Differential Equations Practice Material 5

MAT 142 - COLLEGE MATHEMATICS - Differential

Equations Practice Material - Set 5

1. Question: Solve the following first-order differential equation:

dx =x2+y2

Ans. Step-by-step solution:

1. Rewrite the given differential equation in standard form:

dx =x2+y2

2. Notice that this is a separable differential equation. Separate the variables by moving all

terms involving y to one side and all terms involving x to the other side:

y2=x2dx

3. Integrate both sides with respect to their respective variables:

y2dy =Zx2dx

4. Solve the integrals on each side:

−1

y=x3

3+C

where Cis the constant of integration.

5. Solve for yby multiplying through by −1:

y=−1

3+C=3

x3+ 3C

2. Question:

Consider the differential equation y′+2xy =e−x2. Find the general solution to the differential

equation.

Ans. Step-by-step solution:

1. First, let’s rewrite the differential equation in the standard form of first-order linear ODE:

y′+P(x)y=Q(x). In this case, P(x) = 2xand Q(x) = e−x2.

2. To find the integrating factor, we calculate e∫P(x)dx. In this case, e∫2x dx =ex2.

3. Multiply the integrating factor by the original ODE to get:

ex2y′+ 2xex2y=ex2e−x2

Simplify this to:

(ex2y)′= 1

4. Integrate both sides with respect to xto get:

ex2y=x+C

where Cis the constant of integration.

5. Finally, divide by ex2to solve for y:

y=x

ex2+C

ex2

which simplifies to:

y=xe−x2+Ce−x2

Thus, the general solution to the differential equation is y=xe−x2+Ce−x2, where Cis an

arbitrary constant.

3. Question: Solve the following first-order linear differential equation:

y′+ 2xy = 4x2,

where y(0) = 3.

Ans. Step-by-step solution: 1. To solve the given first-order linear differential equation, we first

need to rearrange it into the standard form y′+P(x)y=Q(x). In this case, P(x) = 2xand

Q(x)=4x2. 2. We can find the integrating factor µ(x)by computing µ(x) = e∫P(x)dx. Thus,

we have:

µ(x) = e∫2xdx =ex2.

3. Multiply both sides of the differential equation by the integrating factor µ(x):

ex2y′+ 2xex2y= 4x2ex2.

4. Notice that the left-hand side of the equation can be written as the derivative of the product

of µ(x)and y:

dx(ex2y) = 4x2ex2.

5. Integrate both sides with respect to x:

dx(ex2y)dx =Z4x2ex2dx.

6. This simplifies to:

ex2y=Z4x2ex2dx +C.

7. Solve the integral on the right-hand side:

Z4x2ex2dx = 2ex2(2x2−1) + K,

where Kis a constant of integration. 8. Substitute this back into the equation:

ex2y= 2ex2(2x2−1) + K.

9. Finally, solve for yby dividing both sides by ex2and simplifying:

y= 2(2x2−1) + K

ex2.

10. Apply the initial condition y(0) = 3 to find the value of K:

3 = 2(0) −1 + K

e0=⇒K= 4.

Therefore, the solution to the initial value problem is:

y= 2(2x2−1) + 4

ex2.

4. Question:

Solve the following differential equation: dy

dx = (x2+y2)1/2 with the initial condition y(0) = 1.

Ans. Solution:

1. Separate variables to get all yterms on one side and all xterms on the other side:

(x2+y2)1/2 =dx

2. Integrate both sides. Let’s perform a substitution where y=xtan(θ):R1

(x2+x2tan2(θ))1/2 xsec(θ)tan(θ)dθ =

Rdx

3. Simplify the integrand: Rxsec(θ)tan(θ)

(x2(1+tan2(θ)))1/2 dθ =Rdx

4. Further simplify the integrand to: Rxsec(θ)

x(1+tan2(θ))1/2 dθ =Rdx

5. After simplifying, we get: Rsec(θ)dθ =Rdx

6. Integrating both sides, we get: ln |sec(θ) + tan(θ)|=x+C

7. Substitute back x=ycot(θ)and simplify: ln |sec(θ) + tan(θ)|=ycot(θ) + C

8. Apply the initial condition y(0) = 1 to find the value of C:ln |1|= 1 ·cot(θ) + C

9. Solve for C:C= 0

10. Thus, the solution to the differential equation is: ln |sec(θ) + tan(θ)|=ycot(θ)

5. Question: Solve the differential equation (y′)2−y′′ = 0.

Ans. Solution: 1. Let v=y′, so the differential equation becomes v2−v′= 0. 2. Rewrite

