MAT 275 - MODERN DIFFERENTIAL EQUATIONS -
Conditions for exactness and methods for finding solutions
Practice Material - Set 5
1. Consider the differential equation:
(2y+x2)dx + (2x)dy = 0
Determine whether the given differential equation is exact and if so, find the general solution.
Ans. To check if the given differential equation is exact, we need to verify if the partial
derivatives satisfy the condition for exactness:
∂M
∂y =∂N
∂x
1. Find the partial derivatives:
Let M= 2y+x2and N= 2x. Compute the partial derivatives:
∂M
∂y = 2
∂N
∂x = 2
Since ∂M
∂y =∂N
∂x , the differential equation is exact.
2. Find the potential function:
Let f(x, y)be the potential function such that ∂f
∂x = 2y+x2and ∂f
∂y = 2x.
Integrating the first equation with respect to x:
f=∫(2y+x2)dx = 2xy +1
3x3+g(y)
Taking the partial derivative of fwith respect to ygives:
∂f
∂y = 2x+g′(y) = 2x
Solving for g(y):
g′(y) = 0 =⇒g(y) = C
Therefore, the potential function is f(x, y) = 2xy +1
3x3+C.
3. General solution:
The solution to the exact differential equation is given by f(x, y) = C, where Cis a constant.
Therefore, the general solution to the given differential equation is:
2xy +1
3x3=C
2. Let M(x, y)dx +N(x, y)dy = 0 be a differential equation where Mand Nare functions of
xand y. Determine the conditions for exactness of the differential equation and find a solution
if it is exact.
Ans. To determine the conditions for exactness of the given differential equation and find a
solution if it is exact, we follow these steps:
1. Check for exactness: The given differential equation is exact if ∂M
∂y =∂N
∂x .
2. If the equation is exact (i.e., ∂M
∂y =∂N
∂x ), then find a function ψ(x, y)such that dψ =
Mdx +Ndy.
3. Solve the exact differential equation by integrating the differential form dψ =M dx +Ndy
to find the solution ψ(x, y) = C.
Let’s consider an example to illustrate this process.
3. Consider the following differential equation given in the form M(x, y)dx +N(x, y)dy = 0:
(2xy + 3)dx + (x2+ 2y)dy = 0
Determine whether the equation is exact and, if so, find the general solution.
Ans. To determine if the given differential equation is exact, we need to check if ∂M
∂y =∂N
∂x .
1. Check for exactness: Calculate the partial derivatives: ∂M
∂y = 2xand ∂N
∂x = 2x
Since ∂M
∂y =∂N
∂x , the given equation is exact.
2. Find the general solution: To find the general solution of an exact differential equation,
we need to solve the following system of equations:
dy
dx =−Mx
My
=x2+ 2y
2xy + 3
Integrating this expression with respect to xgives:
y=∫x2+ 2y
2xy + 3dx
From here, we can substitute back into the original differential equation to solve for y.
4. Consider the following differential equation:
(2xy +y2)dx + (x2+ 2y)dy = 0
Determine whether the given equation is exact. If it is exact, find the general solution. If it
is not exact, find an integrating factor that makes it exact.
Ans. To determine whether the given equation is exact, we will check the mixed partial
derivatives of the coefficients with respect to yand x.
1. Given differential equation:
(2xy +y2)dx + (x2+ 2y)dy = 0
2. Check if the mixed partial derivatives are equal:
∂
∂y (2xy +y2) = 2x+ 2y
∂
∂x(x2+ 2y) = 0
3. Since the mixed partial derivatives are not equal, the equation is not exact. To make it
exact, we need to find an integrating factor.
4. Let the integrating factor be denoted by µ(x, y). We want the equation to be of the form
µ(2xy +y2)dx +µ(x2+ 2y)dy = 0 where the mixed partial derivatives are equal.
5. To find the integrating factor, solve the following partial differential equation:
∂µ
∂y (2xy +y2) = ∂µ
∂x (x2+ 2y)
6. Solve the partial differential equation to find µ(x, y).
7. After finding µ(x, y), multiply both sides of the original equation by µ(x, y)and check if
the equation becomes exact.
8. If the equation becomes exact, find the general solution by integrating both sides with
respect to xand y.
9. If you encounter any challenges or errors during the process, double-check your calculations
and consider alternative methods to solve the problem.
Therefore, the next steps involve finding the integrating factor and proceeding with the solu-
tion based on the updated equation.
5. Determine whether the following first-order differential equation is exact and, if so, find its
general solution:
(ycos(x) + xsin(y))dx + (cos(x)−xy cos(y))dy = 0
Ans. To check if the given equation is exact, we compute the partial derivatives of the
coefficients with respect to yand xand compare them:
1. Let M=ycos(x) + xsin(y)and N=cos(x)−xy cos(y). 2. Compute ∂M
∂y =cos(x) +
xcos(y)and ∂N
∂x =−sin(x) + ycos(y). 3. Since ∂M
∂y =∂N
∂x , the equation is not exact.
To solve the equation, we need to find an integrating factor µ(x, y)such that µMdx+µNdy =
0is exact. We have the equation in the form:
µ(ycos(x) + xsin(y))dx +µ(cos(x)−xy cos(y))dy = 0
Comparing coefficients, we get the following two equations:
∂
∂y (µM) = ∂
∂x(µN)
yµ sin(y) + µcos(x)−µxy sin(y) = −µsin(x) + yµ cos(y)
This leads to solving the first-order partial differential equation for µ:
µ(cos(x)−xy sin(y)) = µcos(x)−µxy sin(y) = −µsin(x) + yµ cos(y)
This is a challenging problem that involves solving a partial differential equation to find an
integrating factor, which can then be used to find the general solution of the differential equation.
6. Question: Consider the following first-order differential equation: (2x2y+sin(xy))dx + (x2+
2ycos(xy))dy = 0.
Determine whether the given equation is exact, and if so, find the general solution.
Ans. Step-by-step solution: 1. To determine if the given differential equation is exact, we check
if ∂M
∂y =∂N
∂x , where M= 2x2y+sin(xy)and N=x2+ 2ycos(xy). 2. Calculate ∂M
∂y :
∂M
∂y = 2x2+xcos(xy)
3. Calculate ∂N
∂x :
∂N
∂x = 2x+ 2ycos(xy)−2ysin(xy) = 2x+ 2y(cos(xy)−sin(xy))
4. Now check if ∂M
∂y =∂N
∂x :
2x2+xcos(xy) = 2x+ 2y(cos(xy)−sin(xy))
5. There does not seem to be a way to simplify this to show exactness by comparing the two
expressions. Therefore, the given differential equation is not exact. 6. To solve this differential
equation, we need to find an integrating factor µ(x, y)such that µM dx +µN dy = 0 is exact.
7. The integrating factor µis given by:
µ=e∫Nx−My
Ndx =e∫2x−(2x2+xcos(xy))
x2+2ycos(xy)dx
8. Simplify the expression inside the integral and then solve for µ. 9. Once you find the
integrating factor µ, multiply both sides of the differential equation by µand proceed to solve the
resulting exact differential equation using the method of integrating factors. 10. After finding the
general solution, don’t forget to check if there are any implicit solutions that need to be solved
for explicitly.
7. Question 7: Consider the differential equation (ytan x+ex)dx + (xsec2y−ey)dy = 0.
Determine whether the given differential equation is exact. If it is exact, find a function
F(x, y)such that ∂F
∂x =ytan x+exand ∂F
∂y =xsec2y−ey.
Ans. To determine if the given differential equation is exact, we check if ∂M
∂y =∂N
∂x .
1. Calculate ∂M
∂y and ∂N
∂x : Given M=ytan x+exand N=xsec2y−ey,∂M
∂y =tan x
and ∂N
∂x =sec2y.
2. Check if ∂M
∂y =∂N
∂x : Since tan x=sec2y, the given differential equation is not exact.
3. Find an integrating factor µ(x)to make the equation exact: To find the integrating factor,
we calculate µ(x) = e
∫
∂N
∂x −∂M
∂y
N
dx
.
4. Calculate µ(x):
µ(x) = e∫
sec2y−tan x
xsec2y−ey
dx
5. Since µ(x)is quite complicated and no straightforward solution exists, we can stop here
without being able to find an integrating factor.
8. Suppose we have a differential equation of the form
M(x, y)dx +N(x, y)dy = 0.
Determine the conditions for exactness of this equation and provide a systematic method for
finding a solution if the equation is exact.
Ans. To determine the conditions for exactness and the method for finding a solution, follow
these steps:
1. Check for exactness: Compute the partial derivatives ∂M
∂y and ∂N
∂x . If ∂M
∂y =∂N
∂x , then the
equation is exact.
2. If the equation is exact: Since the equation is exact, there exists a function F(x, y)such
that ∂F
∂x =Mand ∂F
∂y =N. Then the general solution is given by F(x, y) = C, where Cis a
constant.
3. Find the solution: Integrate Mwith respect to xand set it equal to F(x, y). Solve this
equation for yto find the solution. The general solution is F(x, y) = C.
9. Consider the following differential equation in standard form:
(cos y+x)exdx + (x2−2ysin y)exdy = 0
Determine whether the given differential equation is exact. If it is exact, find a solution
f(x, y) = C.
Ans. To check for exactness, we calculate the partial derivatives ∂M
∂y and ∂N
∂x , where M=
cos yex+xexand N=−2ysin yex+x2ex.
1. Calculate the partial derivatives:
∂M
∂y =−sin yex+ 0 and ∂N
∂x =−2ysin yex+ 2xex
2. Check for exactness by comparing the mixed partial derivatives:
∂M
∂y =∂N
∂x
Since the partial derivatives are equal, the given differential equation is exact.
3. Find the solution f(x, y)by integrating Mwith respect to x:
f(x, y) = ∫cos yex+xexdx =sin yex+xex+g(y)
4. Differentiate f(x, y)with respect to yand compare it with Nto find g(y):
∂f
∂y =sin yex+g′(y) = −2ysin yex+x2ex
5. Solve for g(y)by comparing coefficients:
g′(y) = x2ex=⇒g(y) = x2yex+C
6. Finally, the solution to the differential equation is f(x, y) = sin yex+xex+x2yex+C= 0,
where Cis the constant of integration.
10. Let Mand Nbe functions of xand y. Consider the first-order differential equation given
by the differential equation
M(x, y) + N(x, y)dy
dx = 0.
