MAT 142 - COLLEGE MATHEMATICS - Differential
Equations Practice Material - Set 2
1. Consider the differential equation:
y′=1
x2+y2
Ans. To solve this differential equation, we can first rewrite it as a separable differential
equation. Then we will integrate both sides to find the solution.
1. Rearrange the terms to get the following:
dy
dx =1
x2+y2
2. Separate the variables by multiplying both sides by (x2+y2)dx:
(x2+y2)dy =dx
3. Integrate both sides: ∫(x2+y2)dy =∫dx
4. Integrate the left side with respect to y:
∫x2dy +∫y2dy =∫dx
x2y+1
3y3=x+C
5. The general solution to the differential equation is:
x2y+1
3y3=x+C
where Cis the constant of integration.
2. Consider the following differential equation:
y′(x) = 2x+y(x)
x+y(x)
Find the general solution to the differential equation.
Ans. To solve this differential equation, let’s rewrite it in a more recognizable form:
dy
dx =2x+y
x+y
1. Let’s begin by doing some algebraic manipulation to make the equation separable:
(x+y)dy
2x+y=dx
2. Now we separate the variables and integrate:
∫(x+y)
(2x+y)dy =∫dx
3. Make the substitution u= 2x+y, then du = 2dx +dy:
∫1
udu =∫dx
4. Integrating both sides gives:
ln |u|=x+C
5. Substituting back u= 2x+y:
ln |2x+y|=x+C
6. Exponentiating both sides to remove the natural logarithm:
2x+y=Kex
where K=±eC. This is the general solution to the differential equation.
3. Find the general solution to the differential equation:
dy
dx +y=exsin x
Ans. To solve the differential equation dy
dx +y=exsin x, we can use the method of integrating
factors. Let’s follow these steps:
1. Rewrite the equation in the form dy
dx +P y =Q, where P= 1 and Q=exsin x.
2. Find the integrating factor µ(x) = e∫P dx. In this case, µ(x) = e∫1dx =ex.
3. Multiply both sides of the differential equation by the integrating factor µ(x):
exdy
dx +exy=e2xsin x
4. Recognize the left side as the result of the product rule for differentiation:
d
dx(exy) = e2xsin x
5. Integrate both sides with respect to x:
∫d
dx(exy)dx =∫e2xsin xdx
6. Solve the integrals:
exy=−1
2e2xcos x+1
2e2xsin x+C
7. Solve for y:
y=−1
2excos x+1
2exsin x+Ce−x
Therefore, the general solution to the differential equation is y=−1
2excos x+1
2exsin x+
Ce−x, where Cis an arbitrary constant.
4. Question: Find the general solution to the differential equation y′′ −y′−6y= 0.
Ans. Step-by-step solution: 1. The characteristic equation for the given differential equation
y′′ −y′−6y= 0 is obtained by substituting y=erx into the equation:
r2−r−6 = 0
2. Solve the characteristic equation r2−r−6 = 0 by factoring or using the quadratic formula:
(r−3)(r+ 2) = 0
r1= 3 and r2=−2are the roots of the characteristic equation. 3. The general solution to the
differential equation is then:
y(x) = c1e3x+c2e−2x
where c1and c2are arbitrary constants. Therefore, the general solution to the differential equation
y′′ −y′−6y= 0 is y(x) = c1e3x+c2e−2x, where c1and c2are arbitrary constants.
5. Question:
Solve the initial value problem:
y′′ −4y′+ 4y= 0, y(0) = 4, y′(0) = 1.
Ans. Step-by-step solution:
1. We first write the characteristic equation:
Let y(t) = ert be the proposed solution. Substituting into the differential equation, we get:
r2ert −4rert + 4ert = 0.
This simplifies to:
r2−4r+ 4 = 0.
2. Solve the characteristic equation:
The characteristic equation has a repeated root at r= 2. Therefore, the general solution is
given by:
y(t) = (c1+c2t)e2t.
3. Apply the initial conditions:
Given y(0) = 4 and y′(0) = 1:
y(0) = c1= 4.
y′(0) = 2c1+c2= 1.
Solving these equations simultaneously, we find c1= 4 and c2=−7.
4. Write the final solution:
Therefore, the solution to the initial value problem is:
y(t) = (4 −7t)e2t.
6. Question: Solve the initial value problem dy
dx =y2−1with y(0) = 2.
Ans. Step-by-step solution: 1. Write the differential equation in the form dy
dx =f(x, y)
where f(x, y) = y2−1. 2. This is a separable differential equation, so we can rearrange it as
dy
y2−1=dx. 3. Integrate both sides: ∫dy
y2−1=∫dx 4. We can use partial fractions to integrate
the left side. We can write 1
y2−1as A
y−1+B
y+1 . 5. Solving for Aand Bgives A=1
2and
B=−1
2. 6. Substituting these back into the integral, we have: ∫(1
2(y−1) −1
2(y+1) )dy =∫dx
7. Simplifying and integrating gives: 1
2ln |y−1| − 1
2ln |y+ 1|=x+C8. Combining the
logarithms: ln (y−1
y+1 )= 2x+C9. Exponentiating both sides gives: y−1
y+1 =e2x+C10. Rewrite
the right side as Ce2xwhere C=eC(a new constant). 11. Solve for y:y−1 = Ce2x(y+ 1)
y−1 = Ce2xy+Ce2x(1 −Ce2x)y= 1 + Ce2xy=1+Ce2x
1−Ce2x12. Apply the initial condition
y(0) = 2 to find the value of C.y(0) = 1+C
1−C= 2 Solving for Cgives C=1
3. 13. Substituting
C=1
3back into the expression for y, we get the particular solution: y=3e2x+1
3−e2x
7. Question: Find the general solution to the differential equation: dy
dx + 3y= 6e−3x.
Ans. Solution: 1. Begin by finding the integrating factor e∫3dx:
e∫3dx =e3x
2. Multiply the integrating factor by the given differential equation:
e3xdy
dx + 3e3xy= 6e3xe−3x
e3xdy
dx + 3e3xy= 6
3. Rewrite the left side of the equation as the derivative of the product of the integrating
factor and y, and simplify the equation:
d
dx(e3xy) = 6
4. Integrate both sides with respect to x:
∫d
dx(e3xy)dx =∫6dx
e3xy= 6x+C
5. Solve for yby dividing both sides by e3x:
y=6x
e3x+C
e3x
Therefore, the general solution to the differential equation dy
dx + 3y= 6e−3xis y=6x
e3x+C
e3x,
where Cis an arbitrary constant.
8. Find the general solution of the following differential equation:
d2y
dx2−5dy
dx+ 6y= 0
Ans. Solution: 1. Write down the characteristic equation by assuming y=erx:
r2−5r+ 6 = 0
2. Solve the characteristic equation by factoring or using the quadratic formula:
(r−2)(r−3) = 0 =⇒r= 2,3
3. Write down the general solution using the roots obtained:
y(x) = c1e2x+c2e3x
where c1and c2are constants determined by initial conditions.
9. Question: Solve the following initial value problem: dy
dx = 2x+y,y(0) = 1.
Ans. Step-by-step solution: 1. First, we rewrite the given differential equation in standard form
dy
dx −y= 2x. 2. We then find the integrating factor µ(x) = e∫−1dx =e−x. 3. Multiply the
integrating factor by the standard form equation to get e−xdy
dx −e−xy= 2xe−x. 4. Recognize
the left side as the derivative of e−xy, so the equation becomes d
dx (e−xy) = 2xe−x. 5. Integrate
both sides with respect to xto get e−xy=∫2xe−xdx. 6. Solve the integral on the right side
to obtain e−xy=−2xe−x−2e−x+C, where Cis the constant of integration. 7. Divide by
e−xto find y=−2x−2 + Cex. 8. Use the initial condition y(0) = 1 to determine the value
of C. Substituting x= 0 and y= 1 into the equation gives 1 = −2(0) −2 + C(1), which
simplifies to C= 3. 9. Substitute C= 3 back into the equation to find the particular solution
y=−2x−2+3ex. Therefore, the solution to the initial value problem is y=−2x−2+3ex.
10. Question: Solve the following first-order differential equation using an integrating factor:
dy
dx +2y
x= 4x3
Ans. Step-by-step solution: 1. We can rewrite the given differential equation in the form:
dy
dx +2y
x= 4x3
2. Notice that this equation is not exact, as
∂
∂y (2y
x)=2
x=∂
∂x (4x3)= 12x2
3. To solve this equation, we will use an integrating factor. Multiply both sides of the equation
by the integrating factor µ(x) = e∫2
xdx =e2ln |x|=x2:
x2dy
dx + 2xy = 4x5
4. The left side of the equation can be written in the form of the product rule:
d
dx(x2y) = 4x5
5. Integrate both sides with respect to x:
∫d
dx(x2y)dx =∫4x5dx
6. Integrate to get:
x2y=4
6x6+C
x2y=2
3x6+C
7. Finally, solve for y:
y=2
3x4+C
x2
Therefore, the general solution to the given differential equation is y=2
3x4+C
x2, where Cis
an arbitrary constant.
11. Find the general solution to the differential equation dy
dx −2y=e2x.
