= 0
Expanding this equation gives us:
−1µ+∂µ
∂y −µ+∂µ
∂x = 0
∂µ
∂x +∂µ
∂y −2µ= 0
3. We can solve this partial differential equation by assuming µ=µ(x), which simplifies the
equation to:
dµ
dx −2µ= 0
Solving this first-order linear differential equation gives:
µ=Ae2x
4. Multiply the given differential equation by the integrating factor µ=e2x:
e2x(2x−y)dx +e2x(x−2y)dy = 0
5. We rewrite this equation as the exact differential form:
d(2xe2x−ye2x) = 0
Therefore, the general solution to the given differential equation is:
2xe2x−ye2x=C
where Cis the constant of integration.
32. Question 32: Consider the differential equation
(2x3y−3y2)dx + (x4−6xy)dy = 0
Determine if the given differential equation is exact. If it is exact, find a solution. If it is not
exact, determine an integrating factor that makes it exact, and then find a solution.
Ans. To determine if the given differential equation is exact, we check if
∂M
∂y =∂N
∂x
where M= 2x3y−3y2and N=x4−6xy.
1. Compute the partial derivatives:
∂M
∂y = 2x3−6y
∂N
∂x = 4x3−6y
2. Check if the equation is exact:
∂M
∂y =∂N
∂x if and only if 2x3−6y= 4x3−6y
Since the partial derivatives are equal, the equation is exact.
3. Find the potential function F(x, y)by integrating Mwith respect to x:
F(x, y) = ∫(2x3y−3y2)dx =x4y−3xy2+g(y)
4. To find g(y), differentiate F(x, y)with respect to yand compare with N:
∂F
∂y =x4−6xy +g′(y)
Since ∂F
∂y =N=x4−6xy, we have g′(y) = 0, so g(y)is a constant.
Therefore, the general solution to the differential equation is x4y−3xy2+C= 0, where C
is a constant.
33. Let M= (2xy −y3)dx + (x2−3y2)dy be a differential 1-form. Determine if Mis an exact
form, and if so, find a function fsuch that df =M.
Ans. To determine if Mis exact, we calculate the partial derivatives ∂M
∂y and ∂N
∂x , where
M=P dx and N=Qdy.
1. Calculate ∂M
∂y and ∂N
∂x : Calculating ∂M
∂y :
∂M
∂y = 2x−3y2.
Calculating ∂N
∂x :
∂N
∂x = 2x.
2. Check for exactness by comparing the mixed partial derivatives: Since ∂M
∂y = 2x−3y2and
∂N
∂x = 2x, we have ∂M
∂y =∂N
∂x .
3. Since the mixed partial derivatives are not equal, Mis not an exact form. Thus, there
exists no function fsuch that df =M.
34. Let M(x, y)dx +N(x, y)dy = 0 be a differential equation. Consider the following equation:
(3x2y+ 2xy2)dx + (x3+ 3y2)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:
∂M
∂y = 3x2+ 2x(2y) = 3x2+ 4xy
∂N
∂x = 3x2+ 0 = 3x2
2. Check if ∂M
∂y =∂N
∂x . Since 3x2+ 4xy = 3x2, the differential equation is not exact. 3. To find
an integrating factor, we calculate
∂M
∂y −∂N
∂x
N:
3x2+ 4xy −3x2
x3+ 3y2=4xy
x3+ 3y2
4. Find an integrating factor u(x)as the above expression:
u(x) = e∫4xy
x3+3y2dy
u(x) = e2ln(x2+y3)
u(x) = (x2+y3)2
5. Multiply the differential equation by the integrating factor:
(x2+y3)2((3x2y+ 2xy2)dx + (x3+ 3y2)dy) = 0
6. Rewrite the equation to be exact by expanding and simplifying:
(3x2y(x2+y3)2+ 2xy2(x2+y3)2)dx + (x3(x2+y3)2+ 3y2(x2+y3)2)dy = 0
7. Integrate Mwith respect to xand Nwith respect to yand set the result equal to a constant
value for the general solution. Remember to simplify the expression.
35. Question 35: Consider the differential equation given by (3y−2x2)dx + (4x−6)dy = 0.
Determine whether the equation is exact, and if so, find a function F(x, y)such that ∂F
∂x =
3y−2x2and ∂F
∂y = 4x−6.
Ans. To check for exactness, we compute the mixed partial derivatives:
1. Compute ∂
∂y (3y−2x2):
∂
∂y (3y−2x2) = 3
2. Compute ∂
∂x (4x−6):
∂
∂x(4x−6) = 4
2. Consider the following differential equation:
(2xy −y2)dx + (x2−2y)dy = 0
Determine whether 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 the mixed
partial derivatives of the coefficients with respect to xand yare equal. In this case, we have:
∂
∂y (2xy −y2) = 2x−2y
∂
∂x(x2−2y) = 2x
Since the mixed partial derivatives are not equal, the given differential equation is not exact.
3. Suppose we have a differential equation of the form
∂M
∂y −∂N
∂x = 0.
Determine the conditions for exactness of the equation and outline the method for finding a
solution.
Ans. To determine the conditions for exactness and find a solution, we follow the steps below:
1. The given equation is exact if and only if
∂M
∂y =∂N
∂x .
2. Once we have verified that the equation is exact, we can find a potential function Φ(x, y)
such that
∂Φ
∂x =Mand ∂Φ
∂y =N.
3. To find Φ(x, y), integrate Mwith respect to xand check if the result depends only on y.
If it does, let this result be a function of yand denote it as f(y).
4. Then, integrate Nwith respect to yand check if the result depends only on x. If it does,
let this result be a function of xand denote it as g(x).
5. Finally, let
Φ(x, y) = f(y) + g(x) + C,
where Cis a constant of integration. This function Φ(x, y)is a potential function for the
exact equation.