Given that the differential equation is exact, show that the solutions satisfy the condition
∂M
∂y =∂N
∂x .
Ans. To show that solutions of an exact differential equation satisfy the condition ∂M
∂y =∂N
∂x ,
we can follow these steps:
1. First, assume that the given differential equation M(x, y) + N(x, y)dy
dx = 0 is exact. This
means that there exists a function f(x, y)such that
df
dx =∂f
∂x +∂f
∂y
dy
dx =Mand df
dy =∂f
∂x
dx
dy +∂f
∂y =N.
2. Differentiate the first equation df
dx =Mwith respect to yand the second equation df
dy =N
with respect to xto obtain
∂M
∂y =∂2f
∂x∂y and ∂N
∂x =∂2f
∂y∂x .
3. By Clairaut’s theorem for mixed partial derivatives, we have ∂2f
∂x∂y =∂2f
∂y∂x , which implies
∂M
∂y =∂N
∂x .
Therefore, solutions of an exact differential equation must satisfy the condition ∂M
∂y =∂N
∂x .
11. Consider the first order differential equation given by:
(2xy +y3)dx + (x2+ 3y2)dy = 0
Ans. To determine whether the given first order differential equation is exact, we need to check
if it satisfies the condition of exactness, ∂M
∂y =∂N
∂x , where Mand Nare the coefficients of dx
and dy respectively.
1. Check for exactness: Let M= 2xy +y3and N=x2+3y2. Find the partial derivatives:
∂M
∂y = 2x+ 3y2
∂N
∂x = 2x
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Finding an integrating factor: To make the equation exact, we find an integrating factor
µ(x, y)such that when multiplied by the entire equation, it becomes exact. The integrating factor
µis given by:
µ=e∫My−Nx
Ndx
Calculating My−Nxand substituting into the formula, we get:
My−Nx= (2x+ 3y2)−2x= 3y2
µ=e∫3y2
x2+3y2dx =e3arctan(y
x)
3. Multiply by integrating factor and rearrange: Multiply the given equation by the
integrating factor µ(x, y):
e3arctan(y
x)[(2xy +y3)dx + (x2+ 3y2)dy] = 0
4. Checking for exactness: Check if the new equation is exact by comparing ∂
∂y of the
coefficient of dx with ∂
∂x of the coefficient of dy.
5. Find the general solution: If the new equation is exact, integrate M dx +N dy = 0 to
find the general solution. If not, other methods such as integrating factors or exact differential
equations techniques may be required.
12. Suppose we have the following first-order ordinary differential equation:
(3y2−x)dx + (6xy −2)dy = 0
Determine whether the equation is exact. If so, find the general solution.
Ans. To determine whether the given equation is exact, we check its partial derivatives with
respect to yfor ∂M
∂y and with respect to xfor ∂N
∂x .
1. Calculate ∂M
∂y :
∂M
∂y =∂(3y2−x)
∂y = 6y
2. Calculate ∂N
∂x :
∂N
∂x =∂(6xy −2)
∂x = 6y
Since ∂M
∂y =∂N
∂x , the equation is exact.
3. To find the general solution, we integrate the equation Mdx +Ndy = 0 with respect to
xto obtain a function f(x, y)such that f(x, y) = c, where cis a constant.
Integrating the equation (3y2−x)dx + (6xy −2)dy = 0 with respect to xgives:
∫(3y2−x)dx +∫(6xy −2)dy = 0
⇒3xy2−x2
2+ϕ(y) = c
4. Next, we differentiate the above result with respect to yand set it equal to the given N
to find ϕ(y).
∂
∂y (3xy2−x2
2+ϕ(y)) = 6xy
⇒6xy +ϕ′(y) = 6xy
⇒ϕ′(y) = 0
⇒ϕ(y) = D
Therefore, the general solution of the given differential equation is:
3xy2−x2
2+D=c
where cand Dare arbitrary constants.
13. Let M= (3xy2−2y)dx + (x2−2xy)dy be a differential form. Determine whether Mis
exact and if so, find a function f(x, y)such that df =M.
Ans. To determine whether the given differential form Mis exact, we will check the equality
of mixed partial derivatives. If Mis exact, we will find a function f(x, y)such that df =M.
1. Check for exactness: Compute the partial derivatives of Mwith respect to yand x.
∂M
∂y = 6xy −2and ∂M
∂x = 3y2−2y.
2. Confirm exactness: We need to check if ∂M
∂y =∂N
∂x , where M=P dx +Qdy. Let
N=−2xdx +x2dy. Check if ∂N
∂x =∂M
∂y .
∂N
∂x =−2and ∂M
∂y = 6xy −2.
Since ∂M
∂y =∂N
∂x , the form Mis not exact.
Hence, the differential form Mis not exact.
14. Consider the following differential equation:
(3x2y−2y3)dx + (x3−6xy2)dy = 0
Determine whether the differential equation above is exact or not. If it is exact, find a solution
to the equation.
Ans. To check if the given differential equation is exact, we must verify if it satisfies the
condition:
∂M
∂y =∂N
∂x
1. Calculate the partial derivatives of M and N:
Let M= 3x2y−2y3and N=x3−6xy2.
We have: ∂M
∂y = 3x2−6y2
∂N
∂x = 3x2−6y2
2. Since ∂M
∂y =∂N
∂x , the given differential equation is exact.
3. To find the solution to the exact equation, we integrate the function Mwith respect to x
and the function Nwith respect to y.
Integrating Mwith respect to xgives:
∫(3x2y−2y3)dx =x3y−2xy3+h(y)
where h(y)is an arbitrary function of y.
Differentiating the above expression with respect to yshould give us N.
4. By differentiating x3y−2xy3+h(y)with respect to y, we find:
d
dy (x3y−2xy3+h(y)) = x3−6xy2
Therefore, h(y)must be a constant term. Let h(y) = C.
5. The general solution to the differential equation is:
x3y−2xy3+C= 0
15. Question 15: Consider the differential equation (3x2y+2 cos(y))dx+(x3+2 sin(y))dy = 0.
Determine whether the given differential equation is exact. If it is exact, find a function
F(x, y)such that ∂F
∂x = 3x2y+ 2 cos(y)and ∂F
∂y =x3+ 2 sin(y).
Ans. To determine if the given differential equation is exact, we check if ∂M
∂y =∂N
∂x , where M
and Nare the coefficients of dx and dy respectively.
1. Calculate ∂M
∂y and ∂N
∂x :
∂M
∂y = 3x2
∂N
∂x = 3x2
Since ∂M
∂y =∂N
∂x , the given differential equation is exact.
2. Find F(x, y)by integrating Mwith respect to xand then differentiating the result with
respect to yto ensure it matches N:
∫(3x2y+ 2 cos(y))dx =x3y+ 2xcos(y) + g(y), where g(y)is a constant of integration with
respect to x.
Now differentiate x3y+ 2xcos(y) + g(y)with respect to y:
∂F
∂y =x3−2sin(y) = x3+ 2 sin(y)
Hence, F(x, y) = x3y+ 2xcos(y) + C, where Cis an arbitrary constant.
16. Question 16: Consider the differential equation
(2x3+y)dx + (x+ 2y)dy = 0.
Determine if the given equation is exact. If it is exact, find a solution. If not, 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 the partial derivatives
of the coefficients with respect to yand xare equal. Then, we proceed to find the solution if the
equation is exact, or find an integrating factor to make it exact.
1. Check for exactness:
Let M= 2x3+yand N=x+ 2y. Calculate the partial derivatives:
∂M
∂y = 1 and ∂N
∂x = 1.
Since ∂M
∂y =∂N
∂x , the differential equation is exact.
2. Find the potential function ψ:
To find the potential function ψ, we integrate Mwith respect to x:
ψ(x, y) = ∫(2x3+y)dx =x4+xy +h(y),
where h(y)is an arbitrary function of y.
Taking the partial derivative of ψwith respect to ygives us:
∂ψ
∂y =x+h′(y).
3. Determine h(y)by comparing with N:
Since ∂ψ
∂y =N, we equate x+h′(y)with x+ 2y:
h′(y) = 2y=⇒h(y) = y2+C,
where Cis a constant of integration.
4. Final solution:
Substitute h(y) = y2+Cback into the potential function ψ:
ψ(x, y) = x4+xy +y2+C.
Therefore, the general solution to the differential equation is:
x4+xy +y2=C.
17. Consider the differential equation
(x2+y)dx + (2y−x)dy = 0
Determine if the given equation is exact, and if so, find the general solution.
Ans. To determine if the given equation is exact, we can check if the partial derivatives of the
terms with respect to yare equal. If they are equal, then the equation is exact.
1. Let’s calculate the partial derivatives:
∂
∂y (x2+y) = 1
∂
∂y (2y−x) = 2
Since the partial derivatives are not equal, the given equation is not exact.
2. To make the equation exact, we need to find an integrating factor. The integrating factor
is given by:
µ(x) = e∫My−Nx
Ndx
where M=x2+yand N= 2y−x.
My−Nx
N=1−(−1)
2y−x=2
2y−x
µ(x) = e∫2
2y−xdx =e2ln |2y−x|=|2y−x|2
µ(x) = (2y−x)2
3. Multiply the original equation by the integrating factor (2y−x)2:
(x2+y)(2y−x)2dx + (2y−x)(2y−x)2dy = 0
(x2+y)(2y−x)2dx −(2y−x)(2y−x)2dy = 0
4. We need to check if the equation is exact now. Calculate the partial derivatives again:
∂
∂y [(x2+y)(2y−x)2] = 2(2y−x)(x2+y)+(x2+y)(4y−2x)
∂
∂x[−(2y−x)(2y−x)2] = −2(2y−x)(2y−x)+(x2+y)(2)
Both partial derivatives are equal, so the equation is exact.
5. To find the general solution, integrate (x2+y)(2y−x)2dx with respect to xand −(2y−
x)(2y−x)2with respect to y. This will result in the general solution to the differential equation.
18. Consider the following differential equation:
(2xy2+ 4y)dx + (x2+ 2xy)dy = 0
Find the conditions for exactness and determine whether the given equation is exact. If it is
exact, find the general solution.
Ans. To determine the conditions for exactness and find the solution to the given differential
equation:
1. Check for exactness: A differential equation of the form Mdx +Ndy = 0 is exact if
∂M
∂y =∂N
∂x .