Ans. Solution: 1. First, write the differential equation in the standard form
dy
dx −2y=e2x
2. This is a first-order linear ordinary differential equation. We can solve it by finding an
integrating factor, µ(x), given by
µ(x) = e∫−2dx =e−2x
3. Multiply both sides of the differential equation by the integrating factor to get
e−2xdy
dx −2e−2xy=e2xe−2x
e−2xdy
dx −2e−2xy= 1
4. Rewrite the left side as the derivative of a product
d
dx (e−2xy)= 1
5. Integrate both sides with respect to xto get
e−2xy=∫1dx =x+C
6. Solve for yto find the general solution
y=e2x(x+C)
12. Question: Solve the following differential equation:
y′=1
x+y
Ans. Solution: 1. Rewrite the given differential equation in the standard form for a first-
order linear differential equation: y′−y=1
x. 2. Find the integrating factor µ(x)by solving
the differential equation µ(x) = e∫−1dx. 3. Simplify the integral to find the integrating factor
µ(x) = e−x. 4. Multiply both sides of the differential equation by the integrating factor µ(x):
e−xy′−e−xy=e−x1
x. 5. Recognize that the left side of the equation is the derivative of the
product e−xywith respect to x. 6. Integrate both sides to get the general solution: e−xy=
∫e−x1
xdx +C. 7. Evaluate the integral on the right side using integration by parts: e−xy=
−e−x(1 + x)ln |x|+C. 8. Solve for yto get the final solution: y=−1−x+Cexln |x|.
Therefore, the solution to the differential equation y′=1
x+yis y=−1−x+Cexln |x|, where
Cis an arbitrary constant.
13. Question: Solve the following first-order nonlinear differential equation:
dy
dx =2x+y
x+y
Ans. Solution: 1. Let’s rewrite the given differential equation as:
dy
dx = 1 + y−x
x+y
2. Now, we can substitute v=x+yand rewrite the differential equation in terms of v:
dv
dx =dy
dx + 1 = 1 + y−x
v
3. We can further simplify the equation as:
vdv
dx =v+y−x
4. To solve this separable differential equation, we arrange terms to get:
vdv = (v+x)dx
5. Integrating both sides, we get:
∫vdv =∫(v+x)dx
6. Solving the integrals gives us:
v2
2=v2
2+x2
2+C
7. Simplify the equation to get:
v2=v2+x2+ 2C
8. Rearranging terms yields:
x2= 2C
9. Therefore, the solution to the differential equation is given by:
x2= 2C
14. Question: Solve the differential equation y′′ −4y′+ 4y= 0 with the initial conditions
y(0) = 2 and y′(0) = 1.
Ans. Step-by-step solution:
1. Write down the characteristic equation by substituting y=erx into the differential equation
y′′ −4y′+ 4y= 0:
r2−4r+ 4 = 0
(r−2)2= 0
Since the characteristic equation has a repeated root r= 2, the general solution can be written
as
y(x) = (c1+c2x)e2x
2. Apply the initial condition y(0) = 2:
2 = c1e0⇒c1= 2
3. Differentiate y(x)with respect to xto find y′(x):
y′(x) = (2 + c2)e2x
4. Apply the initial condition y′(0) = 1:
1 = (2 + c2)e0⇒c2=−1
5. Substitute c1= 2 and c2=−1back into the general solution to get the particular solution:
y(x) = (2 −x)e2x
Therefore, the solution to the differential equation with the given initial conditions is y(x) =
(2 −x)e2x.
15. Question 15: Solve the initial value problem
y(4) −8y′′ + 16y= 0, y(0) = 1, y′(0) = 0, y′′(0) = 0, y′′′(0) = 0.
Ans. Solution: We first find the characteristic equation of the differential equation by assuming
the solution has the form y=ert:
r4−8r2+ 16 = 0.
1. Solve the characteristic equation:
r4−8r2+ 16 = 0.
(r2−4)2= 0.
r2−4 = 0.
r=±2.
Hence, the general solution to the differential equation is:
y(t) = c1e2t+c2te2t+c3e−2t+c4te−2t.
2. Substitute the initial conditions to find the specific solution: Given: y(0) = 1, y′(0) =
0, y′′(0) = 0, y′′′(0) = 0.
Plugging in the initial conditions yields the following equations:
y(0) = c1+c3= 1,
y′(0) = 2c1+ 2c2−2c3−2c4= 0,
y′′(0) = 4c1+ 4c2+ 4c3+ 4c4= 0,
y′′′(0) = 8c1−8c2+ 8c3−8c4= 0.
Solving the system of equations gives c1=1
4, c2=−1
4, c3=3
4, c4=−1
4. Therefore, the
solution to the initial value problem is:
y(t) = 1
4e2t−1
4te2t+3
4e−2t−1
4te−2t.
16. Let y(x)be the solution to the differential equation dy
dx = 2xy2with initial condition
y(0) = 1. Find the value of y(1).
Ans. Solution: 1. Given the differential equation dy
dx = 2xy2, we can separate variables to
obtain: dy
y2= 2xdx
2. Integrating both sides gives: ∫1
y2dy =∫2xdx
−1
y=x2+C
3. Use the initial condition y(0) = 1 to find the value of the constant C:
−1
1= 02+C
C=−1
4. Hence, the solution to the differential equation is −1
y=x2−1. 5. To find y(1), substitute
x= 1 into the equation:
−1
y(1) = 12−1
−1
y(1) = 0
y(1) = ∞
Therefore, the value of y(1) is ∞.
17. Question: Solve the following first-order ordinary differential equation:
dy
dx =y
x+x2
Ans. Step-by-step solution:
1. Rewriting the given equation in standard form, we have:
dy
dx =y
x+x2
This can be rewritten as:
dy
dx −y
x=x2
2. We now identify the integrating factor, which is given by:
I(x) = e∫−1
xdx
Calculating the integral, we get:
I(x) = e−ln |x|=1
|x|
So the integrating factor is 1
|x|.
3. Now we multiply both sides of the differential equation by the integrating factor:
1
|x|
dy
dx −1
|x|
y
x=x21
|x|
This simplifies to:
1
|x|
dy
dx −1
x2y=x
|x|
4. Recognizing the left side as the derivative of the product of the integrating factor and y,
we can rewrite the equation as:
d
dx (1
|x|y)=x
|x|
5. Integrate both sides with respect to x to solve for y:
∫d
dx (1
|x|y)dx =∫x
|x|dx
1
|x|y=∫sgn(x)dx =|x|+C
where sgn(x)is the sign function and C is the constant of integration.
6. Finally, solving for y, we have:
y=|x|2+C|x|
where Cis an arbitrary constant.
18. Find the general solution to the differential equation: dy
dx +y=e−x.
Ans. To solve the given differential equation, we can use an integrating factor method. Let
µ(x) = e∫1dx =ex. Multiply both sides of the differential equation by µ(x):
exdy
dx +exy=exe−x
This can be rewritten as: d
dx(exy) = 1
Integrating both sides with respect to xgives:
exy=∫1dx =x+C
Therefore, the general solution to the given differential equation is:
y=e−x(x+C)
where Cis an arbitrary constant.
19. Question: Solve the initial value problem dy
dx =3x2+4y2
2xy with initial condition y(1) = 2.
Ans. Step-by-step solution:
1. Rewrite the given differential equation in a more standard form by multiplying through by
dx and separating variables:
dy
dx =3x2+ 4y2
2xy =⇒2y dy = (3x2+ 4y2)dx
x
2. Next, divide through by 2y(3x2+ 4y2)to separate variables further:
2
3x2+ 4y2dy =dx
x
3. Integrate both sides with respect to their respective variables:
∫2
3x2+ 4y2dy =∫1
xdx
4. For the left integral, use substitution (u= 3x2+ 4y2, du = 8y dy):
1
8∫1
udu =1
8ln |u|=1
8ln |3x2+ 4y2|
5. For the right integral, integrate ∫1
xdx =ln |x|
6. Combining the integrals and adding the constant of integration C, we have:
1
8ln |3x2+ 4y2|=ln |x|+C
7. Applying the initial condition y(1) = 2:
1
8ln |3(1)2+ 4(2)2|=ln |1|+C
1
8ln |19|= 0 + C
C=1
8ln(19)
8. Finally, substitute Cback into the equation to find the particular solution:
1
8ln |3x2+ 4y2|=ln |x|+1
8ln(19)
ln |3x2+ 4y2|= 8 ln |x|+ln(19)
3x2+ 4y2=Ax8·19 (where A is a constant)
20. Question: Find the general solution to the differential equation: y′′ −2y′+y=exsin(x).
Ans. Solution: 1. First, we find the complementary solution by solving the characteristic
equation r2−2r+ 1 = 0. The characteristic equation has a double root at r= 1, so the
complementary solution is of the form yc=C1ex+C2xex.
2. Next, we find a particular solution ypfor the non-homogeneous part exsin(x). Notice
that the right side of the differential equation consists of terms that can be solutions to the
complementary part (polynomial multiplied by ex) and sin(x)which can be part of the general
solution for yp. Since sin(x)is a part of the particular solution, we try a particular solution of
the form:
yp= (Ax +B)exsin(x)
3. We differentiate ypwith respect to xto find y′
pand y′′
p:
y′
p= (A+Ax +B)exsin(x)+(Ax +B)excos(x)
y′′
p= (2A+Ax +B)exsin(x) + (2A+ 3Ax +B−Ax −B)excos(x)
Simplify the expressions:
y′
p= (A+Ax +B)exsin(x)+(Ax +B)excos(x)
y′′
p= (2A+Ax +B)exsin(x)+(A+ 2Ax +B)excos(x)
4. Substitute yp,y′
p, and y′′
pinto the original differential equation y′′ −2y′+y=exsin(x).