4. Suppose we have the following differential equation:
(x2y−2xy2)dx + (x3−3xy)dy = 0
Determine whether the given differential equation is exact. If it is exact, find the general
solution.
Ans. To determine whether the given differential equation is exact, we need to ensure that the
partial derivatives satisfy the condition for exactness: ∂M
∂y =∂N
∂x .
1. Calculate the partial derivatives:
M=x2y−2xy2,N=x3−3xy
∂M
∂y =x2−4xy and ∂N
∂x = 3x−3y
2. Check if ∂M
∂y =∂N
∂x :
x2−4xy = 3x−3y
Since the condition for exactness is not satisfied, we cannot conclude that the given differential
equation is exact. We will now proceed with finding an integrating factor to make it exact.
3. To find the integrating factor, we will solve the following differential equation:
∂M
∂y −∂N
∂x
N=−4x−3
x3−3x
4. Simplify the above expression and integrate it to find the integrating factor:
Integrating factor =e∫−4x−3
x3−3xdx
5. After finding the integrating factor, multiply it by the differential equation to make it
exact.
6. Solve the exact differential equation using standard techniques to find the general solution.
5. Consider the following differential equation:
(2x+ 3y)dx + (x+ 2y)dy = 0
Determine if the equation is exact and, if so, find its general solution.
Ans. To check if the given differential equation is exact, we need to verify if the partial derivatives
of the coefficients with respect to y and x are equal. 1. Compute the partial derivatives:
∂M
∂y = 3
∂N
∂x = 1
2. Check if the equation is exact: Since ∂M
∂y =∂N
∂x , the equation is not exact. 3. Find an
integrating factor: To make the equation exact, we need to find a function µ(x, y)such that:
µ(2x+ 3y)dx +µ(x+ 2y)dy = 0
Comparing coefficients, we get the following equations:
∂
∂y (µ(x+ 2y)) = ∂
∂x(µ(2x+ 3y))
2µ′(x+ 2y) = 3µ′(2x+ 3y)
4. Solve for µ(x, y): Solving the above equation, we find that µ(x, y) = e3x−2y
6=ex
2−y
3. 5.
Multiply the given differential equation by the integrating factor: Multiplying the given differential
equation by µ(x, y), we get:
e3x
6−2y
6((2x+ 3y)dx + (x+ 2y)dy) = 0
2xex
2−y
3dx + 3ye x
2−y
3dx +xex
2−y
3dy + 2ye x
2−y
3dy = 0
6. Integrate the exact differential equation: By recognizing the left-hand side as the total differ-
ential of a particular function, we have:
d(ex
2−y
3(2x+ 3y)) = 0
Integrating both sides, we get:
ex
2−y
3(2x+ 3y) = C
where C is the constant of integration. 7. General solution: Therefore, the general solution of
the given differential equation is:
ex
2−y
3(2x+ 3y) = C
6. Consider the following differential equation:
(2x+y2)dx + (2xy)dy = 0
Determine whether the equation is exact. If it is exact, find a potential function ϕ(x, y)such
that dϕ = (2x+y2)dx + (2xy)dy. If it is not exact, explain why and determine an integrating
factor to make it exact.
Ans. To determine whether the equation is exact, we check the mixed partial derivatives:
∂
∂y (2x+y2) = 0 and ∂
∂x(2xy) = 2y
Since the two mixed partial derivatives are not equal, the equation is not exact. We can find
the integrating factor µ(x)by dividing the partial derivative of Mwith respect to yby the partial
derivative of Nwith respect to x:
µ(y) = e∫(∂M
∂y
−∂N
∂x
N)dy
=e∫(0−2y
2xy )dy =e−∫1
xdy =e−y
x
Multiplying the given differential equation by the integrating factor µ=e−y
x, we get:
(e−y
x(2x+y2))dx + (e−y
x(2xy))dy = 0
The equation is now exact. Let’s find ϕ.
By integrating the first term with respect to x, we get:
∫(e−y
x(2x+y2))dx =e−y
x(x2+xy2) + h(y)
where h(y)is a function of y.
Taking the partial derivative of this expression with respect to y, we find:
∂
∂y (e−y
x(x2+xy2)+h(y)) = −x
x2e−y
x(x2+xy2)+e−y
x(2xy)+h′(y) = 2xy−xy+h′(y) = h′(y)+xy
Comparing this with the original N, we see that h′(y) = 0. Thus, h(y) = Cfor some
constant C.
Therefore, the potential function ϕ(x, y)is given by:
ϕ(x, y) = e−y
x(x2+xy2) + C
7. Determine whether the following first-order ordinary differential equation is exact, and if so,
find the general solution:
(2xy + 3)dx + (x2+ 1)dy = 0
Ans. To determine whether the given differential equation is exact, we will check if the partial
derivatives of the coefficients with respect to the opposite variable are equal. If they are equal,
we will proceed to find the integrating factor and then find the general solution.
1. Calculate the partial derivatives:
∂
∂y (2xy + 3) = 2xand ∂
∂x(x2+ 1) = 2x
Since the partial derivatives are equal, the given differential equation is exact.
2. Find the integrating factor: The integrating factor, denoted by µ, is given by:
µ(x) = e∫
∂M
∂y
−∂N
∂x
Ndx
In this case:
µ(x) = e∫2x−2x
x2+1 dx =e∫0dx =eC=C(where C is a constant)
3. Multiplying the differential equation by the integrating factor:
C(2xy + 3)dx +C(x2+ 1)dy = 0
4. Integrate to find the general solution: Integrating both sides:
C(x2y+ 3x) + C1(x) = 0
where C1(x)is the constant of integration.
8. Question: Consider the differential equation
(2xy cos(x2y)−x2sin(x2y))dx + (x2cos(x2y)−4ysin(x2y))dy = 0
Determine whether the given equation is exact or not. If it is exact, find a potential function
for the equation. If it is not exact, determine an integrating factor to make it exact.