Given equation: (2xy2+ 4y)dx + (x2+ 2xy)dy = 0 Calculate ∂M
∂y :∂
∂y (2xy2+ 4y) = 4xy + 4
Calculate ∂N
∂x :∂
∂x (x2+ 2xy) = 2y+ 2y= 4ySince ∂M
∂y =∂N
∂x , the equation is exact.
2. Find the potential function: Assume there exists a function f(x, y)such that ∂f
∂x =
2xy2+ 4yand ∂f
∂y =x2+ 2xy.
Integrate the partial derivative with respect to x:f=∫(2xy2+ 4y)dx =x2y2+ 4xy +g(y),
where g(y)is a function of yonly. Differentiate f(ignoring the constant of integration) partially
with respect to y:∂f
∂y =x2+ 2xy +dg
dy Equating this to x2+ 2xy, we get dg
dy = 0, which implies
g(y)is a constant.
Therefore, the potential function f=x2y2+ 4xy +Cwhere Cis a constant.
3. General solution: The general solution to the exact differential equation is obtained by
setting the potential function equal to a constant: x2y2+ 4xy +C= 0
So, the general solution to the given differential equation is x2y2+ 4xy +C= 0, where Cis
a constant.
19. Consider the following differential equation:
(2y2−x)dx + (4xy −2)dy = 0
Determine whether the given differential equation is exact. If it is exact, find the general
solution. If it is not exact, find an integrating factor that makes it exact and then find the general
solution.
Solution
1. Determine Exactness:
Let’s check if the given equation is exact by calculating the partial derivatives of the coeffi-
cients: ∂M
∂y =∂
∂y (2y2−x) = 4y
∂N
∂x =∂
∂x(4xy −2) = 4y
Since ∂M
∂y =∂N
∂x , the given differential equation is exact.
2. Find the General Solution:
To find the general solution, we integrate the coefficient of dx with respect to xand the
coefficient of dy with respect to y.
Integrating the coefficient of dx with respect to xgives us:
∫(2y2−x)dx = 2xy2−1
2x2+h(y)
Now, we differentiate h(y)with respect to yto find the integrating factor v(y):
dh
dy = 4xy +v(y)
v(y) = dh
dy −4xy
v(y) = 4y2−4xy
Multiplying the given differential equation by the integrating factor v(y), we get the exact
equation:
(2y2−x)v(y)dx + (4xy −2)v(y)dy = 0
(2y2v(y)−xv(y))dx + (4xyv(y)−2v(y))dy = 0
Integrating the coefficients of dx and dy will give us the general solution.
Therefore, the general solution to the given differential equation is:
2xy2−1
2x2+h(y) = C
where Cis the constant of integration.
20. Consider the following differential equation:
(2x−y)dx + (x2+ 2y)dy = 0
Determine whether the differential equation is exact. If it is exact, find the general solution.
Ans. To determine if the given equation is exact, we need to check if ∂M
∂y =∂N
∂x , where the
given equation can be written as Mdx +Ndy = 0.
1. Check for exactness: Given M= 2x−yand N=x2+ 2y, let’s find the partial
derivatives: ∂M
∂y =−1
∂N
∂x = 2x
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Integrating factor: To find the integrating factor u(x, y), we solve the following:
∂uM
∂y −∂uN
∂x
u= 0
∂uM
∂y =∂uN
∂x
∂u(2x−y)
∂y =∂u(x2+ 2y)
∂x
Solving the above equations will give the integrating factor.
Once the integrating factor is found, we multiply the given equation by it and check if the
equation becomes exact. If it does, we proceed to find the general solution.
21. Suppose we have a differential equation of the form:
M(x, y)dx +N(x, y)dy = 0
where Mand Nare functions of xand y.
Determine the conditions for exactness of the given differential equation, and provide step-
by-step instructions on how to find a potential function ϕ(x, y)such that dϕ =Mdx +Ndy.
Ans. To check for exactness, we need to verify if:
1. ∂M
∂y =∂N
∂x
If the condition for exactness is satisfied, to find a potential function ϕ(x, y):
2. Integrate Mwith respect to xto get F(x, y)where Fx=M. 3. Compute ∂F
∂y and
compare it with N. 4. If they are equal, then ϕ(x, y) = F(x, y) + C, where Cis an arbitrary
constant.
22. Question: Determine whether the following first-order ordinary differential equation is exact,
and if it is, find a solution:
(x2+ 2xy)dx + (x2+y2)dy = 0
Ans. To check for exactness, we compare the mixed partial derivatives of the coefficients. If
the partial derivatives are equal, the equation is exact. If not, we must use an integrating factor
to make it exact.
1. Check for Exactness:
Let M(x, y) = x2+ 2xy and N(x, y) = x2+y2. We calculate the partial derivatives:
∂M
∂y = 2xand ∂N
∂x = 2x
Since ∂M
∂y =∂N
∂x , the equation is exact.
2. Find Solution:
To find a solution, we need to solve the equation dy
dx =−Mx
Nx, where Mxand Nyare the
partial derivatives of Mand Nwith respect to xand yrespectively.
Given our equation:
dy
dx =−Mx
Ny
=−2x
2x=−1
Integrating both sides with respect to x, we have:
∫dy =−∫dx
y=−x+C
Therefore, the general solution to the given differential equation is y=−x+C, where Cis
a constant.
23. Question: Determine whether the following first-order differential equation is exact, and if
so, find its solution: (3x2+y
x)dx + (2y−x)dy = 0
Ans. To determine if the given differential equation is exact, we check if the partial derivatives
of the coefficients with respect to yand xare equal. Then, if the equation is exact, we can find
the integrating factor and proceed to find the solution.
1. Checking for exactness: Let M= 3x2+y
xand N= 2y−x. The partial derivative of
Mwith respect to yis ∂M
∂y =1
x, and the partial derivative of Nwith respect to xis ∂N
∂x =−1.
Since ∂M
∂y =∂N
∂x , the given differential equation is not exact.
2. Finding the integrating factor: To make the equation exact, we multiply by an integrating
factor µ(x, y):
µ(x, y) = e∫(∂N
∂x
−∂M
∂y
M)dx
In this case, we calculate:
∂N
∂x −∂M
∂y
M=−1−1
x
3x2+y
x
=−x−1
x(3x2+y)
µ(x, y) = e∫−x−1
x(3x2+y)dx
3. Solving the integrating factor: At this point, integrating the above expression can be quite
complex and may involve special functions. Depending on the properties of the given problem,
the integrating factor can be simplified further to make the calculations more manageable.
Since the previous step involves complex calculations, let us assume the integrating factor
µ(x, y)is a certain manageable form (e.g., a function of xor yonly) and proceed to find the
solution using this assumption.
24. Question 24: Consider the differential equation given by (3y2−2x)dx −(2y3−2)dy = 0.
Determine whether the given equation is exact. If it is exact, find the solution. If it is not
exact, determine an integrating factor to make it exact and hence find the solution.
Ans. To check for exactness, we calculate the partial derivatives Myand Nx:
My=∂
∂y (3y2−2x) = 6y, Nx=∂
∂x(−2y3+ 2) = −2.
1. Since My=Nx, the given differential equation is not exact. To make it exact, we can
find an integrating factor µ(x, y)such that:
µ(x, y)(3y2−2x)dx −µ(x, y)(2y3−2)dy = 0.
Comparing with the original equation, we have:
∂
∂y (µ(x, y)(3y2−2x)) = ∂
∂x(−µ(x, y)(2y3−2)),
3y2∂µ
∂y −2µ=−6yµ,
which simplifies to:
3y2dµ
dy −2µ=−6yµ.
2. We can now solve the separable differential equation to find µ(x, y):
3y2dµ
dy = 4µ,
dµ
µ=2
3y2dy,
ln |µ|=2
3(1
y)+C, where Cis a constant,
µ=e2
3(1
y)+C,
µ=e2
3yeC,
µ=eC
y2/3 .
3. Multiplying the entire equation by µ=eC
y2/3 , we get the exact equation:
(3eC
y2/3 y2−2xeC
y2/3 )dx −(−2y3eC
y2/3 −2eC
y2/3 )dy = 0.
4. Integrating the above equation, we find the solution to the original differential equation.
25. Let M= 3x2y−2yand N=x3−2xy + 4. Determine whether the following differential
equation is exact, and if it is, find the solution satisfying the initial condition f(1,2) = 3.
Mdx +Ndy = 0
Ans. To check if the given differential equation is exact, we need to verify if ∂M
∂y =∂N
∂x . If
it is exact, we can find the potential function f(x, y)by integrating Mwith respect to xand
comparing it with the derivative of the potential function. Finally, we use the initial condition to
find the specific solution.
1. Check for exactness: Compute the partial derivatives:
∂M
∂y = 3x2
∂N
∂x = 3x2−2y
Since ∂M
∂y =∂N
∂x , the differential equation is not exact.
2. Make it exact: To make the differential equation exact, we need to find an integrating
factor (x, y)such that M dx +N dy is exact. We can find by solving the following equation:
∂M
∂y −∂N
∂x = 0
3. Solve for the integrating factor: First, rewrite as a function of xand y, i.e., = (x, y).
Then, solve the partial differential equation ∂M
∂y −∂N
∂x = (∂M
∂y −∂N
∂x )to find (x, y).
4. Multiply through by the integrating factor: Once you find the integrating factor ,
multiply both Mand Nby and proceed to solve the exact differential equation.
26. Consider the following differential equation:
(3x2+ 2y)dx + (2x+sin(y))dy = 0
Determine if the equation is exact, and if so, find the general solution.
Ans. To determine if the given differential equation is exact, we need to check if the partial
derivatives of the terms with respect to yare equal. If the equation is exact, we can find the
general solution by integrating in terms of xand yseparately.
1. Check for Exactness:
Calculating the partial derivatives of M= 3x2+ 2ywith respect to yand N= 2x+sin(y)
with respect to x:
∂M
∂y = 2
∂N
∂x = 2
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. The integrating factor µ
is given by:
µ=e∫Nx−My
Mdx
Calculating Nx−My:
Nx−My= 2 −0 = 2
Therefore, the integrating factor is µ=e2x.