We have:
(2A+Ax+B)exsin(x)+(A+2Ax+B)excos(x)−2[(A+Ax+B)exsin(x)+(Ax+B)excos(x)]+(Ax+B)exsin(x) = exsin(x)
5. Simplify the equation and equate the coefficients of similar terms on both sides of the
equation to solve for Aand B. After solving, we find A=−1
4and B=3
8.
6. Therefore, the particular solution is: yp=−1
4xexsin(x) + 3
8exsin(x)
7. The general solution to the given differential equation is the sum of the complementary
solution and the particular solution:
y=yc+yp=C1ex+C2xex−1
4xexsin(x) + 3
8exsin(x)
where C1and C2are arbitrary constants.
21. Question 21: Solve the following first-order differential equation with an integrating factor.
y′+2
xy=x3
Ans. To solve the given first-order differential equation with an integrating factor, follow these
steps:
1. Write the given differential equation in the form y′+P(x)y=Q(x). 2. Find the
integrating factor I(x) = e∫P(x)dx. 3. Multiply the integrating factor with both sides of the
differential equation. 4. Rewrite the equation in the form of a product rule. 5. Integrate both
sides and solve for the unknown function.
1. Write the given differential equation: The given differential equation is y′+2
xy=x3.
2. Find the integrating factor: The integrating factor is given by
I(x) = e∫P(x)dx
For the differential equation y′+2
xy=x3,P(x) = 2
xSo, the integrating factor is
I(x) = e∫2
xdx =e2ln |x|=eln(x2)=x2
3. Multiply the integrating factor: Multiply both sides of the differential equation by the
integrating factor x2:
x2y′+ 2xy =x5
4. Rewrite the equation: Rewrite the equation in the form of a product rule:
d
dx(x2y) = x5
5. Solve the equation: Integrate both sides with respect to x:
∫d
dx(x2y)dx =∫x5dx
x2y=1
6x6+C
y=1
6x4+C
x2
Therefore, the solution to the first-order differential equation y′+2
xy=x3is y=1
6x4+C
x2,
where Cis the constant of integration.
22. Find the general solution to the differential equation: y′′ −y′−6y= 0.
Ans. To solve the differential equation y′′ −y′−6y= 0, we first find the characteristic equation
by assuming y=ert: 1. The characteristic equation is r2−r−6 = 0. 2. Solve the characteristic
equation to find the roots: r= 3 and r=−2. 3. The general solution is y(t) = c1e3t+c2e−2t,
where c1and c2are arbitrary constants.
23. Question: Solve the differential equation (1 + x2)y′+xy =x3, given that y(1) = 3.
Ans. Solution:
1. First, let’s rewrite the given differential equation in standard form:
(1 + x2)y′+xy =x3
Divide through by (1 + x2)to get:
y′+x
1+x2y=x3
1+x2.
2. Now, we identify the integrating factor, which is e∫x
1+x2dx.
The integrating factor is µ(x) = e∫x
1+x2dx.
3. Let’s find the integral and simplify the integrating factor:
∫x
1+x2dx =1
2ln(1 + x2) + C.
Thus, the integrating factor becomes µ(x) = e1
2ln(1+x2).
Simplifying the integrating factor gives µ(x) = (1 + x2)1/2.
4. Multiply the differential equation by the integrating factor:
(1 + x2)1/2y′+x(1 + x2)1/2y=x3(1 + x2)1/2.
5. Rewrite the equation in terms of the product rule for differentiation:
d
dx [(1 + x2)1/2y]=x3(1 + x2)1/2.
Integrating both sides gives:
(1 + x2)1/2y=∫x3(1 + x2)1/2dx +C.
6. Simplify the integral on the right side:
∫x3(1 + x2)1/2dx =1
5(1 + x2)5/2 +C.
Therefore, the solution to the differential equation is:
(1 + x2)1/2y=1
5(1 + x2)5/2 +C.
7. Now, apply the initial condition y(1) = 3 to find the value of the constant C:
(1 + 12)1/2 ·3 = 1
5(1 + 12)5/2 +C.
2·3 = 1
5·25+C.
6 = 32
5+C.
C= 6 −32
5=3
5.
8. Finally, substitute the value of Cback into the solution:
(1 + x2)1/2y=1
5(1 + x2)5/2 +3
5.
Therefore, the solution to the differential equation with the initial condition is:
y=1
5(1 + x2)2+3
5(1+x2)1/2 .
24. Question: Solve the following differential equation using the method of undetermined
coefficients:
y′′ −4y′+ 4y= 6e2x
Ans. Step-by-step solution:
1. Find the complementary function by solving the characteristic equation r2−4r+ 4 = 0:
The characteristic equation is (r−2)2= 0, so we have a repeated root r= 2. The complementary
function is then:
yc=c1e2x+c2xe2x
2. Assume the particular solution has the form yp=Axe2x. Calculate the derivatives needed
for substitution:
y′
p= (A+ 2Ax)e2x
y′′
p= (2A+ 4Ax)e2x
3. Substitute yp, y′
p, y′′
pinto the differential equation:
(2A+ 4Ax)e2x−4(A+ 2Ax)e2x+ 4Axe2x= 6e2x
4. Simplify and equate coefficients of e2xon both sides:
2A−4A−4A= 6
−6A= 6
A=−1
5. The particular solution is yp=−xe2x.
6. The general solution is the sum of the complementary and particular solutions:
y=yc+yp=c1e2x+c2xe2x−xe2x
y= (c1−1)e2x+c2xe2x
25. Question 25: Solve the differential equation dy
dx −2y= 5x.
Ans. Solution:
1. First, we identify the differential equation as a first-order linear differential equation in the
form dy
dx +P(x)y=Q(x), where P(x) = −2and Q(x) = 5x.
2. To solve this type of differential equation, we use an integrating factor µ(x) = e∫P(x)dx.
3. In our case, P(x) = −2, so the integrating factor µ(x) = e∫−2dx =e−2x.
4. Multiply the differential equation by the integrating factor: e−2xdy
dx −2e−2xy= 5xe−2x.
5. Rewrite the left side as a derivative of the product of yand the integrating factor using
the product rule: d
dx (ye−2x) = 5xe−2x.
6. Integrate both sides with respect to x:∫d
dx (ye−2x)dx =∫5xe−2xdx.
7. Simplify the integrals: ye−2x=−5
2xe−2x−5
2∫e−2xdx.
8. Integrate ∫e−2xdx =−1
2e−2x. Substitute this back into the equation.
9. Solve for y:y=−5
2x−5
4+Ce2x, where Cis the constant of integration.
Therefore, the general solution to the differential equation is y=−5
2x−5
4+Ce2x.
26. Question 26: Solve the following differential equation using the method of undetermined
coefficients: y′′ −2y′+y=ex.
Ans. Solution: 1. First, we write down the characteristic equation for the homogeneous part
of the differential equation:
r2−2r+ 1 = 0
(r−1)2= 0
This gives us a repeated root of r= 1. So, the solution to the homogeneous equation is:
yh(x) = c1ex+c2xex
2. Next, we guess a particular solution to the given nonhomogeneous equation in the form of
yp(x) = Aex. We substitute this guess back into the differential equation:
y′′ −2y′+y=ex
(Aex)′′ −2(Aex)′+Aex=ex
Aex−2Aex+Aex=ex
Aex=ex
This gives us A= 1.
3. Therefore, the particular solution is:
yp(x) = ex
4. Combining the general solution of the homogeneous part and the particular solution, we
have:
y(x) = yh(x) + yp(x)
y(x) = c1ex+c2xex+ex
y(x) = (c1+ 1)ex+c2xex
where c1and c2are constants.
Thus, the solution to the differential equation y′′ −2y′+y=exis y(x) = (c1+ 1)ex+c2xex.
27. Question 27: Solve the following initial value problem using the method of undetermined
coefficients:
y′′ −y′−6y= 12e3x, y(0) = 0, y′(0) = 0
Ans. Step 1. Find the characteristic equation by setting the homogeneous part of the
differential equation to zero:
y′′ −y′−6y= 0
The characteristic equation is r2−r−6 = 0, which factors as (r−3)(r+ 2) = 0. Therefore,
the characteristic roots are r1= 3 and r2=−2.
Step 2. Write the general solution to the homogeneous equation:
yh(x) = c1e3x+c2e−2x
Step 3. Determine a particular solution for the nonhomogeneous equation using the method
of undetermined coefficients. Since the right-hand side term is 12e3x, assume a particular solution
of the form yp(x) = Ae3x.
Step 4. Substitute the trial particular solution into the differential equation and solve for A:
9A−3A−6A= 12 ⇒0 = 12
Since the equation has no solution, we need to modify our trial particular solution. Try yp(x) =
Axe3x.
Step 5. Substitute the new trial particular solution into the differential equation and solve
for A:
9Ae3x−3Ae3x−6Axe3x= 12e3x
This simplifies to 6A= 12, so A= 2.
Step 6. Write down the particular solution: yp(x) = 2xe3x.