Ans. To determine if the given equation is exact, we check if the partial derivatives of the
coefficients with respect to yand xare equal. Then, we proceed to find the potential function
or integrating factor accordingly.
1. We calculate the partial derivatives of the coefficients with respect to yand x:
∂
∂y (2xy cos(x2y)−x2sin(x2y)) = 2xcos(x2y)−2x3ysin(x2y)
∂
∂x(x2cos(x2y)−4ysin(x2y)) = 2xcos(x2y)−2x3ysin(x2y)
Since the partial derivatives are equal, the given equation is exact.
2. To find the potential function, we integrate the coefficient of dx with respect to xand
then differentiate the result with respect to y.
Let f(x, y)be the potential function. Integrating the coefficient of dx with respect to x:
f(x, y) = ∫(2xy cos(x2y)−x2sin(x2y))dx =x2ysin(x2y)+g(y)
where g(y)is a function of ythat we will determine by differentiating f(x, y)with respect to
yand comparing it to the coefficient of dy:
∂f
∂y =x2cos(x2y)+g′(y) = x2cos(x2y)−4ysin(x2y)
Comparing coefficients, we find:
g′(y) = −4ysin(x2y)
Integrating both sides with respect to y:
g(y) = 2 cos(x2y)+C
Therefore, the potential function for the given equation is:
f(x, y) = x2ysin(x2y)+ 2 cos(x2y)+C
where Cis a constant.
9. Find the solution to the exact differential equation
(2x2y+y2)dx + (2x3+ 2xy)dy = 0
Ans. To determine if the given differential equation is exact, we will compute the partial
derivatives of Mand Nwith respect to yand xrespectively, and check if they are equal. If they
are equal, we can then find the integrating factor µ(x, y)and solve the differential equation.
1. Determine if the equation is exact: Given differential equation is:
M= 2x2y+y2and N= 2x3+ 2xy
Calculate the partial derivatives of Mand N:
∂M
∂y = 2x2+ 2yand ∂N
∂x = 6x+ 2y
2. Check for exactness: For the equation to be exact, we require:
∂M
∂y =∂N
∂x
Comparing the two partial derivatives, we see that they are not equal. Thus, the given differential
equation is not exact.
3. Find the integrating factor: To find the integrating factor, we calculate:
µ(x) = e∫(∂N
∂x
−∂M
∂y )
Mdx
y
4. Solve the differential equation: Multiply the differential equation by the integrating
factor found in step 3 to make it exact, then solve for the solution.
10. Consider the following differential equation:
(3x2y+ 2y2)dx + (x3+ 4xy)dy = 0
Determine if the given differential equation is exact, and if it is, find the general solution.
Ans. To check if the given differential equation is exact, we can compute the partial derivatives
of M= 3x2y+ 2y2with respect to yand N=x3+ 4xy with respect to x:
1. ∂M
∂y = 3x2+ 4y2. ∂N
∂x = 3x2+ 4y
Since ∂M
∂y =∂N
∂x , the differential equation is exact. To find the general solution, we can follow
these steps:
1. Integrate Mwith respect to xto find a potential function f(x, y).
f(x, y) = ∫(3x2y+ 2y2)dx =x3y+ 2xy2+g(y)
2. Compute ∂f
∂y .
∂f
∂y =x3+ 4xy +g′(y)
3. Set ∂f
∂y =Nto find g(y).
x3+ 4xy +g′(y) = x3+ 4xy =
⇒g′(y) = 0
4. Integrate g(y) = 0 with respect to y.
g(y) = C
5. Substitute f(x, y) = x3y+ 2xy2+Cand solve the equation f(x, y) = Cfor the general
solution.
x3y+ 2xy2+C= 0
Therefore, the general solution to the differential equation is x3y+ 2xy2=−Cwhere Cis a
constant.
11. Question 11: Consider the differential equation
(2x2y+ 3y2)dx +(x3+ 6xy)dy = 0
Determine whether the given differential equation is exact. If it is exact, find a solution to
the equation. If it is not exact, determine 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
the partial derivatives of the terms with respect to yand xare equal. Let’s start by checking
exactness.
1. We check for exactness: The given equation is (2x2y+ 3y2)dx + (x3+ 6xy)dy = 0.
Calculating the partial derivatives:
∂M
∂y = 2x2+ 6y
∂N
∂x = 6y
Since ∂M
∂y =∂N
∂x , the equation is exact.
2. Find the potential function: To find the potential function F(x, y), we integrate the term
with respect to xto get F(x, y):
F(x, y) = ∫(2x2y+ 3y2)dx =x2y2+ 3xy2+g(y)
Now, we differentiate F(x, y)with respect to yand set it equal to the Nterm to find g(y):
∂F
∂y =x2+ 6xy +g′(y) = x3+ 6xy
g′(y) = 0
Thus, g(y) = C, where Cis a constant.
3. General solution: Therefore, the general solution to the differential equation is:
x2y2+ 3xy2+C= 0
12. Find the integrating factor for the following differential equation, if possible, so that it
becomes exact:
(2xy + 4y2)dx + (x2+ 8y)dy = 0
Ans. To determine if the given differential equation is exact, we must check if the partial
derivatives of the coefficients with respect to y and x are equal:
∂M
∂y = 2x+ 8y
∂N
∂x = 2y
Since ∂M
∂y =∂N
∂x , the differential equation is not exact. To make it exact, we introduce an
integrating factor µsuch that µ(x, y)M dx +µ(x, y)Ndy = 0 is exact. The integrating factor is
given by:
µ(x, y) = 1
µ(x, y)M−∂µ
∂y N
Plugging in the values for M, N, and their partial derivatives, we have:
µ(x, y) = 1
2xy + 4y2+ 8y−2x−8y=1
2xy −2x=1
2x(y−1)
Therefore, the integrating factor for the given differential equation is µ(x, y) = 1
2x(y−1) .