3. Find the General Solution:
Multiply the given differential equation by the integrating factor µ=e2x:
e2x(3x2+ 2y)dx +e2x(2x+sin(y))dy = 0
Now we check if the equation is exact:
M=e2x(3x2+ 2y)
N=e2x(2x+sin(y))
Calculating the partial derivatives of Mwith respect to yand Nwith respect to x:
∂M
∂y = 2e2x
∂N
∂x = 2e2x
Since ∂M
∂y =∂N
∂x , the equation is exact.
Integrate Mdx with respect to xand Ndy with respect to yto find the general solution. The
general solution of the given differential equation is:
e2xx3+e2xy=C
where Cis the constant of integration.
Therefore, the general solution of the differential equation is e2xx3+e2xy=C.
27. Find the values of kfor which the following differential equation is exact, and then find the
solution if k= 2:
(2kx + 3y)dx + (3x−ky)dy = 0
Ans. The given differential equation is exact for values of kif the following condition is met:
∂M
∂y =∂N
∂x , where M= 2kx + 3yand N= 3x−ky.
1. Calculate the partial derivatives: ∂M
∂y = 3 and ∂N
∂x = 3
Since ∂M
∂y =∂N
∂x , the differential equation is exact for all values of k.
2. To solve the equation, we need to find a function F(x, y)such that:
∂F
∂x = 2kx + 3y
∂F
∂y = 3x−ky
3. Integrating the first equation with respect to x, we get: F(x, y) = kx2+ 3xy +g(y),
where g(y)is the constant of integration.
4. Now, differentiate F(x, y)with respect to yand equate it to N:
∂F
∂y = 3x+g′(y) = 3x−ky
5. Solve g′(y) = −ky to find g(y):
g(y) = −k
2y2+C
6. Substitute g(y)back into the expression for F(x, y):F(x, y) = kx2+ 3xy −k
2y2+C
7. For k= 2, the solution becomes: F(x, y) = 2x2+ 3xy −y2+C
28. Consider the differential equation ∂M
∂y =∂N
∂x .
Find the integrating factor µ(x, y)for the given differential equation.
Ans. To find the integrating factor µ(x, y)for the given differential equation, we follow these
steps:
1. Write the given differential equation in the form µ(x, y)∂M
∂y −µ(x, y)∂N
∂x = 0.
2. Identify µ(x, y)from the equation above and calculate ∂µ
∂y =
∂M
∂y −∂2µ
∂y∂x
N.
3. Integrate ∂µ
∂y with respect to yto find µ(x, y).
29. Consider the following differential equation:
(3x2y+ 2y2)dx + (x3−4xy)dy = 0
Determine whether the given equation is exact, and if so, find the general solution.
Ans. To determine if the given differential equation is exact, we compute the partial derivatives
of Mand Nwith respect to yand x, respectively:
1. Let M= 3x2y+ 2y2and N=x3−4xy. Then,
∂M
∂y = 3x2+ 4yand ∂N
∂x = 3x2−4y
Since ∂M
∂y =∂N
∂x , the given differential equation is not exact.
2. To make the equation exact, we look for an integrating factor µ(x, y)such that
µ(x, y)M dx +µ(x, y)Ndy = 0
is exact. This means we need to find µ(x, y)such that
∂
∂y (µM) = ∂
∂x(µN)
3. Compute the necessary condition for µ:
∂
∂y (µM) = ∂
∂y (µ(3x2y+ 2y2)) = 3x2µ+ 2yµ
∂
∂x(µN) = ∂
∂x(µ(x3−4xy)) = 3x2µ−4yµ
4. Equating the above two expressions, we get:
3x2µ+ 2yµ = 3x2µ−4yµ ⇒6yµ = 0
Since µcannot be zero, the above equation simplifies to y= 0. Therefore, an integrating
factor is µ(x, y) = e0= 1.
5. We multiply the given differential equation by the integrating factor µ= 1 to make it
exact:
(3x2y+ 2y2)dx + (x3−4xy)dy = 0
6. The modified equation is now exact. We can find the general solution by integrating.
Write M dx +N dy = 0 as an exact differential form, d(1
3x3y+2
3y3) = 0, then the general
solution is: 1
3x3y+2
3y3=C
where Cis the constant of integration.
30. Consider the following first-order ordinary differential equation (ODE) in standard form:
(2xy +cos(y)) dx + (x2−sin(y)) dy = 0
Determine whether the ODE is exact. If it is exact, find the general solution.
Ans. To check for exactness, we compute the partial derivatives of the coefficients with respect
to yand x:
1. Determine exactness
Let M= 2xy +cos(y)and N=x2−sin(y). Compute ∂M
∂y = 2x−sin(y)and ∂N
∂x = 2x.
Therefore, the ODE is not exact since ∂M
∂y =∂N
∂x .
2. Integrating factor
The integrating factor µ(x)can be found as:
µ(x) = e∫1
M(∂N
∂x −∂M
∂y )dx
Substitute the given values of Mand Ninto the integral and solve for µ(x).
µ(x) = e∫1
2xy+cos(y)(2x−sin(y)) dx
3. Multiplying by the integrating factor
Multiply both sides of the ODE by the integrating factor µ(x):
µ(x)(2xy +cos(y)) dx +µ(x)(x2−sin(y)) dy = 0
4. Rewriting the equation
Rewrite the equation in the form d(µ(x)F) = 0, where Fis a function of xand y:
d(µ(x)F) = 0
5. Solve for F
Integrate d(µ(x)F) = 0 to find F(x, y). This can be done by integrating each term separately
with respect to xand y.
6. General solution
Once F(x, y)is found, the general solution to the ODE can be written as:
µ(x)F(x, y) = C
where Cis the constant of integration.
31. Consider the following differential equation:
y′=2x
y
Determine if the equation is exact, and if so, find the general solution. If the equation is not
exact, find an integrating factor that makes it exact.
Ans. To determine if the given differential equation is exact, we check if ∂M
∂y =∂N
∂x .
1. Check for exactness: Let M= 2xand N=2x
y. Then,
∂M
∂y = 0
∂N
∂x =2
y
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Find the integrating factor: To make the equation exact, we find the integrating factor
µ(x)by the equation:
µ(x) = e∫My−Nx
Ndx
Here, My= 0,Nx=−2x
y2. Therefore, the integrating factor is:
µ(x) = e∫−2x
y3dx =e
−x2
y2+C
3. Multiply the equation by the integrating factor: Multiply the given equation by the
integrating factor found:
e
−x2
y2y′−2xe
−x2
y2
y= 0
4. Rewrite the equation in exact form: The equation is now exact and can be written in
the form: d
dx (e
−x2
y2y)= 0
5. Solve the exact equation: Integrating both sides, we get:
e
−x2
y2y=C
which can be rearranged to:
y=Cex2
y2
where Cis the constant of integration.
Therefore, the general solution to the differential equation is y=Cex2
y2.
32. Question 32: Consider the differential equation:
(2x+y2)dx + (2y+xex
y)dy = 0
Determine whether the given differential equation is exact or not. If the equation is exact,
find the general solution.
Ans. To determine if the given differential equation is exact, we check if the partial derivatives
of the coefficients with respect to the opposite variable are equal. If the equation is exact, we
proceed to find the general solution.
1. Check for exactness: Let’s calculate the partial derivatives of the coefficients:
∂M
∂y = 2yand ∂N
∂x = 2 + x
yex
y
Since ∂M
∂y =∂N
∂x , the given differential equation is not exact.
2. Make the Equation Exact: To make the equation exact, we need to find an integrating
factor µ(x, y)such that:
µMdx +µNdy =∂
∂y (µM)dx +∂
∂x(µN)dy
In our case, we need to find µ(x, y)such that:
µ(2x+y2) = ∂(µM)
∂y and µ(2y+xex
y) = ∂(µN )
∂x
Solving these two equations would give us the integrating factor µ(x, y).
3. Find the General Solution: If we were able to find the integrating factor µ(x, y)in the
previous step, then we would multiply the given differential equation by µ(x, y)to make it exact.
After that, we would find the general solution by integrating the exact equation and including an
arbitrary constant.
Since we were not able to find an integrating factor in this case, we cannot proceed with
finding the general solution.
33. Let M(x, y)dx +N(x, y)dy = 0 be a first-order differential equation. Determine conditions
for exactness and methods for finding solutions for the differential equation.
Ans. To determine if the differential equation M(x, y)dx +N(x, y)dy = 0 is exact, we check
if the partial derivatives satisfy the condition ∂M
∂y =∂N
∂x . If the equation is exact, we find the
integrating factor µ(x)by solving the equation
∂M
∂y −∂N
∂x
N=dµ(x)
µ(x)dx . Then, we multiply the equation
by the integrating factor and try to find a function ϕ(x, y)such that ∂ϕ
∂x =M µ and ∂ϕ
∂y =N µ.
Finally, the general solution to the exact differential equation will be given by ϕ(x, y) = C, where
Cis an arbitrary constant.
34. Let Mand Nbe functions of xand ysuch that the differential equation
M(x, y)dx +N(x, y)dy = 0
is exact. Determine conditions for exactness and outline the method for finding a solution if
these conditions are satisfied.
Ans. To determine exactness of the given differential equation, we need to check the equality
of mixed partial derivatives of Mand Nwith respect to xand y. If the equation is exact, we
can use integrating factors to find a solution.
Conditions for exactness: 1. Verify that ∂M
∂y =∂N
∂x .
Method for finding solutions: 1. If the equation is exact: a. Integrate Mwith respect to
xand add a constant function f(y)that only depends on y. b. Take the partial derivative of the
result with respect to y. c. Set the result equal to N. d. Solve the resulting differential equation
for the constant function f(y)to find the solution.
35. Consider the following differential equation:
(3x2+ 2yx)dx + (x2+ 2y)dy = 0
Determine whether the equation is exact or not. If it is exact, find the general solution.
Ans. To determine if the given differential equation is exact, we will check if ∂M
∂y =∂N
∂x .
1. Check for exactness:
Let M= 3x2+ 2yx and N=x2+ 2y. Then,
∂M
∂y = 2x=∂N
∂x = 2x
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Integrate to find integrating factor:
To make the equation exact, we need to find the integrating factor (x, y)such that (x, y)M
and (x, y)Nwill have exact partial derivatives.
Since the integrating factor depends only on xor only on y, we will try the form (x)first:
(x) = e∫P(x)dx
where P(x) =
∂M
∂y −∂N
∂x
N.
Calculating P(x)gives:
P(x) = 2x−2x
x2+ 2y=0
x2+ 2y= 0
So, the integrating factor (x)is 1. This implies we should try the form (y).