the equation as v(v−1) = 0. 3. Solve for vby setting each factor to zero: v= 0 or v= 1. 4.

For v= 0, integrate vwith respect to xto find y:y=R0dx =C1, where C1is an arbitrary

constant. 5. For v= 1, integrate vwith respect to xto find y:y=R1dx =x+C2, where C2

is an arbitrary constant. 6. Therefore, the general solution to the differential equation is y=C1

or y=x+C2, where C1and C2are arbitrary constants.

6. Question: Solve the differential equation y′=x

y2.

Ans. Step-by-step solution:

1. Rewrite the given differential equation as:

y2dy

dx =x

2. Separate variables by moving all terms involving yto the left side and all terms involving

xto the right side:

y2dy =x dx

3. Integrate both sides:

Zy2dy =Zx dx

4. Evaluate the integrals:

3=x2

2+C

where Cis the constant of integration.

5. Solve for y:

y3=3x2

2+C

6. Finally, solve for yby taking the cube root of both sides:

y=3

r3x2

2+C

So the solution to the differential equation y′=x

y2is y=3

q3x2

2+C, where Cis the constant

of integration.

7. Question: Find the general solution to the differential equation y′′ −4y′+ 4y= 0.

Ans. Step 1. Write down the characteristic equation by assuming a solution of the form

y=ert. The characteristic equation is: r2−4r+ 4 = 0.

Step 2. Solve the characteristic equation to find the roots r1and r2. Factorizing the

characteristic equation gives: (r−2)2= 0. Therefore, the repeated root is r= 2.

Step 3. Write down the general solution to the differential equation. Since we have a repeated

root, the general solution is: y(t) = (c1+c2t)e2t, where c1and c2are arbitrary constants.

8. Question: Solve the differential equation y′′ −5y′+ 6y=e3x.

Ans. Step-by-step solution: 1. First, we solve the homogenous equation y′′ −5y′+ 6y= 0.

The characteristic equation is r2−5r+ 6 = 0, which factors as (r−2)(r−3) = 0. Therefore,

the homogenous solution is given by yh=c1e2x+c2e3x. 2. To find a particular solution for

the non-homogenous equation, we use the method of undetermined coeﬀicients and assume a

particular solution in the form yp=Ae3x. Plugging this into the differential equation, we get

9A−15A+ 6Ae3x=e3x. Solving for Agives A=1

6. 3. Therefore, the general solution is

y=yh+yp=c1e2x+c2e3x+1

6e3x.

9. Find the general solution to the differential equation: y′′ −4y′+ 4y=e2x.

Ans. Solution: 1. First, we find the characteristic equation by solving r2−4r+ 4 = 0:

r2−4r+ 4 = (r−2)2= 0

This gives us a repeated root of r= 2.

2. Therefore, the complementary function ycis given by:

yc(x) = c1e2x+c2xe2x

3. Next, we find a particular solution for the non-homogeneous part yp. Since e2xis of the

form f(x) = emx where mis equal to the repeated root 2, we use the method of undetermined

coeﬀicients with yp=Ax2e2x.

4. Substitute ypinto the differential equation and solve for A:

y′′

p−4y′

p+ 4yp= 2Ae2x−8Axe2x+ 4Ax2e2x=e2x

4Ax2−8Ax + 2A= 1

5. Equate coeﬀicients and solve for A, we have:

4A= 0

−8A= 0

2A= 1

Therefore, A=1

6. Thus, the particular solution is yp(x) = 1

2x2e2x.