13. Consider the following differential equation:
(2x3y2+cos(y))dx + (3x2y+sin(x))dy = 0
Determine if the given differential equation is exact. If it is exact, find the general solution.
If not, determine an integrating factor to make it exact, and then find the solution.
Ans. To determine if the given differential equation is exact, we first check for exactness by
computing the partial derivatives of the terms involving xand y.
1. We check for exactness by computing ∂M
∂y and ∂N
∂x :
∂M
∂y = 4x3y, ∂N
∂x = 6x2y
Since ∂M
∂y =∂N
∂x , the differential equation is not exact. To make it exact, we need to find an
integrating factor µ(x, y)such that
µ(x, y)(2x3y2+cos(y))dx +µ(x, y)(3x2y+sin(x))dy = 0
2. We find the integrating factor µ(x, y):
∂(µM)
∂y =∂(µN)
∂x
2x3y2dµ
dy +µ(4x3y) = 3x2ydµ
dx +µ(6x2y)
3. Equating the coefficients of dx and dy, we get:
2x3y2dµ
dy = 6x2yµ and µ(4x3y) = 3x2ydµ
dx
Solving for µ(x, y)will give us our integrating factor.
4. After finding the integrating factor, the differential equation should become exact. We
can then solve it by finding the potential function, which will give the general solution.
14. Let M= 2xy2+cos(x)and N=x2y−sin(y). Determine whether the differential form
ω=Mdx +Ndy is exact. If it is exact, find a function f(x, y)such that df =ω.
Ans. The given differential form ω=M dx +Ndy is exact because ∂M
∂y = 4xy =∂N
∂x = 2xy.
1. To find f(x, y)such that df =ω, we integrate Mwith respect to xand then differentiate
the result with respect to y(or integrate Nwith respect to yand differentiate with respect to
x).
Integrating M= 2xy2+cos(x)with respect to xgives f(x, y) = x2y2+sin(x) + g(y), where
g(y)is the constant of integration.
2. Next, we differentiate f(x, y)with respect to yand set it equal to N=x2y−sin(y).
∂f
∂y = 2xy2+g′(y)
N=x2y−sin(y)
Matching the yterms on both sides, we get g′(y) = −sin(y)which implies g(y) = cos(y)+C
for some constant C.
Therefore, the function f(x, y) = x2y2+sin(x)+cos(y)+Cwhere Cis an arbitrary constant.
15. Let ω= (3x2+ 2y)dx + (2x+ 3y2)dy. Determine whether the differential form is exact,
and if so, find the potential function F(x, y)such that ω=dF .
Ans. To determine if the given differential form is exact, we check if ∂M
∂y =∂N
∂x , where
ω=Mdx +Ndy.
1. Calculate the partial derivatives: ∂M
∂y = 2 and ∂N
∂x = 2
Since ∂M
∂y =∂N
∂x , the differential form is exact.
2. Find the potential function F(x, y): We integrate Fx=Mto find Fup to a constant:
F(x, y) = ∫(3x2+ 2y)dx =x3+ 2xy +g(y)
3. Find the remaining term of Fusing Fy=N:∂F
∂y =x2+2+g′(y) = 2x+ 3y2From this,
we find g′(y) = 3y2−2Integrate g′(y)to find g(y):g(y) = y3−2y+C
Therefore, the potential function F(x, y) = x3+ 2xy +y3−2y+C
16. Consider the following differential equation:
(2xy2+ 3)dx + (x2y+ 2)dy = 0
Determine if the equation is exact. If it is exact, find its solution.
Ans. To check if the given differential equation is exact, we compute the partial derivatives of
each term with respect to yand x.
1. Partial derivatives: Let M= 2xy2+ 3 and N=x2y+ 2. Calculate ∂M
∂y and ∂N
∂x .
∂M
∂y = 4xy and ∂N
∂x = 2xy
2. Check for exactness: The equation is exact if ∂M
∂y =∂N
∂x . Since 4xy = 2xy, the equation
is not exact.
3. Integrating factor: To make the equation exact, we need to find an integrating factor
µ(x, y)such that multiplying both sides of the equation by µwill make it exact. The integrating
factor µis given by µ=e∫
∂N
∂x
−∂M
∂y
Ndx.
Calculate
∂N
∂x −∂M
∂y
Nand integrate.
∂N
∂x −∂M
∂y
N=2xy −4xy
x2y+ 2 =−2xy
x2y+ 2
4. Solution: Now, we find the integrating factor:
µ=e∫−2xy
x2y+2 dx
To complete the solution, integrate µwith respect to xand then use it to multiply the original
equation in order to find the solution.
17. Consider the following differential equation:
(2x2y2+e2xy)dx + (4xy3+e2xy −1)dy = 0
Determine if the differential equation is exact, and if so, find the general solution.
Ans. To determine if the given equation is exact, we check if the partial derivatives of the
coefficients of dx and dy are equal. If so, we can find a potential function whose total derivative
yields the given equation.
1. Check for exactness: We have the differential equation:
(2x2y2+e2xy)dx + (4xy3+e2xy −1)dy = 0
Calculating the partial derivatives:
∂
∂y (2x2y2+e2xy ) = 4xy + 2xe2xy
∂
∂x(4xy3+e2xy −1) = 4y3+ (2y)e2xy
Since the mixed partial derivatives are equal (2xe2xy = 2ye2xy), the equation is exact.
2. Find potential function: We need to find a function fsuch that:
∂f
∂x = 2x2y2+e2xy
∂f
∂y = 4xy3+e2xy −1
Integrating the first equation with respect to xgives us:
f=∫(2x2y2+e2xy)dx =x2y2+1
2e2xy +h(y)
where h(y)is a constant of integration depending on y.