Step 7. Combine the general solution and the particular solution to get the final solution:
y(x) = yh(x) + yp(x) = c1e3x+c2e−2x+ 2xe3x
Step 8. Apply the initial conditions to solve for c1and c2:
y(0) = c1+c2= 0
y′(0) = 3c1−2c2+ 6 = 0
Solving these equations simultaneously, we find c1=4
5and c2=−4
5.
Step 9. Therefore, the solution to the initial value problem is
y(x) = 4
5e3x−4
5e−2x+ 2xe3x
28. Question: Solve the following initial value problem:
y′+ 2xy = 4x, y(0) = 3.
Ans. Step-by-step solution: 1. First, we need to find the integrating factor. The integrating
factor is given by e∫2x dx =ex2. 2. Multiplying both sides of the given differential equation by
the integrating factor ex2, we get:
ex2y′+ 2xex2y= 4xex2.
3. Recognize that the left side is the derivative of the product ex2y. So, we rewrite the equation
as:
d(ex2y)
dx = 4xex2.
4. Integrate both sides with respect to xto solve for y:
∫d(ex2y) = ∫4xex2dx.
⇒ex2y= 2ex2+C,
where Cis the constant of integration. 5. Now, solve for yby dividing by ex2:
y= 2 + Ce−x2.
6. To find the value of C, use the initial condition y(0) = 3:
3 = 2 + C·e−0,
3 = 2 + C.
Therefore, C= 1. 7. The solution to the initial value problem is:
y= 2 + e−x2.
29. Question: Solve the initial value problem y′=y2+cos(x)with y(0) = 1.
Ans. Step-by-step solution: 1. First, we note that the given differential equation is separable
since all terms involving yare on one side and terms involving xare on the other side. So, we can
rewrite the equation as dy
dx =y2+cos(x). 2. Next, we separate variables by dividing both sides
of the equation by (y2+cos(x)) to get 1
y2+cos(x)dy =dx. 3. Integrate both sides with respect
to x:∫1
y2+cos(x)dy =∫dx
4. The integral on the left side can be evaluated using a trigonometric substitution. Let u=
sin(x). Then, du =cos(x)dx. Substituting these in, we get:
∫1
y2+cos(x)dy =∫1
y2+ 1 −u2dy =∫1
y2+ 1 −sin2(x)dy =∫1
y2+cos2(x)dy =∫1
y2+ (1 −u2)dy =∫1
y2+u2dy
5. This integral can be evaluated using a standard trigonometric substitution. Let u=ytan(v),
then du =tan(v)dv. Substituting these in, we get:
∫1
y2+u2dy =∫1
y2+y2tan2(v)dy =∫1
y2(1 + tan2(v))dy =∫1
y2sec2(v)dy =∫cos2(v)dv
6. The integral on the right side evaluates to v
2+sin(2v)
4+C, where Cis the constant of
integration. 7. Substitute back u=sin(x)and v=arctan (y
sin(x)). Therefore, the integral
simplifies to arctan(y
sin(x))
2+sin(2arctan(y
sin(x)))
4+C. 8. Finally, substitute back y(0) = 1 to determine
the value of C. We find C=π
4. Hence, the solution to the initial value problem is y(x) =
arctan(y
sin(x))
2+sin(2arctan(y
sin(x)))
4+π
4.
30. Question:
Solve the following first-order differential equation:
dy
dx =3y2
x
Ans. Solution:
1. Rewrite the differential equation in separable form by bringing all terms involving yto one
side and all terms involving xto the other side:
dy
y2=3
xdx
2. Integrate both sides with respect to their respective variables:
∫1
y2dy =∫3
xdx
3. The left-hand side integrates to −1
yand the right-hand side integrates to 3ln |x|+C,
where Cis the constant of integration.
4. Combining the results of the integrations, we get:
−1
y= 3 ln |x|+C
5. Solving for y, we have:
y=−1
3ln |x|+C
where Cis the constant of integration.
31. Question: Solve the following differential equation:
y′′ −2y′+ 2y=excos x
Ans. Step-by-step solution: 1. First, find the complementary function by solving the homoge-
neous equation y′′ −2y′+ 2y= 0. The characteristic equation is r2−2r+ 2 = 0. The roots of
the characteristic equation are r= 1 ±i. Therefore, the complementary function is given by
yc(x) = ex(Acos x+Bsin x)
where Aand Bare constants. 2. Next, find a particular solution for the given non-homogeneous
equation using the method of undetermined coefficients. Since excos xis a product of exponential
and trigonometric functions, assume
yp(x) = (C1ex+C2ex)cos x+ (C3ex+C4ex)sin x
where C1,C2,C3, and C4are unknown coefficients to be determined. 3. Substitute yp(x)into
the original differential equation and solve for the unknown coefficients by equating coefficients
of like terms. 4. Plug the values of the coefficients back into yp(x)to get the particular solution.
5. The general solution of the differential equation is the sum of the complementary function
yc(x)and the particular solution yp(x):
y(x) = yc(x) + yp(x)
y(x) = ex(Acos x+Bsin x)+(C1ex+C2ex)cos x+ (C3ex+C4ex)sin x
where A,B,C1,C2,C3, and C4are constants determined from the initial conditions or boundary
conditions.
32. Question 32: Solve the following initial value problem: dy
dx =xe−y, where y(0) = 1.
Ans. Solution: 1. Separate the variables by moving all terms involving yto the left side and
all terms involving xto the right side:
dy
ey=xdx
2. Integrate both sides with respect to their respective variables:
∫dy
ey=∫xdx
3. Integrate 1
eywith respect to yusing substitution (u=ey, du =eydy):
∫1
eydy =∫1
udu =ln |u|+C=ln |ey|+C=y+C
4. Integrate xwith respect to x:
∫xdx =1
2x2+C
5. Combining both results, we get:
y=1
2x2+C
6. To find the value of C, use the initial condition y(0) = 1:
1 = 1
2(0)2+C=⇒C= 1
7. Therefore, the solution to the initial value problem is y=1
2x2+ 1.
33. Find the general solution to the differential equation: dy
dx =y2+1
x.
Ans. Solution: 1. Rearrange the differential equation so that all terms involving yare on one
side and all terms involving xare on the other side:
dy
y2+ 1 =dx
x
2. Integrate both sides with respect to x:
∫dy
y2+ 1 =∫dx
x
3. The integral on the left side can be solved using a trigonometric substitution. Let y=
tan(θ)so that dy =sec2(θ)dθ:
∫1
tan2(θ)+1sec2(θ)dθ =∫1
sec2(θ)sec2(θ)dθ
∫sec2(θ)dθ =∫dθ =θ
4. Re-substitute y=tan(θ)back in:
θ=arctan(y)
5. Therefore, the left integral becomes:
∫dy
y2+ 1 =arctan(y) + C1
6. The right integral is a simple logarithmic function:
∫dx
x=ln |x|+C2
7. Combining both sides, we have:
arctan(y) + C1=ln |x|+C2
8. Solving for y, the general solution to the differential equation is:
y=tan(ln |x|+˜
C)
where ˜
C=C2−C1.
34. Find the general solution to the differential equation:
dy
dx =x2+ 2y
x
Ans. Solution: 1. Rewrite the given differential equation in the standard form by moving all
terms involving yto one side:
xdy
dx −2y=x2
2. Now, we observe that this is a first-order linear differential equation in the form of:
dy
dx +P(x)y=Q(x)
where P(x) = −2and Q(x) = x2.
3. To solve this linear differential equation, we first find the integrating factor I(x)defined
as:
I(x) = e∫P(x)dx
4. Substituting P(x) = −2into the formula, we get:
I(x) = e∫−2dx =e−2x
5. We then multiply the integrating factor to the original equation to get:
e−2x·(xdy
dx −2y) = e−2x·x2
6. By the product rule for differentiation, the left-hand side simplifies to:
d
dx(e−2xy) = x2e−2x
7. Integrating both sides with respect to x, we obtain:
e−2xy=∫x2e−2xdx
e−2xy=−1
2xe−2x−1
4e−2x+C
8. Multiplying by e2xgives us the general solution to the differential equation:
y=−1
2x−1
4+Ce2x
where Cis an arbitrary constant.
35. Question: Solve the following initial value problem:
y′+2x
x2+ 1y=x2, y(0) = 3
Ans. Solution: 1. This is a first-order linear differential equation in the form y′+P(x)y=Q(x),
where P(x) = 2x
x2+1 and Q(x) = x2. 2. To solve the differential equation, we first find the
integrating factor. The integrating factor is given by µ(x) = e∫P(x)dx. 3. Calculate ∫P(x)dx:
∫2x
x2+ 1dx =∫2x
x2+ 1 ·x2+ 1
x2+ 1dx =∫2x(x2+ 1)
x2+ 1 dx =∫2xdx =x2+C
4. Therefore, the integrating factor is µ(x) = ex2. 5. Multiply the differential equation by the
integrating factor µ(x):
ex2y′+2x
x2+ 1ex2y=x2ex2
6. Notice that the left side is the derivative of the product ex2y, so we rewrite the equation as:
(ex2y)′=x2ex2
7. Integrate both sides with respect to x:
∫(ex2y)′dx =∫x2ex2dx
ex2y=∫x2ex2dx +C
8. Calculate ∫x2ex2dx using substitution (u=x2, du = 2xdx):
∫x2ex2dx =1
2∫eudu =1
2eu+C=1
2ex2+C
9. Substitute this result back into the previous equation:
ex2y=1
2ex2+C
10. Solve for y:
y=1
2+Ce−x2
11. Apply the initial condition y(0) = 3 to find the value of C:
3 = 1
2+Ce0=1
2+C
C=5
2
y′(x) = 2x+y(x)
x+y(x)
Find the general solution to the differential equation.