7. The general solution to the differential equation is:

y(x) = yc(x) + yp(x) = c1e2x+c2xe2x+1

2x2e2x

where c1and c2are arbitrary constants.

10. Question: Solve the following first-order separable differential equation:

dx =2x+ 3y

x+ 2y

Ans. Solution: 1. Rewrite the differential equation in the form dy

dx =M(x,y)

N(x,y):

dx =2x+ 3y

x+ 2y

2. Rearrange the equation to separate variables:

2y−dx

x=3

3. Integrate both sides of the equation with respect to x:

2ydy −Z1

xdx =Z3

2dx

4. Solve the integrals: 1

2ln |y| − ln |x|=3

2x+C

5. Combine the natural logarithms and simplify:

ln 



√x



= 3x+C

6. Exponentiate both sides of the equation:

√x=e3x+C=Ce3x

7. Solve for y:

y=Ce3x√x

Therefore, the solution to the differential equation is y=Ce3x√x, where Cis an arbitrary

constant.

11. Question: Solve the following differential equation using the method of undetermined

coeﬀicients:

y′′ −5y′+ 6y= 4e2x

Ans. Step-by-step solution: 1. First, we find the complementary function yc(x)by solving the

homogeneous equation y′′ −5y′+ 6y= 0. The characteristic equation is r2−5r+ 6 = 0, which

factors as (r−2)(r−3) = 0. So the complementary function is yc(x) = c1e2x+c2e3x, where c1

and c2are arbitrary constants.

2. To find the particular solution, we assume yp(x) = Ae2x, where Ais a constant to

be determined. Since 2xis a solution in yc, we multiply by xto avoid overlap. Substitute

yp(x) = Axe2xinto the differential equation:

(4A−10A)e2x+ 6Axe2x= 4e2x

Simplify to get −6A= 4 and solve for Ato get A=−2

3. Therefore, yp(x) = −2

3xe2x.

3. The general solution is the sum of the complementary function and the particular solution:

y(x) = yc(x) + yp(x) = c1e2x+c2e3x−2

3xe2x

12. Question: Solve the following initial value problem: dy

dx =sin(2x+y),y(0) = 0.

Ans. Step-by-step solution: 1. Let’s first separate variables by writing:

dx =sin(2x+y)

sin(2x+y)=dx

2. Perform a substitution to simplify the integration. Let u= 2x+y. Then du = 2dx which

means 1

2du =dx. We can rewrite the above equation as:

sin u=1

2du

3. Integrate both sides with respect to their respective variables:

Zdy

sin u=1

2Zdu

−ln |csc u−cot u|=u

2+C

4. Return to xand yvariables by substituting back u= 2x+y:

−ln |csc(2x+y)−cot(2x+y)|=2x+y

2+C

−ln |csc(2x+y)−cot(2x+y)|=x+y

2+C

5. Apply the initial condition y(0) = 0 to solve for the constant C:

−ln |csc(0) −cot(0)|= 0 + 0

2+C

−ln |1|=C

C= 0

6. Plug this back into the solution to obtain the final answer:

−ln |csc(2x+y)−cot(2x+y)|=x+y

ln |csc(2x+y)−cot(2x+y)|=−x−y

|csc(2x+y)−cot(2x+y)|=e−xe−y

csc(2x+y)−cot(2x+y) = ±e−xe−y

This is the general solution to the given initial value problem.

13. Question: Solve the following differential equation: dy

dx =x2+y2

xy .

Ans. Solution: 1. Rearrange the equation to separate variables: dy

dx =x2+y2

xy ⇒dy

dx =x

y+y

x. 2.

Rewrite the equation in terms of dy and dx:dy

y=dx

x+dx

y. 3. Integrate both sides with respect

to x:R1

ydy =R1

xdx +R1

ydx. 4. Solve the integrals: ln |y|=ln |x|+ln |y|+C. 5. Combine

the logarithms: ln |y|=ln |x|+ln |y|+ln |y2|=ln |x|+ln |y3|. 6. Remove the logarithms by

taking the exponential of both sides: |y|=|x·y3|. 7. Rewrite the absolute value as two cases:

y=x·y3or y=−x·y3. 8. Solve the two separable equations: a) y=x·y3:1 = y2,y=±1.

b) y=−x·y3:−1 = y2,y=±i. 9. The general solution consists of all possible solutions:

y=x,y=−x,y=i,y=−i. Therefore, the solutions to the differential equation are y=x,

y=−x,y=i, and y=−i.