Now we differentiate fwith respect to y:
∂f
∂y = 4xy3+e2xy +h′(y)
Comparing with the given ∂f
∂y , we get:
h′(y) = −1
h(y) = −y+C
Therefore, the potential function is:
f=x2y2+1
2e2xy −y+C
3. General solution: Since the equation is exact, the general solution is obtained by setting
the potential function fequal to a constant:
x2y2+1
2e2xy −y+C= 0
Thus, the general solution to the given differential equation is:
x2y2+1
2e2xy −y=−C
where Cis a constant.
18. Suppose we have a differential equation given by (y2−x3) + (2xy −3y2)dy = 0.
Determine whether the equation is exact, and if so, find its general solution.
Ans. To check if the given differential equation is exact, we compute the partial derivatives of
the expression F(x, y) = y2−x3with respect to yand ∂(2xy −3y2) /∂x.
1. Compute ∂F /∂y:
∂F
∂y = 2y
2. Compute ∂(2xy −3y2)/∂x:
∂
∂x (2xy −3y2)= 2y
Since ∂F
∂y =∂
∂x (2xy −3y2), the given differential equation is exact.
3. To find the general solution, solve the equation Mdx +Ndy = 0 where M=y2−x3and
N= 2xy −3y2.
4. Find potential function U(x, y)such that dU =Mdx +Ndy: Integrating M=∂U/∂x:
U=∫(y2−x3)dx =−x3y+1
3y3+h(y)
where h(y)is the constant of integration with respect to x.
5. Partial derivative of Uwith respect to y:
∂U
∂y =−x3+y2+h′(y)
6. Equate ∂U
∂y to N= 2xy −3y2to find h′(y):
−x3+y2+h′(y) = 2xy −3y2
h′(y) = 2xy −3y2+x3−y2
h′(y) = 2xy −4y2+x3
7. Integrate h′(y)to find h(y):
h(y) = ∫(2xy −4y2+x3)dy
h(y) = x3y−4
3y3+x3y+c
where cis the constant of integration.
8. Thus, the general solution is given by:
−x3y+1
3y3+x3y−4
3y3+x3y+c= 0
2x3y−y3+c= 0
19. Consider the following differential equation:
(3y2+ 2x)dx + (6xy + 4)dy = 0
Determine whether the given differential equation is exact or not. If it is exact, find a function
u(x, y)such that the given differential equation can be written in the form ∂u
∂x dx +∂u
∂y dy = 0.
Ans. To check if the given differential equation is exact, we need to calculate the mixed partial
derivatives. If ∂
∂y (3y2+ 2x) = ∂
∂x (6xy + 4), then the equation is exact.
1. Check for exactness:
∂
∂y (3y2+ 2x) = 6y, ∂
∂x(6xy + 4) = 6y
Since the mixed partial derivatives are equal, the given differential equation is exact.
2. Find the function u(x, y): The function u(x, y)can be found by integrating the
expression with respect to xand setting it equal to a function of y.
u(x, y) = ∫(3y2+ 2x)dx = 3y2x+x2+h(y)
3. Find ∂u ∂y and compare coefficients to find h(y):
∂u
∂y = 6yx +dh
dy
Comparing coefficients with (6xy + 4), we get:
6x= 6xand dh
dy = 4
Integrating 4with respect to ygives h(y) = 4y+C, where C is a constant.
Therefore, the function u(x, y) = 3y2x+x2+ 4y+Csatisfies the condition for the exact
differential equation.
20. Consider the following differential equation:
(3x2y4+ 4y3)dx +(12x3y3+ 3xy2)dy = 0
Determine whether the given differential equation is exact or not. If it is exact, find the
general solution.
Ans. To check for exactness, we verify if ∂P
∂y =∂Q
∂x . 1. Calculate the partial derivatives:
∂P
∂y = 12x2y3+ 12y2and ∂Q
∂x = 36x2y3+ 3y2
Since ∂P
∂y =∂Q
∂x , the differential equation is not exact. We proceed to find an integrating factor
µ(x, y).
2. To find µ(x, y)such that µP dx +µQdy = 0 becomes exact, we solve ∂(µQ)
∂x =∂(µP )
∂y .
Expanding, we have:
µ(12x3y3+ 3xy2)=∂(µP )
∂y = 12x2y3+ 12y2
µ(12x3y3+ 3xy2)=∂(µQ)
∂x = 36x2y3+ 3y2
Equating the two expressions, we have:
dµ
µ=dx
x
3. Integrating both sides, we find µ(x, y) = xα, where αis a constant. Substituting this into
the differential equation, we have:
3x2+αy4+ 4xαy3dx + 12x3+αy3+ 3xα+1y2dy = 0
4. Matching coefficients of dx and dy, we find that 2 + α= 3 + αand 1 = 0, which is not
possible. Hence, an integrating factor does not exist, and the differential equation is not exact.
One possible way to solve this equation is using Exact Differential Equation methods.
21. Consider the following first-order ordinary differential equation (ODE):
(x3y−3xy + 5)dx + (x2−3y)dy = 0
Determine whether the ODE is exact. If so, find the general solution.
Ans. To determine whether the ODE is exact, we check the partial derivatives of the coefficients
with respect to yin the order Pyand with respect to xin the order Qx, where P(x, y)and Q(x, y)
are the coefficients of dx and dy respectively.
1. Determine exactness:
Let P(x, y) = x3y−3xy + 5 and Q(x, y) = x2−3y. Calculate Py=x3−3xand Qx= 2x.
Since Pydoes not equal Qx, the ODE is not exact.
2. Make exact:
To make the ODE exact, we need to find a function u(x, y)such that multiplying the entire
equation by umakes it exact. We use the integrating factor method. Our goal is to find a
function u(x, y)such that when we multiply each term by u, the equation becomes exact, that
is, it should satisfy the condition ∂
∂y (uP ) = ∂
∂x (uQ).