Ans. To solve this differential equation, let’s rewrite it in a more recognizable form:
dy
dx =2x+y
x+y
1. Let’s begin by doing some algebraic manipulation to make the equation separable:
(x+y)dy
2x+y=dx
2. Now we separate the variables and integrate:
∫(x+y)
(2x+y)dy =∫dx
3. Make the substitution u= 2x+y, then du = 2dx +dy:
∫1
udu =∫dx
4. Integrating both sides gives:
ln |u|=x+C
5. Substituting back u= 2x+y:
ln |2x+y|=x+C
6. Exponentiating both sides to remove the natural logarithm:
2x+y=Kex
where K=±eC. This is the general solution to the differential equation.
3. Find the general solution to the differential equation:
dy
dx +y=exsin x
Ans. To solve the differential equation dy
dx +y=exsin x, we can use the method of integrating
factors. Let’s follow these steps:
1. Rewrite the equation in the form dy
dx +P y =Q, where P= 1 and Q=exsin x.
2. Find the integrating factor µ(x) = e∫P dx. In this case, µ(x) = e∫1dx =ex.
3. Multiply both sides of the differential equation by the integrating factor µ(x):
exdy
dx +exy=e2xsin x
4. Recognize the left side as the result of the product rule for differentiation:
d
dx(exy) = e2xsin x
5. Integrate both sides with respect to x:
∫d
dx(exy)dx =∫e2xsin xdx
6. Solve the integrals:
exy=−1
2e2xcos x+1
2e2xsin x+C
7. Solve for y:
y=−1
2excos x+1
2exsin x+Ce−x
Therefore, the general solution to the differential equation is y=−1
2excos x+1
2exsin x+
Ce−x, where Cis an arbitrary constant.
4. Question: Find the general solution to the differential equation y′′ −y′−6y= 0.
Ans. Step-by-step solution: 1. The characteristic equation for the given differential equation
y′′ −y′−6y= 0 is obtained by substituting y=erx into the equation:
r2−r−6 = 0
2. Solve the characteristic equation r2−r−6 = 0 by factoring or using the quadratic formula:
(r−3)(r+ 2) = 0
r1= 3 and r2=−2are the roots of the characteristic equation. 3. The general solution to the
differential equation is then:
y(x) = c1e3x+c2e−2x
where c1and c2are arbitrary constants. Therefore, the general solution to the differential equation
y′′ −y′−6y= 0 is y(x) = c1e3x+c2e−2x, where c1and c2are arbitrary constants.
5. Question:
Solve the initial value problem:
y′′ −4y′+ 4y= 0, y(0) = 4, y′(0) = 1.
Ans. Step-by-step solution:
1. We first write the characteristic equation:
Let y(t) = ert be the proposed solution. Substituting into the differential equation, we get:
r2ert −4rert + 4ert = 0.
This simplifies to:
r2−4r+ 4 = 0.
2. Solve the characteristic equation:
The characteristic equation has a repeated root at r= 2. Therefore, the general solution is
given by:
y(t) = (c1+c2t)e2t.
3. Apply the initial conditions:
Given y(0) = 4 and y′(0) = 1:
y(0) = c1= 4.
y′(0) = 2c1+c2= 1.
Solving these equations simultaneously, we find c1= 4 and c2=−7.
4. Write the final solution:
Therefore, the solution to the initial value problem is:
y(t) = (4 −7t)e2t.
6. Question: Solve the initial value problem dy
dx =y2−1with y(0) = 2.
Ans. Step-by-step solution: 1. Write the differential equation in the form dy
dx =f(x, y)
where f(x, y) = y2−1. 2. This is a separable differential equation, so we can rearrange it as
dy
y2−1=dx. 3. Integrate both sides: ∫dy
y2−1=∫dx 4. We can use partial fractions to integrate
the left side. We can write 1
y2−1as A
y−1+B
y+1 . 5. Solving for Aand Bgives A=1
2and
B=−1
2. 6. Substituting these back into the integral, we have: ∫(1
2(y−1) −1
2(y+1) )dy =∫dx
7. Simplifying and integrating gives: 1
2ln |y−1| − 1
2ln |y+ 1|=x+C8. Combining the
logarithms: ln (y−1
y+1 )= 2x+C9. Exponentiating both sides gives: y−1
y+1 =e2x+C10. Rewrite
the right side as Ce2xwhere C=eC(a new constant). 11. Solve for y:y−1 = Ce2x(y+ 1)
y−1 = Ce2xy+Ce2x(1 −Ce2x)y= 1 + Ce2xy=1+Ce2x
1−Ce2x12. Apply the initial condition
y(0) = 2 to find the value of C.y(0) = 1+C
1−C= 2 Solving for Cgives C=1
3. 13. Substituting
C=1
3back into the expression for y, we get the particular solution: y=3e2x+1
3−e2x
7. Question: Find the general solution to the differential equation: dy
dx + 3y= 6e−3x.
Ans. Solution: 1. Begin by finding the integrating factor e∫3dx:
e∫3dx =e3x
2. Multiply the integrating factor by the given differential equation:
e3xdy
dx + 3e3xy= 6e3xe−3x
e3xdy
dx + 3e3xy= 6
3. Rewrite the left side of the equation as the derivative of the product of the integrating
factor and y, and simplify the equation:
d
dx(e3xy) = 6
4. Integrate both sides with respect to x:
∫d
dx(e3xy)dx =∫6dx
e3xy= 6x+C
5. Solve for yby dividing both sides by e3x:
y=6x
e3x+C
e3x
Therefore, the general solution to the differential equation dy
dx + 3y= 6e−3xis y=6x
e3x+C
e3x,
where Cis an arbitrary constant.
8. Find the general solution of the following differential equation:
d2y
dx2−5dy
dx+ 6y= 0
Ans. Solution: 1. Write down the characteristic equation by assuming y=erx:
r2−5r+ 6 = 0
2. Solve the characteristic equation by factoring or using the quadratic formula:
(r−2)(r−3) = 0 =⇒r= 2,3
3. Write down the general solution using the roots obtained:
y(x) = c1e2x+c2e3x
where c1and c2are constants determined by initial conditions.
9. Question: Solve the following initial value problem: dy
dx = 2x+y,y(0) = 1.
Ans. Step-by-step solution: 1. First, we rewrite the given differential equation in standard form
dy
dx −y= 2x. 2. We then find the integrating factor µ(x) = e∫−1dx =e−x. 3. Multiply the
integrating factor by the standard form equation to get e−xdy
dx −e−xy= 2xe−x. 4. Recognize
the left side as the derivative of e−xy, so the equation becomes d
dx (e−xy) = 2xe−x. 5. Integrate
both sides with respect to xto get e−xy=∫2xe−xdx. 6. Solve the integral on the right side
to obtain e−xy=−2xe−x−2e−x+C, where Cis the constant of integration. 7. Divide by
e−xto find y=−2x−2 + Cex. 8. Use the initial condition y(0) = 1 to determine the value
of C. Substituting x= 0 and y= 1 into the equation gives 1 = −2(0) −2 + C(1), which
simplifies to C= 3. 9. Substitute C= 3 back into the equation to find the particular solution
y=−2x−2+3ex. Therefore, the solution to the initial value problem is y=−2x−2+3ex.
10. Question: Solve the following first-order differential equation using an integrating factor:
dy
dx +2y
x= 4x3
Ans. Step-by-step solution: 1. We can rewrite the given differential equation in the form:
dy
dx +2y
x= 4x3
2. Notice that this equation is not exact, as
∂
∂y (2y
x)=2
x=∂
∂x (4x3)= 12x2
3. To solve this equation, we will use an integrating factor. Multiply both sides of the equation
by the integrating factor µ(x) = e∫2
xdx =e2ln |x|=x2:
x2dy
dx + 2xy = 4x5
4. The left side of the equation can be written in the form of the product rule:
d
dx(x2y) = 4x5
5. Integrate both sides with respect to x:
∫d
dx(x2y)dx =∫4x5dx
6. Integrate to get:
x2y=4
6x6+C
x2y=2
3x6+C
7. Finally, solve for y:
y=2
3x4+C
x2
Therefore, the general solution to the given differential equation is y=2
3x4+C
x2, where Cis
an arbitrary constant.
11. Find the general solution to the differential equation dy
dx −2y=e2x.