3. Find the integrating factor:
Let’s try the form (y):
(y) = e∫Q(y)dy
where Q(y) =
∂N
∂x −∂M
∂y
M.
Calculating Q(y)gives:
Q(y) = 2x−2x
3x2+ 2yx =0
3x2+ 2yx = 0
So the integrating factor (y)is also 1, indicating an error in our calculations.
Conclusion:
Since we were unable to find a suitable integrating factor, the given differential equation
(3x2+ 2yx)dx + (x2+ 2y)dy = 0 is not exact and we need to use other methods for solving it.
2xy +1
3x3=C
2. Let M(x, y)dx +N(x, y)dy = 0 be a differential equation where Mand Nare functions of
xand y. Determine the conditions for exactness of the differential equation and find a solution
if it is exact.
Ans. To determine the conditions for exactness of the given differential equation and find a
solution if it is exact, we follow these steps:
1. Check for exactness: The given differential equation is exact if ∂M
∂y =∂N
∂x .
2. If the equation is exact (i.e., ∂M
∂y =∂N
∂x ), then find a function ψ(x, y)such that dψ =
Mdx +Ndy.
3. Solve the exact differential equation by integrating the differential form dψ =Mdx +Ndy
to find the solution ψ(x, y) = C.
Let’s consider an example to illustrate this process.
3. Consider the following differential equation given in the form M(x, y)dx +N(x, y)dy = 0:
(2xy + 3)dx + (x2+ 2y)dy = 0
Determine whether the equation is exact and, if so, find the general solution.
Ans. To determine if the given differential equation is exact, we need to check if ∂M
∂y =∂N
∂x .
1. Check for exactness: Calculate the partial derivatives: ∂M
∂y = 2xand ∂N
∂x = 2x
Since ∂M
∂y =∂N
∂x , the given equation is exact.
2. Find the general solution: To find the general solution of an exact differential equation,
we need to solve the following system of equations:
dy
dx =−Mx
My
=x2+ 2y
2xy + 3
Integrating this expression with respect to xgives:
y=∫x2+ 2y
2xy + 3dx
From here, we can substitute back into the original differential equation to solve for y.
4. Consider the following differential equation:
(2xy +y2)dx + (x2+ 2y)dy = 0
Determine whether the given equation is exact. If it is exact, find the general solution. If it
is not exact, find an integrating factor that makes it exact.
Ans. To determine whether the given equation is exact, we will check the mixed partial
derivatives of the coefficients with respect to yand x.
1. Given differential equation:
(2xy +y2)dx + (x2+ 2y)dy = 0
2. Check if the mixed partial derivatives are equal:
∂
∂y (2xy +y2) = 2x+ 2y
∂
∂x(x2+ 2y) = 0
3. Since the mixed partial derivatives are not equal, the equation is not exact. To make it
exact, we need to find an integrating factor.
4. Let the integrating factor be denoted by µ(x, y). We want the equation to be of the form
µ(2xy +y2)dx +µ(x2+ 2y)dy = 0 where the mixed partial derivatives are equal.
5. To find the integrating factor, solve the following partial differential equation:
∂µ
∂y (2xy +y2) = ∂µ
∂x (x2+ 2y)
6. Solve the partial differential equation to find µ(x, y).
7. After finding µ(x, y), multiply both sides of the original equation by µ(x, y)and check if
the equation becomes exact.
8. If the equation becomes exact, find the general solution by integrating both sides with
respect to xand y.
9. If you encounter any challenges or errors during the process, double-check your calculations
and consider alternative methods to solve the problem.
Therefore, the next steps involve finding the integrating factor and proceeding with the solu-
tion based on the updated equation.
5. Determine whether the following first-order differential equation is exact and, if so, find its
general solution:
(ycos(x) + xsin(y))dx + (cos(x)−xy cos(y))dy = 0
Ans. To check if the given equation is exact, we compute the partial derivatives of the
coefficients with respect to yand xand compare them:
1. Let M=ycos(x) + xsin(y)and N=cos(x)−xy cos(y). 2. Compute ∂M
∂y =cos(x) +
xcos(y)and ∂N
∂x =−sin(x) + ycos(y). 3. Since ∂M
∂y =∂N
∂x , the equation is not exact.
To solve the equation, we need to find an integrating factor µ(x, y)such that µMdx+µNdy =
0is exact. We have the equation in the form:
µ(ycos(x) + xsin(y))dx +µ(cos(x)−xy cos(y))dy = 0
Comparing coefficients, we get the following two equations:
∂
∂y (µM) = ∂
∂x(µN)
yµ sin(y) + µcos(x)−µxy sin(y) = −µsin(x) + yµ cos(y)
This leads to solving the first-order partial differential equation for µ:
µ(cos(x)−xy sin(y)) = µcos(x)−µxy sin(y) = −µsin(x) + yµ cos(y)
This is a challenging problem that involves solving a partial differential equation to find an
integrating factor, which can then be used to find the general solution of the differential equation.
6. Question: Consider the following first-order differential equation: (2x2y+sin(xy))dx + (x2+
2ycos(xy))dy = 0.
Determine whether the given equation is exact, and if so, find the general solution.
Ans. Step-by-step solution: 1. To determine if the given differential equation is exact, we check
if ∂M
∂y =∂N
∂x , where M= 2x2y+sin(xy)and N=x2+ 2ycos(xy). 2. Calculate ∂M
∂y :
∂M
∂y = 2x2+xcos(xy)
3. Calculate ∂N
∂x :
∂N
∂x = 2x+ 2ycos(xy)−2ysin(xy) = 2x+ 2y(cos(xy)−sin(xy))
4. Now check if ∂M
∂y =∂N
∂x :
2x2+xcos(xy) = 2x+ 2y(cos(xy)−sin(xy))
5. There does not seem to be a way to simplify this to show exactness by comparing the two
expressions. Therefore, the given differential equation is not exact. 6. To solve this differential
equation, we need to find an integrating factor µ(x, y)such that µM dx +µN dy = 0 is exact.
7. The integrating factor µis given by:
µ=e∫Nx−My
Ndx =e∫2x−(2x2+xcos(xy))
x2+2ycos(xy)dx
8. Simplify the expression inside the integral and then solve for µ. 9. Once you find the
integrating factor µ, multiply both sides of the differential equation by µand proceed to solve the
resulting exact differential equation using the method of integrating factors. 10. After finding the
general solution, don’t forget to check if there are any implicit solutions that need to be solved
for explicitly.
7. Question 7: Consider the differential equation (ytan x+ex)dx + (xsec2y−ey)dy = 0.
Determine whether the given differential equation is exact. If it is exact, find a function
F(x, y)such that ∂F
∂x =ytan x+exand ∂F
∂y =xsec2y−ey.
Ans. To determine if the given differential equation is exact, we check if ∂M
∂y =∂N
∂x .
1. Calculate ∂M
∂y and ∂N
∂x : Given M=ytan x+exand N=xsec2y−ey,∂M
∂y =tan x
and ∂N
∂x =sec2y.
2. Check if ∂M
∂y =∂N
∂x : Since tan x=sec2y, the given differential equation is not exact.
3. Find an integrating factor µ(x)to make the equation exact: To find the integrating factor,
we calculate µ(x) = e
∫
∂N
∂x −∂M
∂y
N
dx
.
4. Calculate µ(x):
µ(x) = e∫
sec2y−tan x
xsec2y−ey
dx
5. Since µ(x)is quite complicated and no straightforward solution exists, we can stop here
without being able to find an integrating factor.
8. Suppose we have a differential equation of the form
M(x, y)dx +N(x, y)dy = 0.
Determine the conditions for exactness of this equation and provide a systematic method for
finding a solution if the equation is exact.
Ans. To determine the conditions for exactness and the method for finding a solution, follow
these steps:
1. Check for exactness: Compute the partial derivatives ∂M
∂y and ∂N
∂x . If ∂M
∂y =∂N
∂x , then the
equation is exact.
2. If the equation is exact: Since the equation is exact, there exists a function F(x, y)such
that ∂F
∂x =Mand ∂F
∂y =N. Then the general solution is given by F(x, y) = C, where Cis a
constant.
3. Find the solution: Integrate Mwith respect to xand set it equal to F(x, y). Solve this
equation for yto find the solution. The general solution is F(x, y) = C.
9. Consider the following differential equation in standard form:
(cos y+x)exdx + (x2−2ysin y)exdy = 0
Determine whether the given differential equation is exact. If it is exact, find a solution
f(x, y) = C.
Ans. To check for exactness, we calculate the partial derivatives ∂M
∂y and ∂N
∂x , where M=
cos yex+xexand N=−2ysin yex+x2ex.
1. Calculate the partial derivatives:
∂M
∂y =−sin yex+ 0 and ∂N
∂x =−2ysin yex+ 2xex
2. Check for exactness by comparing the mixed partial derivatives:
∂M
∂y =∂N
∂x
Since the partial derivatives are equal, the given differential equation is exact.
3. Find the solution f(x, y)by integrating Mwith respect to x:
f(x, y) = ∫cos yex+xexdx =sin yex+xex+g(y)
4. Differentiate f(x, y)with respect to yand compare it with Nto find g(y):
∂f
∂y =sin yex+g′(y) = −2ysin yex+x2ex
5. Solve for g(y)by comparing coefficients:
g′(y) = x2ex=⇒g(y) = x2yex+C
6. Finally, the solution to the differential equation is f(x, y) = sin yex+xex+x2yex+C= 0,
where Cis the constant of integration.
10. Let Mand Nbe functions of xand y. Consider the first-order differential equation given
by the differential equation
M(x, y) + N(x, y)dy
dx = 0.
Given that the differential equation is exact, show that the solutions satisfy the condition
∂M
∂y =∂N
∂x .
Ans. To show that solutions of an exact differential equation satisfy the condition ∂M
∂y =∂N
∂x ,
we can follow these steps:
1. First, assume that the given differential equation M(x, y) + N(x, y)dy
dx = 0 is exact. This
means that there exists a function f(x, y)such that
df
dx =∂f
∂x +∂f
∂y
dy
dx =Mand df
dy =∂f
∂x
dx
dy +∂f
∂y =N.
2. Differentiate the first equation df
dx =Mwith respect to yand the second equation df
dy =N
with respect to xto obtain
∂M
∂y =∂2f
∂x∂y and ∂N
∂x =∂2f
∂y∂x .