Thus, we need to solve for uin the following equation:
∂
∂y (uP ) = ∂
∂x(uQ)
3. Find integrating factor:
We have ∂
∂y [(x3y−3xy + 5)u] = ∂
∂x[(x2−3y)u]
This becomes:
x3u+ (x3−3x)uy = 2xu
Letting M(x, y) = x3and N(x, y) = (x3−3x), we get the nonlinear first-order partial
differential equation (PDE):
dy
M=dx
N=du
Q−Mu
4. Solve the PDE:
Solving the PDE, we equate the first two expressions and solve for y(x)in terms of xand a
constant. We then integrate the third expression to find u(x, y).
Finally, we substitute uback into the ODE and solve for yto obtain the general solution.
22. Let M=x2y2and N= 2xy3−1. Determine if the differential form Mdx +N dy is exact,
and if so, find the general solution f(x, y) = Cwhere Cis a constant.
Ans. To determine if the given form is exact, we need to check if ∂M
∂y =∂N
∂x . 1. Calculate the
partial derivatives:
∂M
∂y = 2x2y, ∂N
∂x = 2y3.
2. Check if the derivatives are equal: Since ∂M
∂y =∂N
∂x , the form is not exact. 3. To find an
integrating factor, calculate I(x) = eP(x), where P(x) = ∫∂N
∂x −∂M
∂y
Mdx.
∂N
∂x −∂M
∂y
M=2y3−2x2y
x2y2=2y(y2−x2)
x2y2=2(y2−x2)
x2y.
4. Solve for P(x):
P(x) = ∫2(y2−x2)
x2ydx = 2 ∫(y2
x−x
y)dx = 2y2ln |x| − 2x2
y+h(y).
5. Now, multiply the original form by the integrating factor
I(x)Mdx +I(x)Ndy =(e2y2ln |x|−2x2/y)(x2y2dx + (2xy3−1)dy).
6. Simplify and integrate the equation
e2y2ln |x|−2x2/y(x2y2dx + (2xy3−1)dy) = 0.
This is the general solution.
23. Question 23: Consider the differential equation
(2xy +1
y)dx + (x2−1
y2)dy = 0
1. Determine the conditions for exactness for the given differential equation. 2. If the
equation is exact, find the solution by finding a potential function.
Ans. 1. To check for exactness, we need to verify that
∂M
∂y =∂N
∂x
where M= 2xy +1
yand N=x2−1
y2.
Calculating the partial derivatives:
∂M
∂y = 2x−1
y2
∂N
∂x = 2x
Thus, the equation is exact if and only if
2x−1
y2= 2x
−1
y2= 0
This is false, so the equation is not exact.
2. To find an integrating factor, we can examine the function µ(x, y), which is a function of
xand ythat makes the equation exact when multiplied by both terms. We assume that µ(x, y)
is a function of xonly, i.e., µ(x).
The integrating factor is given by
µ(x) = e∫
∂M
∂y
−∂N
∂x
Ndx
Plugging in the values,
µ(x) = e∫(−1
y2−2x)
x2−1
y2
dx
Solving the integral and simplifying gives
µ(x) = e−2ln |x|
µ(x) = eln(1
x2)
µ(x) = 1
x2
Now, by multiplying the given equation by the integrating factor µ(x) = 1
x2, we can obtain
an exact equation to solve.
24. Question: Consider the differential equation
(1 + 2xy)dx + (x2+y2)dy = 0
Determine whether the given differential equation is exact or not. If it is exact, find the
general solution.
Ans. Step-by-step solution: 1. To determine if the given differential equation is exact, we need
to check if
∂M
∂y =∂N
∂x
where M= 1 + 2xy and N=x2+y2.
2. Calculate ∂M
∂y and ∂N
∂x :
∂M
∂y = 2x
∂N
∂x = 2x
Since ∂M
∂y =∂N
∂x , the given differential equation is exact.
3. To find the potential function f(x, y), we integrate Mwith respect to xand set it equal
to N:
∫(1 + 2xy)dx =x+x2y=f(x, y) + C(y)
Since f(x, y)is a function of xand C(y)is a function of y, the partial derivative with respect
to yof f(x, y)will give the missing term:
∂f
∂y =x2
Integrating this with respect to ygives us the potential function f(x, y):
f(x, y) = 1
3x3+g(y)
4. The general solution of the exact differential equation is given by f(x, y) = C, where C
is the constant of integration. Combining the expressions for f(x, y):
1
3x3+g(y) = C
where g(y)is a constant of integration. Thus, the general solution of the given differential
equation is
1
3x3+g(y) = C
25. Question: Determine whether the following first-order differential equation is exact. If it
is exact, find its solution.
(2xy +y)dx + (x+ 2x2)dy = 0
Ans. Solution: 1. To check for exactness, we need to calculate the partial derivatives of the
terms with respect to yin the equation.
2. Taking the partial derivative of the term with dx with respect to y, we get:
∂
∂y (2xy +y) = 2x+ 1
3. Taking the partial derivative of the term with dy with respect to x, we get:
∂
∂x(x+ 2x2) = 1 + 4x
4. Since the mixed partial derivatives are not equal (2x+ 1 = 1 + 4x), the given differential
equation is not exact.
5. To make it exact, we need to find an integrating factor µ(x, y), such that
µ((2xy +y)dx + (x+ 2x2)dy)= 0
6. We find the integrating factor µ(x, y)by solving the following equation:
∂
∂y (µ(2xy +y)) = ∂
∂x (µ(x+ 2x2))
7. Simplifying the above equation, we have:
2xµy+µ=µ+ 4xµ
8. Solving the above equation, we find that µ(x, y) = e3yis the integrating factor.
9. Multiplying the differential equation by the integrating factor, we get:
e3y(2xy +y)dx +e3y(x+ 2x2)dy = 0
10. Now, we can check whether this new equation is exact. After verifying that it is exact,
we can find the solution by integrating with respect to xand y.
(Alternatively, if the problem requires showing the step-by-step solution, it can be continued
from here.)