Ans. Solution: 1. First, write the differential equation in the standard form
dy
dx −2y=e2x
2. This is a first-order linear ordinary differential equation. We can solve it by finding an
integrating factor, µ(x), given by
µ(x) = e∫−2dx =e−2x
3. Multiply both sides of the differential equation by the integrating factor to get
e−2xdy
dx −2e−2xy=e2xe−2x
e−2xdy
dx −2e−2xy= 1
4. Rewrite the left side as the derivative of a product
d
dx (e−2xy)= 1
5. Integrate both sides with respect to xto get
e−2xy=∫1dx =x+C
6. Solve for yto find the general solution
y=e2x(x+C)
12. Question: Solve the following differential equation:
y′=1
x+y
Ans. Solution: 1. Rewrite the given differential equation in the standard form for a first-
order linear differential equation: y′−y=1
x. 2. Find the integrating factor µ(x)by solving
the differential equation µ(x) = e∫−1dx. 3. Simplify the integral to find the integrating factor
µ(x) = e−x. 4. Multiply both sides of the differential equation by the integrating factor µ(x):
e−xy′−e−xy=e−x1
x. 5. Recognize that the left side of the equation is the derivative of the
product e−xywith respect to x. 6. Integrate both sides to get the general solution: e−xy=
∫e−x1
xdx +C. 7. Evaluate the integral on the right side using integration by parts: e−xy=
−e−x(1 + x)ln |x|+C. 8. Solve for yto get the final solution: y=−1−x+Cexln |x|.
Therefore, the solution to the differential equation y′=1
x+yis y=−1−x+Cexln |x|, where
Cis an arbitrary constant.
13. Question: Solve the following first-order nonlinear differential equation:
dy
dx =2x+y
x+y
Ans. Solution: 1. Let’s rewrite the given differential equation as:
dy
dx = 1 + y−x
x+y
2. Now, we can substitute v=x+yand rewrite the differential equation in terms of v:
dv
dx =dy
dx + 1 = 1 + y−x
v
3. We can further simplify the equation as:
vdv
dx =v+y−x
4. To solve this separable differential equation, we arrange terms to get:
vdv = (v+x)dx
5. Integrating both sides, we get:
∫vdv =∫(v+x)dx
6. Solving the integrals gives us:
v2
2=v2
2+x2
2+C
7. Simplify the equation to get:
v2=v2+x2+ 2C
8. Rearranging terms yields:
x2= 2C
9. Therefore, the solution to the differential equation is given by:
x2= 2C
14. Question: Solve the differential equation y′′ −4y′+ 4y= 0 with the initial conditions
y(0) = 2 and y′(0) = 1.
Ans. Step-by-step solution:
1. Write down the characteristic equation by substituting y=erx into the differential equation
y′′ −4y′+ 4y= 0:
r2−4r+ 4 = 0
(r−2)2= 0
Since the characteristic equation has a repeated root r= 2, the general solution can be written
as
y(x) = (c1+c2x)e2x
2. Apply the initial condition y(0) = 2:
2 = c1e0⇒c1= 2
3. Differentiate y(x)with respect to xto find y′(x):
y′(x) = (2 + c2)e2x
4. Apply the initial condition y′(0) = 1:
1 = (2 + c2)e0⇒c2=−1
5. Substitute c1= 2 and c2=−1back into the general solution to get the particular solution:
y(x) = (2 −x)e2x
Therefore, the solution to the differential equation with the given initial conditions is y(x) =
(2 −x)e2x.
15. Question 15: Solve the initial value problem
y(4) −8y′′ + 16y= 0, y(0) = 1, y′(0) = 0, y′′(0) = 0, y′′′(0) = 0.
Ans. Solution: We first find the characteristic equation of the differential equation by assuming
the solution has the form y=ert:
r4−8r2+ 16 = 0.
1. Solve the characteristic equation:
r4−8r2+ 16 = 0.
(r2−4)2= 0.
r2−4 = 0.
r=±2.
Hence, the general solution to the differential equation is:
y(t) = c1e2t+c2te2t+c3e−2t+c4te−2t.
2. Substitute the initial conditions to find the specific solution: Given: y(0) = 1, y′(0) =
0, y′′(0) = 0, y′′′(0) = 0.
Plugging in the initial conditions yields the following equations:
y(0) = c1+c3= 1,
y′(0) = 2c1+ 2c2−2c3−2c4= 0,
y′′(0) = 4c1+ 4c2+ 4c3+ 4c4= 0,
y′′′(0) = 8c1−8c2+ 8c3−8c4= 0.
Solving the system of equations gives c1=1
4, c2=−1
4, c3=3
4, c4=−1
4. Therefore, the
solution to the initial value problem is:
y(t) = 1
4e2t−1
4te2t+3
4e−2t−1
4te−2t.
16. Let y(x)be the solution to the differential equation dy
dx = 2xy2with initial condition
y(0) = 1. Find the value of y(1).
Ans. Solution: 1. Given the differential equation dy
dx = 2xy2, we can separate variables to
obtain: dy
y2= 2xdx
2. Integrating both sides gives: ∫1
y2dy =∫2xdx
−1
y=x2+C
3. Use the initial condition y(0) = 1 to find the value of the constant C:
−1
1= 02+C
C=−1
4. Hence, the solution to the differential equation is −1
y=x2−1. 5. To find y(1), substitute
x= 1 into the equation:
−1
y(1) = 12−1
−1
y(1) = 0
y(1) = ∞
Therefore, the value of y(1) is ∞.
17. Question: Solve the following first-order ordinary differential equation:
dy
dx =y
x+x2
Ans. Step-by-step solution:
1. Rewriting the given equation in standard form, we have:
dy
dx =y
x+x2
This can be rewritten as:
dy
dx −y
x=x2
2. We now identify the integrating factor, which is given by:
I(x) = e∫−1
xdx
Calculating the integral, we get:
I(x) = e−ln |x|=1
|x|
So the integrating factor is 1
|x|.
3. Now we multiply both sides of the differential equation by the integrating factor:
1
|x|
dy
dx −1
|x|
y
x=x21
|x|
This simplifies to:
1
|x|
dy
dx −1
x2y=x
|x|
4. Recognizing the left side as the derivative of the product of the integrating factor and y,
we can rewrite the equation as:
d
dx (1
|x|y)=x
|x|
5. Integrate both sides with respect to x to solve for y:
∫d
dx (1
|x|y)dx =∫x
|x|dx
1
|x|y=∫sgn(x)dx =|x|+C
where sgn(x)is the sign function and C is the constant of integration.
6. Finally, solving for y, we have:
y=|x|2+C|x|
where Cis an arbitrary constant.
18. Find the general solution to the differential equation: dy
dx +y=e−x.
Ans. To solve the given differential equation, we can use an integrating factor method. Let
µ(x) = e∫1dx =ex. Multiply both sides of the differential equation by µ(x):
exdy
dx +exy=exe−x
This can be rewritten as: d
dx(exy) = 1
Integrating both sides with respect to xgives:
exy=∫1dx =x+C
Therefore, the general solution to the given differential equation is:
y=e−x(x+C)
where Cis an arbitrary constant.
19. Question: Solve the initial value problem dy
dx =3x2+4y2
2xy with initial condition y(1) = 2.
Ans. Step-by-step solution:
1. Rewrite the given differential equation in a more standard form by multiplying through by
dx and separating variables:
dy
dx =3x2+ 4y2
2xy =⇒2y dy = (3x2+ 4y2)dx
x
2. Next, divide through by 2y(3x2+ 4y2)to separate variables further:
2
3x2+ 4y2dy =dx
x
3. Integrate both sides with respect to their respective variables:
∫2
3x2+ 4y2dy =∫1
xdx
4. For the left integral, use substitution (u= 3x2+ 4y2, du = 8y dy):
1
8∫1
udu =1
8ln |u|=1
8ln |3x2+ 4y2|
5. For the right integral, integrate ∫1
xdx =ln |x|
6. Combining the integrals and adding the constant of integration C, we have:
1
8ln |3x2+ 4y2|=ln |x|+C
7. Applying the initial condition y(1) = 2:
1
8ln |3(1)2+ 4(2)2|=ln |1|+C
1
8ln |19|= 0 + C
C=1
8ln(19)
8. Finally, substitute Cback into the equation to find the particular solution:
1
8ln |3x2+ 4y2|=ln |x|+1
8ln(19)
ln |3x2+ 4y2|= 8 ln |x|+ln(19)
3x2+ 4y2=Ax8·19 (where A is a constant)
20. Question: Find the general solution to the differential equation: y′′ −2y′+y=exsin(x).
Ans. Solution: 1. First, we find the complementary solution by solving the characteristic
equation r2−2r+ 1 = 0. The characteristic equation has a double root at r= 1, so the
complementary solution is of the form yc=C1ex+C2xex.
2. Next, we find a particular solution ypfor the non-homogeneous part exsin(x). Notice
that the right side of the differential equation consists of terms that can be solutions to the
complementary part (polynomial multiplied by ex) and sin(x)which can be part of the general
solution for yp. Since sin(x)is a part of the particular solution, we try a particular solution of
the form:
yp= (Ax +B)exsin(x)
3. We differentiate ypwith respect to xto find y′
pand y′′
p:
y′
p= (A+Ax +B)exsin(x)+(Ax +B)excos(x)
y′′
p= (2A+Ax +B)exsin(x) + (2A+ 3Ax +B−Ax −B)excos(x)
Simplify the expressions:
y′
p= (A+Ax +B)exsin(x)+(Ax +B)excos(x)
y′′
p= (2A+Ax +B)exsin(x)+(A+ 2Ax +B)excos(x)
4. Substitute yp,y′
p, and y′′
pinto the original differential equation y′′ −2y′+y=exsin(x).