3. By Clairaut’s theorem for mixed partial derivatives, we have ∂2f
∂x∂y =∂2f
∂y∂x , which implies
∂M
∂y =∂N
∂x .
Therefore, solutions of an exact differential equation must satisfy the condition ∂M
∂y =∂N
∂x .
11. Consider the first order differential equation given by:
(2xy +y3)dx + (x2+ 3y2)dy = 0
Ans. To determine whether the given first order differential equation is exact, we need to check
if it satisfies the condition of exactness, ∂M
∂y =∂N
∂x , where Mand Nare the coefficients of dx
and dy respectively.
1. Check for exactness: Let M= 2xy +y3and N=x2+3y2. Find the partial derivatives:
∂M
∂y = 2x+ 3y2
∂N
∂x = 2x
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Finding an integrating factor: To make the equation exact, we find an integrating factor
µ(x, y)such that when multiplied by the entire equation, it becomes exact. The integrating factor
µis given by:
µ=e∫My−Nx
Ndx
Calculating My−Nxand substituting into the formula, we get:
My−Nx= (2x+ 3y2)−2x= 3y2
µ=e∫3y2
x2+3y2dx =e3arctan(y
x)
3. Multiply by integrating factor and rearrange: Multiply the given equation by the
integrating factor µ(x, y):
e3arctan(y
x)[(2xy +y3)dx + (x2+ 3y2)dy] = 0
4. Checking for exactness: Check if the new equation is exact by comparing ∂
∂y of the
coefficient of dx with ∂
∂x of the coefficient of dy.
5. Find the general solution: If the new equation is exact, integrate M dx +N dy = 0 to
find the general solution. If not, other methods such as integrating factors or exact differential
equations techniques may be required.
12. Suppose we have the following first-order ordinary differential equation:
(3y2−x)dx + (6xy −2)dy = 0
Determine whether the equation is exact. If so, find the general solution.
Ans. To determine whether the given equation is exact, we check its partial derivatives with
respect to yfor ∂M
∂y and with respect to xfor ∂N
∂x .
1. Calculate ∂M
∂y :
∂M
∂y =∂(3y2−x)
∂y = 6y
2. Calculate ∂N
∂x :
∂N
∂x =∂(6xy −2)
∂x = 6y
Since ∂M
∂y =∂N
∂x , the equation is exact.
3. To find the general solution, we integrate the equation Mdx +Ndy = 0 with respect to
xto obtain a function f(x, y)such that f(x, y) = c, where cis a constant.
Integrating the equation (3y2−x)dx + (6xy −2)dy = 0 with respect to xgives:
∫(3y2−x)dx +∫(6xy −2)dy = 0
⇒3xy2−x2
2+ϕ(y) = c
4. Next, we differentiate the above result with respect to yand set it equal to the given N
to find ϕ(y).
∂
∂y (3xy2−x2
2+ϕ(y)) = 6xy
⇒6xy +ϕ′(y) = 6xy
⇒ϕ′(y) = 0
⇒ϕ(y) = D
Therefore, the general solution of the given differential equation is:
3xy2−x2
2+D=c
where cand Dare arbitrary constants.
13. Let M= (3xy2−2y)dx + (x2−2xy)dy be a differential form. Determine whether Mis
exact and if so, find a function f(x, y)such that df =M.
Ans. To determine whether the given differential form Mis exact, we will check the equality
of mixed partial derivatives. If Mis exact, we will find a function f(x, y)such that df =M.
1. Check for exactness: Compute the partial derivatives of Mwith respect to yand x.
∂M
∂y = 6xy −2and ∂M
∂x = 3y2−2y.
2. Confirm exactness: We need to check if ∂M
∂y =∂N
∂x , where M=P dx +Qdy. Let
N=−2xdx +x2dy. Check if ∂N
∂x =∂M
∂y .
∂N
∂x =−2and ∂M
∂y = 6xy −2.
Since ∂M
∂y =∂N
∂x , the form Mis not exact.
Hence, the differential form Mis not exact.
14. Consider the following differential equation:
(3x2y−2y3)dx + (x3−6xy2)dy = 0
Determine whether the differential equation above is exact or not. If it is exact, find a solution
to the equation.
Ans. To check if the given differential equation is exact, we must verify if it satisfies the
condition:
∂M
∂y =∂N
∂x
1. Calculate the partial derivatives of M and N:
Let M= 3x2y−2y3and N=x3−6xy2.
We have: ∂M
∂y = 3x2−6y2
∂N
∂x = 3x2−6y2
2. Since ∂M
∂y =∂N
∂x , the given differential equation is exact.
3. To find the solution to the exact equation, we integrate the function Mwith respect to x
and the function Nwith respect to y.
Integrating Mwith respect to xgives:
∫(3x2y−2y3)dx =x3y−2xy3+h(y)
where h(y)is an arbitrary function of y.
Differentiating the above expression with respect to yshould give us N.
4. By differentiating x3y−2xy3+h(y)with respect to y, we find:
d
dy (x3y−2xy3+h(y)) = x3−6xy2
Therefore, h(y)must be a constant term. Let h(y) = C.
5. The general solution to the differential equation is:
x3y−2xy3+C= 0
15. Question 15: Consider the differential equation (3x2y+2 cos(y))dx+(x3+2 sin(y))dy = 0.
Determine whether the given differential equation is exact. If it is exact, find a function
F(x, y)such that ∂F
∂x = 3x2y+ 2 cos(y)and ∂F
∂y =x3+ 2 sin(y).
Ans. To determine if the given differential equation is exact, we check if ∂M
∂y =∂N
∂x , where M
and Nare the coefficients of dx and dy respectively.
1. Calculate ∂M
∂y and ∂N
∂x :
∂M
∂y = 3x2
∂N
∂x = 3x2
Since ∂M
∂y =∂N
∂x , the given differential equation is exact.
2. Find F(x, y)by integrating Mwith respect to xand then differentiating the result with
respect to yto ensure it matches N:
∫(3x2y+ 2 cos(y))dx =x3y+ 2xcos(y) + g(y), where g(y)is a constant of integration with
respect to x.
Now differentiate x3y+ 2xcos(y) + g(y)with respect to y:
∂F
∂y =x3−2sin(y) = x3+ 2 sin(y)
Hence, F(x, y) = x3y+ 2xcos(y) + C, where Cis an arbitrary constant.
16. Question 16: Consider the differential equation
(2x3+y)dx + (x+ 2y)dy = 0.
Determine if the given equation is exact. If it is exact, find a solution. If not, 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 the partial derivatives
of the coefficients with respect to yand xare equal. Then, we proceed to find the solution if the
equation is exact, or find an integrating factor to make it exact.
1. Check for exactness:
Let M= 2x3+yand N=x+ 2y. Calculate the partial derivatives:
∂M
∂y = 1 and ∂N
∂x = 1.
Since ∂M
∂y =∂N
∂x , the differential equation is exact.
2. Find the potential function ψ:
To find the potential function ψ, we integrate Mwith respect to x:
ψ(x, y) = ∫(2x3+y)dx =x4+xy +h(y),
where h(y)is an arbitrary function of y.
Taking the partial derivative of ψwith respect to ygives us:
∂ψ
∂y =x+h′(y).
3. Determine h(y)by comparing with N:
Since ∂ψ
∂y =N, we equate x+h′(y)with x+ 2y:
h′(y) = 2y=⇒h(y) = y2+C,
where Cis a constant of integration.
4. Final solution:
Substitute h(y) = y2+Cback into the potential function ψ:
ψ(x, y) = x4+xy +y2+C.
Therefore, the general solution to the differential equation is:
x4+xy +y2=C.
17. Consider the differential equation
(x2+y)dx + (2y−x)dy = 0
Determine if the given equation is exact, and if so, find the general solution.
Ans. To determine if the given equation is exact, we can check if the partial derivatives of the
terms with respect to yare equal. If they are equal, then the equation is exact.
1. Let’s calculate the partial derivatives:
∂
∂y (x2+y) = 1
∂
∂y (2y−x) = 2
Since the partial derivatives are not equal, the given equation is not exact.
2. To make the equation exact, we need to find an integrating factor. The integrating factor
is given by:
µ(x) = e∫My−Nx
Ndx
where M=x2+yand N= 2y−x.
My−Nx
N=1−(−1)
2y−x=2
2y−x
µ(x) = e∫2
2y−xdx =e2ln |2y−x|=|2y−x|2
µ(x) = (2y−x)2
3. Multiply the original equation by the integrating factor (2y−x)2:
(x2+y)(2y−x)2dx + (2y−x)(2y−x)2dy = 0
(x2+y)(2y−x)2dx −(2y−x)(2y−x)2dy = 0
4. We need to check if the equation is exact now. Calculate the partial derivatives again:
∂
∂y [(x2+y)(2y−x)2] = 2(2y−x)(x2+y)+(x2+y)(4y−2x)
∂
∂x[−(2y−x)(2y−x)2] = −2(2y−x)(2y−x)+(x2+y)(2)
Both partial derivatives are equal, so the equation is exact.
5. To find the general solution, integrate (x2+y)(2y−x)2dx with respect to xand −(2y−
x)(2y−x)2with respect to y. This will result in the general solution to the differential equation.
18. Consider the following differential equation:
(2xy2+ 4y)dx + (x2+ 2xy)dy = 0
Find the conditions for exactness and determine whether the given equation is exact. If it is
exact, find the general solution.
Ans. To determine the conditions for exactness and find the solution to the given differential
equation:
1. Check for exactness: A differential equation of the form Mdx +Ndy = 0 is exact if
∂M
∂y =∂N
∂x .
Given equation: (2xy2+ 4y)dx + (x2+ 2xy)dy = 0 Calculate ∂M
∂y :∂
∂y (2xy2+ 4y) = 4xy + 4
Calculate ∂N
∂x :∂
∂x (x2+ 2xy) = 2y+ 2y= 4ySince ∂M
∂y =∂N
∂x , the equation is exact.
2. Find the potential function: Assume there exists a function f(x, y)such that ∂f
∂x =
2xy2+ 4yand ∂f
∂y =x2+ 2xy.