26. Let M(x, y) = ey−xand N(x, y) = x2−2xy. Determine if the following differential form
is exact, and if so, find its solution:
(x2−2xy)dx + (ey−x)dy = 0
Ans. To check if the given differential form is exact, we need to verify if the partial derivatives
satisfy the condition:
1. Compute the partial derivative of Mw.r.t. y:∂M
∂y =ey.
2. Compute the partial derivative of Nw.r.t. x:∂N
∂x = 2x−2y.
3. Check if ∂M
∂y =∂N
∂x .
Since ey= 2x−2y, the given differential form is not exact.
To find an integrating factor µ, we divide
∂N
∂x −∂M
∂y
Nand set it equal to a function of xand y:
1
N(∂N
∂x −∂M
∂y ) = 2x−2y−ey
x2−2xy .
This expression can be written as a function of xand ysolely, say f(x, y). In this case,
f(x, y) = 2x−2y−ey
x2−2xy . Now, we can multiply the given differential form by the integrating factor
and then check for exactness again.
27. Question 27: Consider the differential equation
(2xy +cos(x))dx + (x2−sin(y))dy = 0
Determine whether the equation is exact. If it is exact, find the general solution.
Ans. To check for exactness, we will verify if
∂M
∂y =∂N
∂x
where M= 2xy +cos(x)and N=x2−sin(y).
1. Calculate the partial derivatives:
∂M
∂y = 2x
∂N
∂x = 2x
Since ∂M
∂y =∂N
∂x , the equation is exact.
2. Find the potential function F: Integrating Mwith respect to xand Nwith respect to y,
we get
F=∫(2xy +cos(x))dx =x2y+sin(x) + h(y)
F=∫(x2−sin(y))dy =x2y+cos(y) + g(x)
3. Equate the two expressions for Fand solve for h(y)and g(x):
x2y+sin(x) + h(y) = x2y+cos(y) + g(x)
Since the two sides are equal to each other, we have
h(y) = cos(y)
g(x) = sin(x)
4. The general solution is given by F=x2y+sin(x) + cos(y) = C, where Cis a constant.
28. Consider the differential equation given by:
(3x2y−y3)dx + (x3−3xy2)dy = 0
Determine whether the equation is exact or not, and 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 :
M= 3x2y−y3, N =x3−3xy2
∂M
∂y = 3x2−3y2,∂N
∂x = 3x2−3y2
Since ∂M
∂y =∂N
∂x , the equation is exact.
1. To find the potential function F(x, y), we integrate Mwith respect to dx:
F(x, y) = ∫(3x2y−y3)dx =x3y−xy3+g(y)
2. We differentiate F(x, y)partially with respect to yand set it equal to N:
∂F
∂y =x3−3xy2+g′(y) = x3−3xy2
g′(y) = 0 ⇒g(y) = C
3. Therefore, the potential function F(x, y)is:
F(x, y) = x3y−xy3+C
4. The general solution to the exact differential equation is then:
x3y−xy3+C= 0
where Cis an arbitrary constant.
29. Let P(x, y)and Q(x, y)be continuous functions such that the partial derivatives Py=∂P
∂y
and Qx=∂Q
∂x exist.
Consider the first-order ordinary differential equation
P(x, y)dx +Q(x, y)dy = 0.
If the equation is exact, find the general solution.
Ans. To check if the given equation is exact, we must verify if
∂P
∂y =∂Q
∂x .
1. Check exactness: ∂P
∂y =Pyand ∂Q
∂x =Qx.
2. If ∂P
∂y =∂Q
∂x , then the equation is exact. In this case, we can find a function F(x, y)such
that ∂F
∂x =Pand ∂F
∂y =Q.
3. Integrate ∂F
∂x =Pwith respect to x:
F=∫P dx +g(y),
where g(y)is a constant of integration that may depend on y.
4. Differentiate Fwith respect to yand set it equal to Qto solve for g(y):
∂F
∂y =∂
∂y (∫P dx)+dg
dy =Q.
5. Solve for dg
dy :
dg
dy =Q−∂
∂y (∫P dx).
6. Integrate dg
dy with respect to yto find g(y):
g(y) = ∫(Q−∂
∂y (∫P dx)) dy.
7. Substitute g(y)back into Fto find the general solution:
F(x, y) = ∫P dx +∫(Q−∂
∂y (∫P dx)) dy.
Therefore, the general solution to the exact ODE is given by F(x, y) = C, where Cis a
constant.
30. Question 30: Determine whether the following first-order differential equation is exact, and
if so, find its general solution:
(2x+ 4y)dx + (x+ 4)dy = 0
Ans. To determine whether the given differential equation is exact, we will check if the partial
derivatives of the coefficients with respect to yare equal. If the equation is exact, we will then
find its general solution by integrating the terms.
1. Check exactness: Let M= 2x+ 4yand N=x+ 4. Compute the partial derivatives:
∂M
∂y = 4
∂N
∂x = 1
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Make the equation exact: To make the equation exact, we need to find an integrating
factor µ(x)such that:
µ(x)Mdx +µ(x)Ndy = 0
is an exact differential equation.
We can find µ(x)by the formula:
µ(x) = e∫
∂N
∂x
−∂M
∂y
Ndx
Calculating the values, we have:
∂N
∂x −∂M
∂y
N=1−4
x+ 4 =−3
x+ 4
Therefore, the integrating factor µ(x)is µ(x) = e−3ln(x+4) =1
(x+4)3.
3. Multiply the given equation by the integrating factor: Multiply both sides of the given
equation by µ(x):2x+ 4y
(x+ 4)3dx +x+ 4
(x+ 4)3dy = 0
4. Check for exactness and find the general solution: Now that the equation is exact, we can
find the general solution by integrating. The solution will be of the form F(x, y) = C, where C
is a constant.
Integrating the terms with respect to xand yand setting the result equal to F(x, y), we find
the general solution.