We have:
(2A+Ax+B)exsin(x)+(A+2Ax+B)excos(x)−2[(A+Ax+B)exsin(x)+(Ax+B)excos(x)]+(Ax+B)exsin(x) = exsin(x)
5. Simplify the equation and equate the coefficients of similar terms on both sides of the
equation to solve for Aand B. After solving, we find A=−1
4and B=3
8.
6. Therefore, the particular solution is: yp=−1
4xexsin(x) + 3
8exsin(x)
7. The general solution to the given differential equation is the sum of the complementary
solution and the particular solution:
y=yc+yp=C1ex+C2xex−1
4xexsin(x) + 3
8exsin(x)
where C1and C2are arbitrary constants.
21. Question 21: Solve the following first-order differential equation with an integrating factor.
y′+2
xy=x3
Ans. To solve the given first-order differential equation with an integrating factor, follow these
steps:
1. Write the given differential equation in the form y′+P(x)y=Q(x). 2. Find the
integrating factor I(x) = e∫P(x)dx. 3. Multiply the integrating factor with both sides of the
differential equation. 4. Rewrite the equation in the form of a product rule. 5. Integrate both
sides and solve for the unknown function.
1. Write the given differential equation: The given differential equation is y′+2
xy=x3.
2. Find the integrating factor: The integrating factor is given by
I(x) = e∫P(x)dx
For the differential equation y′+2
xy=x3,P(x) = 2
xSo, the integrating factor is
I(x) = e∫2
xdx =e2ln |x|=eln(x2)=x2
3. Multiply the integrating factor: Multiply both sides of the differential equation by the
integrating factor x2:
x2y′+ 2xy =x5
4. Rewrite the equation: Rewrite the equation in the form of a product rule:
d
dx(x2y) = x5
5. Solve the equation: Integrate both sides with respect to x:
∫d
dx(x2y)dx =∫x5dx
x2y=1
6x6+C
y=1
6x4+C
x2
Therefore, the solution to the first-order differential equation y′+2
xy=x3is y=1
6x4+C
x2,
where Cis the constant of integration.
22. Find the general solution to the differential equation: y′′ −y′−6y= 0.
Ans. To solve the differential equation y′′ −y′−6y= 0, we first find the characteristic equation
by assuming y=ert: 1. The characteristic equation is r2−r−6 = 0. 2. Solve the characteristic
equation to find the roots: r= 3 and r=−2. 3. The general solution is y(t) = c1e3t+c2e−2t,
where c1and c2are arbitrary constants.
23. Question: Solve the differential equation (1 + x2)y′+xy =x3, given that y(1) = 3.
Ans. Solution:
1. First, let’s rewrite the given differential equation in standard form:
(1 + x2)y′+xy =x3
Divide through by (1 + x2)to get:
y′+x
1+x2y=x3
1+x2.
2. Now, we identify the integrating factor, which is e∫x
1+x2dx.
The integrating factor is µ(x) = e∫x
1+x2dx.
3. Let’s find the integral and simplify the integrating factor:
∫x
1+x2dx =1
2ln(1 + x2) + C.
Thus, the integrating factor becomes µ(x) = e1
2ln(1+x2).
Simplifying the integrating factor gives µ(x) = (1 + x2)1/2.
4. Multiply the differential equation by the integrating factor:
(1 + x2)1/2y′+x(1 + x2)1/2y=x3(1 + x2)1/2.
5. Rewrite the equation in terms of the product rule for differentiation:
d
dx [(1 + x2)1/2y]=x3(1 + x2)1/2.
Integrating both sides gives:
(1 + x2)1/2y=∫x3(1 + x2)1/2dx +C.
6. Simplify the integral on the right side:
∫x3(1 + x2)1/2dx =1
5(1 + x2)5/2 +C.
Therefore, the solution to the differential equation is:
(1 + x2)1/2y=1
5(1 + x2)5/2 +C.
7. Now, apply the initial condition y(1) = 3 to find the value of the constant C:
(1 + 12)1/2 ·3 = 1
5(1 + 12)5/2 +C.
2·3 = 1
5·25+C.
6 = 32
5+C.
C= 6 −32
5=3
5.
8. Finally, substitute the value of Cback into the solution:
(1 + x2)1/2y=1
5(1 + x2)5/2 +3
5.
Therefore, the solution to the differential equation with the initial condition is:
y=1
5(1 + x2)2+3
5(1+x2)1/2 .
24. Question: Solve the following differential equation using the method of undetermined
coefficients:
y′′ −4y′+ 4y= 6e2x
Ans. Step-by-step solution:
1. Find the complementary function by solving the characteristic equation r2−4r+ 4 = 0:
The characteristic equation is (r−2)2= 0, so we have a repeated root r= 2. The complementary
function is then:
yc=c1e2x+c2xe2x
2. Assume the particular solution has the form yp=Axe2x. Calculate the derivatives needed
for substitution:
y′
p= (A+ 2Ax)e2x
y′′
p= (2A+ 4Ax)e2x
3. Substitute yp, y′
p, y′′
pinto the differential equation:
(2A+ 4Ax)e2x−4(A+ 2Ax)e2x+ 4Axe2x= 6e2x
4. Simplify and equate coefficients of e2xon both sides:
2A−4A−4A= 6
−6A= 6
A=−1
5. The particular solution is yp=−xe2x.
6. The general solution is the sum of the complementary and particular solutions:
y=yc+yp=c1e2x+c2xe2x−xe2x
y= (c1−1)e2x+c2xe2x
25. Question 25: Solve the differential equation dy
dx −2y= 5x.
Ans. Solution:
1. First, we identify the differential equation as a first-order linear differential equation in the
form dy
dx +P(x)y=Q(x), where P(x) = −2and Q(x) = 5x.
2. To solve this type of differential equation, we use an integrating factor µ(x) = e∫P(x)dx.
3. In our case, P(x) = −2, so the integrating factor µ(x) = e∫−2dx =e−2x.
4. Multiply the differential equation by the integrating factor: e−2xdy
dx −2e−2xy= 5xe−2x.
5. Rewrite the left side as a derivative of the product of yand the integrating factor using
the product rule: d
dx (ye−2x) = 5xe−2x.
6. Integrate both sides with respect to x:∫d
dx (ye−2x)dx =∫5xe−2xdx.
7. Simplify the integrals: ye−2x=−5
2xe−2x−5
2∫e−2xdx.
8. Integrate ∫e−2xdx =−1
2e−2x. Substitute this back into the equation.
9. Solve for y:y=−5
2x−5
4+Ce2x, where Cis the constant of integration.
Therefore, the general solution to the differential equation is y=−5
2x−5
4+Ce2x.
26. Question 26: Solve the following differential equation using the method of undetermined
coefficients: y′′ −2y′+y=ex.
Ans. Solution: 1. First, we write down the characteristic equation for the homogeneous part
of the differential equation:
r2−2r+ 1 = 0
(r−1)2= 0
This gives us a repeated root of r= 1. So, the solution to the homogeneous equation is:
yh(x) = c1ex+c2xex
2. Next, we guess a particular solution to the given nonhomogeneous equation in the form of
yp(x) = Aex. We substitute this guess back into the differential equation:
y′′ −2y′+y=ex
(Aex)′′ −2(Aex)′+Aex=ex
Aex−2Aex+Aex=ex
Aex=ex
This gives us A= 1.
3. Therefore, the particular solution is:
yp(x) = ex
4. Combining the general solution of the homogeneous part and the particular solution, we
have:
y(x) = yh(x) + yp(x)
y(x) = c1ex+c2xex+ex
y(x) = (c1+ 1)ex+c2xex
where c1and c2are constants.
Thus, the solution to the differential equation y′′ −2y′+y=exis y(x) = (c1+ 1)ex+c2xex.
27. Question 27: Solve the following initial value problem using the method of undetermined
coefficients:
y′′ −y′−6y= 12e3x, y(0) = 0, y′(0) = 0
Ans. Step 1. Find the characteristic equation by setting the homogeneous part of the
differential equation to zero:
y′′ −y′−6y= 0
The characteristic equation is r2−r−6 = 0, which factors as (r−3)(r+ 2) = 0. Therefore,
the characteristic roots are r1= 3 and r2=−2.
Step 2. Write the general solution to the homogeneous equation:
yh(x) = c1e3x+c2e−2x
Step 3. Determine a particular solution for the nonhomogeneous equation using the method
of undetermined coefficients. Since the right-hand side term is 12e3x, assume a particular solution
of the form yp(x) = Ae3x.
Step 4. Substitute the trial particular solution into the differential equation and solve for A:
9A−3A−6A= 12 ⇒0 = 12
Since the equation has no solution, we need to modify our trial particular solution. Try yp(x) =
Axe3x.
Step 5. Substitute the new trial particular solution into the differential equation and solve
for A:
9Ae3x−3Ae3x−6Axe3x= 12e3x
This simplifies to 6A= 12, so A= 2.
Step 6. Write down the particular solution: yp(x) = 2xe3x.
Step 7. Combine the general solution and the particular solution to get the final solution:
y(x) = yh(x) + yp(x) = c1e3x+c2e−2x+ 2xe3x
Step 8. Apply the initial conditions to solve for c1and c2:
y(0) = c1+c2= 0
y′(0) = 3c1−2c2+ 6 = 0
Solving these equations simultaneously, we find c1=4
5and c2=−4
5.
Step 9. Therefore, the solution to the initial value problem is
y(x) = 4
5e3x−4
5e−2x+ 2xe3x
28. Question: Solve the following initial value problem:
y′+ 2xy = 4x, y(0) = 3.