Integrate the partial derivative with respect to x:f=∫(2xy2+ 4y)dx =x2y2+ 4xy +g(y),
where g(y)is a function of yonly. Differentiate f(ignoring the constant of integration) partially
with respect to y:∂f
∂y =x2+ 2xy +dg
dy Equating this to x2+ 2xy, we get dg
dy = 0, which implies
g(y)is a constant.
Therefore, the potential function f=x2y2+ 4xy +Cwhere Cis a constant.
3. General solution: The general solution to the exact differential equation is obtained by
setting the potential function equal to a constant: x2y2+ 4xy +C= 0
So, the general solution to the given differential equation is x2y2+ 4xy +C= 0, where Cis
a constant.
19. Consider the following differential equation:
(2y2−x)dx + (4xy −2)dy = 0
Determine whether the given differential equation is exact. If it is exact, find the general
solution. If it is not exact, find an integrating factor that makes it exact and then find the general
solution.
Solution
1. Determine Exactness:
Let’s check if the given equation is exact by calculating the partial derivatives of the coeffi-
cients: ∂M
∂y =∂
∂y (2y2−x) = 4y
∂N
∂x =∂
∂x(4xy −2) = 4y
Since ∂M
∂y =∂N
∂x , the given differential equation is exact.
2. Find the General Solution:
To find the general solution, we integrate the coefficient of dx with respect to xand the
coefficient of dy with respect to y.
Integrating the coefficient of dx with respect to xgives us:
∫(2y2−x)dx = 2xy2−1
2x2+h(y)
Now, we differentiate h(y)with respect to yto find the integrating factor v(y):
dh
dy = 4xy +v(y)
v(y) = dh
dy −4xy
v(y) = 4y2−4xy
Multiplying the given differential equation by the integrating factor v(y), we get the exact
equation:
(2y2−x)v(y)dx + (4xy −2)v(y)dy = 0
(2y2v(y)−xv(y))dx + (4xyv(y)−2v(y))dy = 0
Integrating the coefficients of dx and dy will give us the general solution.
Therefore, the general solution to the given differential equation is:
2xy2−1
2x2+h(y) = C
where Cis the constant of integration.
20. Consider the following differential equation:
(2x−y)dx + (x2+ 2y)dy = 0
Determine whether the differential equation is exact. If it is exact, find the general solution.
Ans. To determine if the given equation is exact, we need to check if ∂M
∂y =∂N
∂x , where the
given equation can be written as Mdx +Ndy = 0.
1. Check for exactness: Given M= 2x−yand N=x2+ 2y, let’s find the partial
derivatives: ∂M
∂y =−1
∂N
∂x = 2x
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Integrating factor: To find the integrating factor u(x, y), we solve the following:
∂uM
∂y −∂uN
∂x
u= 0
∂uM
∂y =∂uN
∂x
∂u(2x−y)
∂y =∂u(x2+ 2y)
∂x
Solving the above equations will give the integrating factor.
Once the integrating factor is found, we multiply the given equation by it and check if the
equation becomes exact. If it does, we proceed to find the general solution.
21. Suppose we have a differential equation of the form:
M(x, y)dx +N(x, y)dy = 0
where Mand Nare functions of xand y.
Determine the conditions for exactness of the given differential equation, and provide step-
by-step instructions on how to find a potential function ϕ(x, y)such that dϕ =Mdx +Ndy.
Ans. To check for exactness, we need to verify if:
1. ∂M
∂y =∂N
∂x
If the condition for exactness is satisfied, to find a potential function ϕ(x, y):
2. Integrate Mwith respect to xto get F(x, y)where Fx=M. 3. Compute ∂F
∂y and
compare it with N. 4. If they are equal, then ϕ(x, y) = F(x, y) + C, where Cis an arbitrary
constant.
22. Question: Determine whether the following first-order ordinary differential equation is exact,
and if it is, find a solution:
(x2+ 2xy)dx + (x2+y2)dy = 0
Ans. To check for exactness, we compare the mixed partial derivatives of the coefficients. If
the partial derivatives are equal, the equation is exact. If not, we must use an integrating factor
to make it exact.
1. Check for Exactness:
Let M(x, y) = x2+ 2xy and N(x, y) = x2+y2. We calculate the partial derivatives:
∂M
∂y = 2xand ∂N
∂x = 2x
Since ∂M
∂y =∂N
∂x , the equation is exact.
2. Find Solution:
To find a solution, we need to solve the equation dy
dx =−Mx
Nx, where Mxand Nyare the
partial derivatives of Mand Nwith respect to xand yrespectively.
Given our equation:
dy
dx =−Mx
Ny
=−2x
2x=−1
Integrating both sides with respect to x, we have:
∫dy =−∫dx
y=−x+C
Therefore, the general solution to the given differential equation is y=−x+C, where Cis
a constant.
23. Question: Determine whether the following first-order differential equation is exact, and if
so, find its solution: (3x2+y
x)dx + (2y−x)dy = 0
Ans. To determine if the given differential equation is exact, we check if the partial derivatives
of the coefficients with respect to yand xare equal. Then, if the equation is exact, we can find
the integrating factor and proceed to find the solution.
1. Checking for exactness: Let M= 3x2+y
xand N= 2y−x. The partial derivative of
Mwith respect to yis ∂M
∂y =1
x, and the partial derivative of Nwith respect to xis ∂N
∂x =−1.
Since ∂M
∂y =∂N
∂x , the given differential equation is not exact.
2. Finding the integrating factor: To make the equation exact, we multiply by an integrating
factor µ(x, y):
µ(x, y) = e∫(∂N
∂x
−∂M
∂y
M)dx
In this case, we calculate:
∂N
∂x −∂M
∂y
M=−1−1
x
3x2+y
x
=−x−1
x(3x2+y)
µ(x, y) = e∫−x−1
x(3x2+y)dx
3. Solving the integrating factor: At this point, integrating the above expression can be quite
complex and may involve special functions. Depending on the properties of the given problem,
the integrating factor can be simplified further to make the calculations more manageable.
Since the previous step involves complex calculations, let us assume the integrating factor
µ(x, y)is a certain manageable form (e.g., a function of xor yonly) and proceed to find the
solution using this assumption.
24. Question 24: Consider the differential equation given by (3y2−2x)dx −(2y3−2)dy = 0.
Determine whether the given equation is exact. If it is exact, find the solution. If it is not
exact, determine an integrating factor to make it exact and hence find the solution.
Ans. To check for exactness, we calculate the partial derivatives Myand Nx:
My=∂
∂y (3y2−2x) = 6y, Nx=∂
∂x(−2y3+ 2) = −2.
1. Since My=Nx, the given differential equation is not exact. To make it exact, we can
find an integrating factor µ(x, y)such that:
µ(x, y)(3y2−2x)dx −µ(x, y)(2y3−2)dy = 0.
Comparing with the original equation, we have:
∂
∂y (µ(x, y)(3y2−2x)) = ∂
∂x(−µ(x, y)(2y3−2)),
3y2∂µ
∂y −2µ=−6yµ,
which simplifies to:
3y2dµ
dy −2µ=−6yµ.
2. We can now solve the separable differential equation to find µ(x, y):
3y2dµ
dy = 4µ,
dµ
µ=2
3y2dy,
ln |µ|=2
3(1
y)+C, where Cis a constant,
µ=e2
3(1
y)+C,
µ=e2
3yeC,
µ=eC
y2/3 .
3. Multiplying the entire equation by µ=eC
y2/3 , we get the exact equation:
(3eC
y2/3 y2−2xeC
y2/3 )dx −(−2y3eC
y2/3 −2eC
y2/3 )dy = 0.
4. Integrating the above equation, we find the solution to the original differential equation.
25. Let M= 3x2y−2yand N=x3−2xy + 4. Determine whether the following differential
equation is exact, and if it is, find the solution satisfying the initial condition f(1,2) = 3.
Mdx +Ndy = 0
Ans. To check if the given differential equation is exact, we need to verify if ∂M
∂y =∂N
∂x . If
it is exact, we can find the potential function f(x, y)by integrating Mwith respect to xand
comparing it with the derivative of the potential function. Finally, we use the initial condition to
find the specific solution.
1. Check for exactness: Compute the partial derivatives:
∂M
∂y = 3x2
∂N
∂x = 3x2−2y
Since ∂M
∂y =∂N
∂x , the differential equation is not exact.
2. Make it exact: To make the differential equation exact, we need to find an integrating
factor (x, y)such that M dx +N dy is exact. We can find by solving the following equation:
∂M
∂y −∂N
∂x = 0
3. Solve for the integrating factor: First, rewrite as a function of xand y, i.e., = (x, y).
Then, solve the partial differential equation ∂M
∂y −∂N
∂x = (∂M
∂y −∂N
∂x )to find (x, y).
4. Multiply through by the integrating factor: Once you find the integrating factor ,
multiply both Mand Nby and proceed to solve the exact differential equation.
26. Consider the following differential equation:
(3x2+ 2y)dx + (2x+sin(y))dy = 0
Determine if the equation is exact, and if so, find the general solution.
Ans. To determine if the given differential equation is exact, we need to check if the partial
derivatives of the terms with respect to yare equal. If the equation is exact, we can find the
general solution by integrating in terms of xand yseparately.
1. Check for Exactness:
Calculating the partial derivatives of M= 3x2+ 2ywith respect to yand N= 2x+sin(y)
with respect to x:
∂M
∂y = 2
∂N
∂x = 2
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. The integrating factor µ
is given by:
µ=e∫Nx−My
Mdx
Calculating Nx−My:
Nx−My= 2 −0 = 2
Therefore, the integrating factor is µ=e2x.
3. Find the General Solution:
Multiply the given differential equation by the integrating factor µ=e2x:
e2x(3x2+ 2y)dx +e2x(2x+sin(y))dy = 0
Now we check if the equation is exact:
M=e2x(3x2+ 2y)
N=e2x(2x+sin(y))
Calculating the partial derivatives of Mwith respect to yand Nwith respect to x:
∂M
∂y = 2e2x
∂N
∂x = 2e2x
Since ∂M
∂y =∂N
∂x , the equation is exact.
Integrate Mdx with respect to xand Ndy with respect to yto find the general solution. The
general solution of the given differential equation is:
e2xx3+e2xy=C
where Cis the constant of integration.
Therefore, the general solution of the differential equation is e2xx3+e2xy=C.
27. Find the values of kfor which the following differential equation is exact, and then find the
solution if k= 2:
(2kx + 3y)dx + (3x−ky)dy = 0
Ans. The given differential equation is exact for values of kif the following condition is met:
∂M
∂y =∂N
∂x , where M= 2kx + 3yand N= 3x−ky.