General Solution: 1
2(x+ 4)2+y
(x+ 4)3=C
31. Determine whether the following first-order ordinary differential equation is exact:
(2x−y)dx + (x−2y)dy = 0
Ans. To determine if the given 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. Calculate ∂M
∂y and ∂N
∂x :
∂M
∂y =−1and ∂N
∂x = 1
2. Since ∂M
∂y =∂N
∂x , the given differential equation is not exact.
To find an integrating factor µ(x, y), we need to use the formula:
1
µ(∂(µM)
∂y −∂(µN)
∂x )= 0
Expanding this equation gives us:
−1µ+∂µ
∂y −µ+∂µ
∂x = 0
∂µ
∂x +∂µ
∂y −2µ= 0
3. We can solve this partial differential equation by assuming µ=µ(x), which simplifies the
equation to:
dµ
dx −2µ= 0
Solving this first-order linear differential equation gives:
µ=Ae2x
4. Multiply the given differential equation by the integrating factor µ=e2x:
e2x(2x−y)dx +e2x(x−2y)dy = 0
5. We rewrite this equation as the exact differential form:
d(2xe2x−ye2x) = 0
Therefore, the general solution to the given differential equation is:
2xe2x−ye2x=C
where Cis the constant of integration.
32. Question 32: Consider the differential equation
(2x3y−3y2)dx + (x4−6xy)dy = 0
Determine if the given differential equation is exact. If it is exact, find a solution. If it is not
exact, determine an integrating factor that makes it exact, and then find a solution.
Ans. To determine if the given differential equation is exact, we check if
∂M
∂y =∂N
∂x
where M= 2x3y−3y2and N=x4−6xy.
1. Compute the partial derivatives:
∂M
∂y = 2x3−6y
∂N
∂x = 4x3−6y
2. Check if the equation is exact:
∂M
∂y =∂N
∂x if and only if 2x3−6y= 4x3−6y
Since the partial derivatives are equal, the equation is exact.
3. Find the potential function F(x, y)by integrating Mwith respect to x:
F(x, y) = ∫(2x3y−3y2)dx =x4y−3xy2+g(y)
4. To find g(y), differentiate F(x, y)with respect to yand compare with N:
∂F
∂y =x4−6xy +g′(y)
Since ∂F
∂y =N=x4−6xy, we have g′(y) = 0, so g(y)is a constant.
Therefore, the general solution to the differential equation is x4y−3xy2+C= 0, where C
is a constant.
33. Let M= (2xy −y3)dx + (x2−3y2)dy be a differential 1-form. Determine if Mis an exact
form, and if so, find a function fsuch that df =M.
Ans. To determine if Mis exact, we calculate the partial derivatives ∂M
∂y and ∂N
∂x , where
M=P dx and N=Qdy.
1. Calculate ∂M
∂y and ∂N
∂x : Calculating ∂M
∂y :
∂M
∂y = 2x−3y2.
Calculating ∂N
∂x :
∂N
∂x = 2x.
2. Check for exactness by comparing the mixed partial derivatives: Since ∂M
∂y = 2x−3y2and
∂N
∂x = 2x, we have ∂M
∂y =∂N
∂x .
3. Since the mixed partial derivatives are not equal, Mis not an exact form. Thus, there
exists no function fsuch that df =M.
34. Let M(x, y)dx +N(x, y)dy = 0 be a differential equation. Consider the following equation:
(3x2y+ 2xy2)dx + (x3+ 3y2)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:
∂M
∂y = 3x2+ 2x(2y) = 3x2+ 4xy
∂N
∂x = 3x2+ 0 = 3x2
2. Check if ∂M
∂y =∂N
∂x . Since 3x2+ 4xy = 3x2, the differential equation is not exact. 3. To find
an integrating factor, we calculate
∂M
∂y −∂N
∂x
N:
3x2+ 4xy −3x2
x3+ 3y2=4xy
x3+ 3y2
4. Find an integrating factor u(x)as the above expression:
u(x) = e∫4xy
x3+3y2dy
u(x) = e2ln(x2+y3)
u(x) = (x2+y3)2
5. Multiply the differential equation by the integrating factor:
(x2+y3)2((3x2y+ 2xy2)dx + (x3+ 3y2)dy) = 0
6. Rewrite the equation to be exact by expanding and simplifying:
(3x2y(x2+y3)2+ 2xy2(x2+y3)2)dx + (x3(x2+y3)2+ 3y2(x2+y3)2)dy = 0
7. Integrate Mwith respect to xand Nwith respect to yand set the result equal to a constant
value for the general solution. Remember to simplify the expression.
35. Question 35: Consider the differential equation given by (3y−2x2)dx + (4x−6)dy = 0.
Determine whether the equation is exact, and if so, find a function F(x, y)such that ∂F
∂x =
3y−2x2and ∂F
∂y = 4x−6.
Ans. To check for exactness, we compute the mixed partial derivatives:
1. Compute ∂
∂y (3y−2x2):
∂
∂y (3y−2x2) = 3
2. Compute ∂
∂x (4x−6):
∂
∂x(4x−6) = 4
Since the mixed partial derivatives are not equal, the equation is not exact.
To find an integrating factor, we first examine the equation (3y−2x2)dx + (4x−6)dy = 0.
We notice that the coefficient of dx is (3y−2x2)and the coefficient of dy is (4x−6).
3. To find the integrating factor, we compute:
My=∂
∂y (3y−2x2) = 3
Nx=∂
∂x(4x−6) = 4
The integrating factor is given by:
µ(x) = e∫My−Nx
Ndx =e∫3−4
4x−6dx =e−∫1
4x−6dx =e−1
4ln |4x−6|
Simplifying the integrating factor, we get:
µ(x) = 1
|4x−6|1
4
=1
4
√|4x−6|
Thus, the integrating factor is µ(x) = 1
4
√|4x−6|.