Ans. Step-by-step solution: 1. First, we need to find the integrating factor. The integrating
factor is given by e∫2x dx =ex2. 2. Multiplying both sides of the given differential equation by
the integrating factor ex2, we get:
ex2y′+ 2xex2y= 4xex2.
3. Recognize that the left side is the derivative of the product ex2y. So, we rewrite the equation
as:
d(ex2y)
dx = 4xex2.
4. Integrate both sides with respect to xto solve for y:
∫d(ex2y) = ∫4xex2dx.
⇒ex2y= 2ex2+C,
where Cis the constant of integration. 5. Now, solve for yby dividing by ex2:
y= 2 + Ce−x2.
6. To find the value of C, use the initial condition y(0) = 3:
3 = 2 + C·e−0,
3 = 2 + C.
Therefore, C= 1. 7. The solution to the initial value problem is:
y= 2 + e−x2.
29. Question: Solve the initial value problem y′=y2+cos(x)with y(0) = 1.
Ans. Step-by-step solution: 1. First, we note that the given differential equation is separable
since all terms involving yare on one side and terms involving xare on the other side. So, we can
rewrite the equation as dy
dx =y2+cos(x). 2. Next, we separate variables by dividing both sides
of the equation by (y2+cos(x)) to get 1
y2+cos(x)dy =dx. 3. Integrate both sides with respect
to x:∫1
y2+cos(x)dy =∫dx
4. The integral on the left side can be evaluated using a trigonometric substitution. Let u=
sin(x). Then, du =cos(x)dx. Substituting these in, we get:
∫1
y2+cos(x)dy =∫1
y2+ 1 −u2dy =∫1
y2+ 1 −sin2(x)dy =∫1
y2+cos2(x)dy =∫1
y2+ (1 −u2)dy =∫1
y2+u2dy
5. This integral can be evaluated using a standard trigonometric substitution. Let u=ytan(v),
then du =tan(v)dv. Substituting these in, we get:
∫1
y2+u2dy =∫1
y2+y2tan2(v)dy =∫1
y2(1 + tan2(v))dy =∫1
y2sec2(v)dy =∫cos2(v)dv
6. The integral on the right side evaluates to v
2+sin(2v)
4+C, where Cis the constant of
integration. 7. Substitute back u=sin(x)and v=arctan (y
sin(x)). Therefore, the integral
simplifies to arctan(y
sin(x))
2+sin(2arctan(y
sin(x)))
4+C. 8. Finally, substitute back y(0) = 1 to determine
the value of C. We find C=π
4. Hence, the solution to the initial value problem is y(x) =
arctan(y
sin(x))
2+sin(2arctan(y
sin(x)))
4+π
4.
30. Question:
Solve the following first-order differential equation:
dy
dx =3y2
x
Ans. Solution:
1. Rewrite the differential equation in separable form by bringing all terms involving yto one
side and all terms involving xto the other side:
dy
y2=3
xdx
2. Integrate both sides with respect to their respective variables:
∫1
y2dy =∫3
xdx
3. The left-hand side integrates to −1
yand the right-hand side integrates to 3ln |x|+C,
where Cis the constant of integration.
4. Combining the results of the integrations, we get:
−1
y= 3 ln |x|+C
5. Solving for y, we have:
y=−1
3ln |x|+C
where Cis the constant of integration.
31. Question: Solve the following differential equation:
y′′ −2y′+ 2y=excos x
Ans. Step-by-step solution: 1. First, find the complementary function by solving the homoge-
neous equation y′′ −2y′+ 2y= 0. The characteristic equation is r2−2r+ 2 = 0. The roots of
the characteristic equation are r= 1 ±i. Therefore, the complementary function is given by
yc(x) = ex(Acos x+Bsin x)
where Aand Bare constants. 2. Next, find a particular solution for the given non-homogeneous
equation using the method of undetermined coefficients. Since excos xis a product of exponential
and trigonometric functions, assume
yp(x) = (C1ex+C2ex)cos x+ (C3ex+C4ex)sin x
where C1,C2,C3, and C4are unknown coefficients to be determined. 3. Substitute yp(x)into
the original differential equation and solve for the unknown coefficients by equating coefficients
of like terms. 4. Plug the values of the coefficients back into yp(x)to get the particular solution.
5. The general solution of the differential equation is the sum of the complementary function
yc(x)and the particular solution yp(x):
y(x) = yc(x) + yp(x)
y(x) = ex(Acos x+Bsin x)+(C1ex+C2ex)cos x+ (C3ex+C4ex)sin x
where A,B,C1,C2,C3, and C4are constants determined from the initial conditions or boundary
conditions.
32. Question 32: Solve the following initial value problem: dy
dx =xe−y, where y(0) = 1.
Ans. Solution: 1. Separate the variables by moving all terms involving yto the left side and
all terms involving xto the right side:
dy
ey=xdx
2. Integrate both sides with respect to their respective variables:
∫dy
ey=∫xdx
3. Integrate 1
eywith respect to yusing substitution (u=ey, du =eydy):
∫1
eydy =∫1
udu =ln |u|+C=ln |ey|+C=y+C
4. Integrate xwith respect to x:
∫xdx =1
2x2+C
5. Combining both results, we get:
y=1
2x2+C
6. To find the value of C, use the initial condition y(0) = 1:
1 = 1
2(0)2+C=⇒C= 1
7. Therefore, the solution to the initial value problem is y=1
2x2+ 1.
33. Find the general solution to the differential equation: dy
dx =y2+1
x.
Ans. Solution: 1. Rearrange the differential equation so that all terms involving yare on one
side and all terms involving xare on the other side:
dy
y2+ 1 =dx
x
2. Integrate both sides with respect to x:
∫dy
y2+ 1 =∫dx
x
3. The integral on the left side can be solved using a trigonometric substitution. Let y=
tan(θ)so that dy =sec2(θ)dθ:
∫1
tan2(θ)+1sec2(θ)dθ =∫1
sec2(θ)sec2(θ)dθ
∫sec2(θ)dθ =∫dθ =θ
4. Re-substitute y=tan(θ)back in:
θ=arctan(y)
5. Therefore, the left integral becomes:
∫dy
y2+ 1 =arctan(y) + C1
6. The right integral is a simple logarithmic function:
∫dx
x=ln |x|+C2
7. Combining both sides, we have:
arctan(y) + C1=ln |x|+C2
8. Solving for y, the general solution to the differential equation is:
y=tan(ln |x|+˜
C)
where ˜
C=C2−C1.
34. Find the general solution to the differential equation:
dy
dx =x2+ 2y
x
Ans. Solution: 1. Rewrite the given differential equation in the standard form by moving all
terms involving yto one side:
xdy
dx −2y=x2
2. Now, we observe that this is a first-order linear differential equation in the form of:
dy
dx +P(x)y=Q(x)
where P(x) = −2and Q(x) = x2.
3. To solve this linear differential equation, we first find the integrating factor I(x)defined
as:
I(x) = e∫P(x)dx
4. Substituting P(x) = −2into the formula, we get:
I(x) = e∫−2dx =e−2x
5. We then multiply the integrating factor to the original equation to get:
e−2x·(xdy
dx −2y) = e−2x·x2
6. By the product rule for differentiation, the left-hand side simplifies to:
d
dx(e−2xy) = x2e−2x
7. Integrating both sides with respect to x, we obtain:
e−2xy=∫x2e−2xdx
e−2xy=−1
2xe−2x−1
4e−2x+C
8. Multiplying by e2xgives us the general solution to the differential equation:
y=−1
2x−1
4+Ce2x
where Cis an arbitrary constant.
35. Question: Solve the following initial value problem:
y′+2x
x2+ 1y=x2, y(0) = 3
Ans. Solution: 1. This is a first-order linear differential equation in the form y′+P(x)y=Q(x),
where P(x) = 2x
x2+1 and Q(x) = x2. 2. To solve the differential equation, we first find the
integrating factor. The integrating factor is given by µ(x) = e∫P(x)dx. 3. Calculate ∫P(x)dx:
∫2x
x2+ 1dx =∫2x
x2+ 1 ·x2+ 1
x2+ 1dx =∫2x(x2+ 1)
x2+ 1 dx =∫2xdx =x2+C
4. Therefore, the integrating factor is µ(x) = ex2. 5. Multiply the differential equation by the
integrating factor µ(x):
ex2y′+2x
x2+ 1ex2y=x2ex2
6. Notice that the left side is the derivative of the product ex2y, so we rewrite the equation as:
(ex2y)′=x2ex2
7. Integrate both sides with respect to x:
∫(ex2y)′dx =∫x2ex2dx
ex2y=∫x2ex2dx +C
8. Calculate ∫x2ex2dx using substitution (u=x2, du = 2xdx):
∫x2ex2dx =1
2∫eudu =1
2eu+C=1
2ex2+C
9. Substitute this result back into the previous equation:
ex2y=1
2ex2+C
10. Solve for y:
y=1
2+Ce−x2
11. Apply the initial condition y(0) = 3 to find the value of C:
3 = 1
2+Ce0=1
2+C
C=5
2
12.
Therefore,
the
solution
to
the
initial
value
problem
is
_
12. Therefore, the solution to the initial value problem is y=1
2+5
2e−x2.