1. Calculate the partial derivatives: ∂M
∂y = 3 and ∂N
∂x = 3
Since ∂M
∂y =∂N
∂x , the differential equation is exact for all values of k.
2. To solve the equation, we need to find a function F(x, y)such that:
∂F
∂x = 2kx + 3y
∂F
∂y = 3x−ky
3. Integrating the first equation with respect to x, we get: F(x, y) = kx2+ 3xy +g(y),
where g(y)is the constant of integration.
4. Now, differentiate F(x, y)with respect to yand equate it to N:
∂F
∂y = 3x+g′(y) = 3x−ky
5. Solve g′(y) = −ky to find g(y):
g(y) = −k
2y2+C
6. Substitute g(y)back into the expression for F(x, y):F(x, y) = kx2+ 3xy −k
2y2+C
7. For k= 2, the solution becomes: F(x, y) = 2x2+ 3xy −y2+C
28. Consider the differential equation ∂M
∂y =∂N
∂x .
Find the integrating factor µ(x, y)for the given differential equation.
Ans. To find the integrating factor µ(x, y)for the given differential equation, we follow these
steps:
1. Write the given differential equation in the form µ(x, y)∂M
∂y −µ(x, y)∂N
∂x = 0.
2. Identify µ(x, y)from the equation above and calculate ∂µ
∂y =
∂M
∂y −∂2µ
∂y∂x
N.
3. Integrate ∂µ
∂y with respect to yto find µ(x, y).
29. Consider the following differential equation:
(3x2y+ 2y2)dx + (x3−4xy)dy = 0
Determine whether the given equation is exact, and if so, find the general solution.
Ans. To determine if the given differential equation is exact, we compute the partial derivatives
of Mand Nwith respect to yand x, respectively:
1. Let M= 3x2y+ 2y2and N=x3−4xy. Then,
∂M
∂y = 3x2+ 4yand ∂N
∂x = 3x2−4y
Since ∂M
∂y =∂N
∂x , the given differential equation is not exact.
2. To make the equation exact, we look for an integrating factor µ(x, y)such that
µ(x, y)M dx +µ(x, y)Ndy = 0
is exact. This means we need to find µ(x, y)such that
∂
∂y (µM) = ∂
∂x(µN)
3. Compute the necessary condition for µ:
∂
∂y (µM) = ∂
∂y (µ(3x2y+ 2y2)) = 3x2µ+ 2yµ
∂
∂x(µN) = ∂
∂x(µ(x3−4xy)) = 3x2µ−4yµ
4. Equating the above two expressions, we get:
3x2µ+ 2yµ = 3x2µ−4yµ ⇒6yµ = 0
Since µcannot be zero, the above equation simplifies to y= 0. Therefore, an integrating
factor is µ(x, y) = e0= 1.
5. We multiply the given differential equation by the integrating factor µ= 1 to make it
exact:
(3x2y+ 2y2)dx + (x3−4xy)dy = 0
6. The modified equation is now exact. We can find the general solution by integrating.
Write M dx +N dy = 0 as an exact differential form, d(1
3x3y+2
3y3) = 0, then the general
solution is: 1
3x3y+2
3y3=C
where Cis the constant of integration.
30. Consider the following first-order ordinary differential equation (ODE) in standard form:
(2xy +cos(y)) dx + (x2−sin(y)) dy = 0
Determine whether the ODE is exact. If it is exact, find the general solution.
Ans. To check for exactness, we compute the partial derivatives of the coefficients with respect
to yand x:
1. Determine exactness
Let M= 2xy +cos(y)and N=x2−sin(y). Compute ∂M
∂y = 2x−sin(y)and ∂N
∂x = 2x.
Therefore, the ODE is not exact since ∂M
∂y =∂N
∂x .
2. Integrating factor
The integrating factor µ(x)can be found as:
µ(x) = e∫1
M(∂N
∂x −∂M
∂y )dx
Substitute the given values of Mand Ninto the integral and solve for µ(x).
µ(x) = e∫1
2xy+cos(y)(2x−sin(y)) dx
3. Multiplying by the integrating factor
Multiply both sides of the ODE by the integrating factor µ(x):
µ(x)(2xy +cos(y)) dx +µ(x)(x2−sin(y)) dy = 0
4. Rewriting the equation
Rewrite the equation in the form d(µ(x)F) = 0, where Fis a function of xand y:
d(µ(x)F) = 0
5. Solve for F
Integrate d(µ(x)F) = 0 to find F(x, y). This can be done by integrating each term separately
with respect to xand y.
6. General solution
Once F(x, y)is found, the general solution to the ODE can be written as:
µ(x)F(x, y) = C
where Cis the constant of integration.
31. Consider the following differential equation:
y′=2x
y
Determine if the equation is exact, and if so, find the general solution. If the equation is not
exact, find an integrating factor that makes it exact.
Ans. To determine if the given differential equation is exact, we check if ∂M
∂y =∂N
∂x .
1. Check for exactness: Let M= 2xand N=2x
y. Then,
∂M
∂y = 0
∂N
∂x =2
y
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Find the integrating factor: To make the equation exact, we find the integrating factor
µ(x)by the equation:
µ(x) = e∫My−Nx
Ndx
Here, My= 0,Nx=−2x
y2. Therefore, the integrating factor is:
µ(x) = e∫−2x
y3dx =e
−x2
y2+C
3. Multiply the equation by the integrating factor: Multiply the given equation by the
integrating factor found:
e
−x2
y2y′−2xe
−x2
y2
y= 0
4. Rewrite the equation in exact form: The equation is now exact and can be written in
the form: d
dx (e
−x2
y2y)= 0
5. Solve the exact equation: Integrating both sides, we get:
e
−x2
y2y=C
which can be rearranged to:
y=Cex2
y2
where Cis the constant of integration.
Therefore, the general solution to the differential equation is y=Cex2
y2.
32. Question 32: Consider the differential equation:
(2x+y2)dx + (2y+xex
y)dy = 0
Determine whether the given differential equation is exact or not. If the equation is exact,
find the general solution.
Ans. To determine if the given differential equation is exact, we check if the partial derivatives
of the coefficients with respect to the opposite variable are equal. If the equation is exact, we
proceed to find the general solution.
1. Check for exactness: Let’s calculate the partial derivatives of the coefficients:
∂M
∂y = 2yand ∂N
∂x = 2 + x
yex
y
Since ∂M
∂y =∂N
∂x , the given differential equation is not exact.
2. Make the Equation Exact: To make the equation exact, we need to find an integrating
factor µ(x, y)such that:
µMdx +µNdy =∂
∂y (µM)dx +∂
∂x(µN)dy
In our case, we need to find µ(x, y)such that:
µ(2x+y2) = ∂(µM)
∂y and µ(2y+xex
y) = ∂(µN )
∂x
Solving these two equations would give us the integrating factor µ(x, y).
3. Find the General Solution: If we were able to find the integrating factor µ(x, y)in the
previous step, then we would multiply the given differential equation by µ(x, y)to make it exact.
After that, we would find the general solution by integrating the exact equation and including an
arbitrary constant.
Since we were not able to find an integrating factor in this case, we cannot proceed with
finding the general solution.
33. Let M(x, y)dx +N(x, y)dy = 0 be a first-order differential equation. Determine conditions
for exactness and methods for finding solutions for the differential equation.
Ans. To determine if the differential equation M(x, y)dx +N(x, y)dy = 0 is exact, we check
if the partial derivatives satisfy the condition ∂M
∂y =∂N
∂x . If the equation is exact, we find the
integrating factor µ(x)by solving the equation
∂M
∂y −∂N
∂x
N=dµ(x)
µ(x)dx . Then, we multiply the equation
by the integrating factor and try to find a function ϕ(x, y)such that ∂ϕ
∂x =M µ and ∂ϕ
∂y =N µ.
Finally, the general solution to the exact differential equation will be given by ϕ(x, y) = C, where
Cis an arbitrary constant.
34. Let Mand Nbe functions of xand ysuch that the differential equation
M(x, y)dx +N(x, y)dy = 0
is exact. Determine conditions for exactness and outline the method for finding a solution if
these conditions are satisfied.
Ans. To determine exactness of the given differential equation, we need to check the equality
of mixed partial derivatives of Mand Nwith respect to xand y. If the equation is exact, we
can use integrating factors to find a solution.
Conditions for exactness: 1. Verify that ∂M
∂y =∂N
∂x .
Method for finding solutions: 1. If the equation is exact: a. Integrate Mwith respect to
xand add a constant function f(y)that only depends on y. b. Take the partial derivative of the
result with respect to y. c. Set the result equal to N. d. Solve the resulting differential equation
for the constant function f(y)to find the solution.
35. Consider the following differential equation:
(3x2+ 2yx)dx + (x2+ 2y)dy = 0
Determine whether the equation is exact or not. If it is exact, find the general solution.
Ans. To determine if the given differential equation is exact, we will check if ∂M
∂y =∂N
∂x .
1. Check for exactness:
Let M= 3x2+ 2yx and N=x2+ 2y. Then,
∂M
∂y = 2x=∂N
∂x = 2x
Since ∂M
∂y =∂N
∂x , the equation is not exact.
2. Integrate to find integrating factor:
To make the equation exact, we need to find the integrating factor (x, y)such that (x, y)M
and (x, y)Nwill have exact partial derivatives.
Since the integrating factor depends only on xor only on y, we will try the form (x)first:
(x) = e∫P(x)dx
where P(x) =
∂M
∂y −∂N
∂x
N.
Calculating P(x)gives:
P(x) = 2x−2x
x2+ 2y=0
x2+ 2y= 0
So, the integrating factor (x)is 1. This implies we should try the form (y).
3. Find the integrating factor:
Let’s try the form (y):
(y) = e∫Q(y)dy
where Q(y) =
∂N
∂x −∂M
∂y
M.
Calculating Q(y)gives:
Q(y) = 2x−2x
3x2+ 2yx =0
3x2+ 2yx = 0
So the integrating factor (y)is also 1, indicating an error in our calculations.
Conclusion:
Since we were unable to find a suitable integrating factor, the given differential equation
(3x2+ 2yx)dx + (x2+ 2y)dy = 0 is not exact and we need to use other methods for solving it.