14. Question 14: Solve the following second-order differential equation with non-constant

coeﬀicients:

y′′ −2xy′+ 2y= 0

Ans. Solution: 1. Write the given differential equation in the standard form:

y′′ −2xy′+ 2y= 0

2. Let’s assume a solution of the form y(x) = P∞

n=0 anxn.

3. Compute the first and second derivatives of y(x):

y′(x) =

∞

n=0

annxn−1

y′′(x) =

∞

n=0

ann(n−1)xn−2

4. Substitute y,y′, and y′′ into the differential equation:

∞

n=0

ann(n−1)xn−2−2x

∞

n=0

annxn−1+ 2

∞

n=0

anxn= 0

5. Rearrange terms and combine like terms:

∞

n=0

ann(n−1)xn−2−2

∞

n=0

annxn+ 2

∞

n=0

anxn= 0

6. Shift indices in the second term by letting m=n−1:

∞

n=2

ann(n−1)xn−2−2

∞

m=1

am+1(m+ 1)xm+1 + 2

∞

n=0

anxn= 0

7. Combine all terms into a single series:

∞

n=2

ann(n−1)xn−2−2

∞

m=1

am+1(m+1)xm+1+2

∞

n=0

anxn=

∞

n=0

(ann(n−1)−2an+1(n+1)+2an)xn= 0

8. This gives us the indicial equation:

n(n−1) −2(n+ 1)an+1 + 2an= 0

9. Solve the indicial equation to find the recursion relation for the coeﬀicients an:

an+1 =n(n−1) + 2

2(n+ 1) an

10. Using the recursion relation, find the first few coeﬀicients a0, a1, a2, ... and thus the

general solution for y(x).

15. Question: Solve the following differential equation:

(y3+ 1)dx −xdy = 0

Ans. Solution: 1. First, we rewrite the given differential equation in terms of yand dy:

(y3+ 1)dx =xdy

2. Next, we divide by x(y3+ 1) to separate the variables:

x=dy

y3+ 1

3. Now, we integrate both sides with respect to their respective variables:

xdx =Z1

y3+ 1 dy

4. Integrating the left side gives us:

ln |x|=ln |x|+c1

where c1is the constant of integration.

5. Integrating the right side involves a bit more work. We can use partial fractions to

decompose 1

y3+1 into simpler fractions:

y3+ 1 =A

y+ 1 +By +C

y2−y+ 1

6. Solving for A,B, and C, we find that A=1

3,B=−1

3, and C=2

7. Substituting the partial fraction decomposition back into the integral, we get:

Z1

3(y+ 1) −1

3(y2−y+ 1)dy

8. Integrating each term separately, we obtain:

3ln |y+ 1| − 1

6ln |y2−y+ 1|+c2

where c2is another constant of integration.

9. Therefore, the complete solution to the differential equation is:

ln |x|=1

3ln |y+ 1| − 1

6ln |y2−y+ 1|+c

where c=c1+c2is the combined constant of integrations.

16. Question:

Solve the differential equation y′=1

xy+1

x2.

Ans. Step-by-step solution:

1. This differential equation is a first-order linear differential equation in the form y′+P(x)y=

Q(x), where P(x) = 1

xand Q(x) = 1

x2. We will solve this using an integrating factor.

2. Multiply through by the integrating factor e∫P(x)dx =e∫1

xdx =eln |x|=|x|. This gives us:

|x|y′+1

x|x|y=1

x2|x|

|x|y′+y= 1

3. Recognize that the equation |x|y′+y= 1 can be separated by variable. Rewrite it as:

|x|y′= 1 −y

y′=1−y

|x|

4. Now we can solve the separated differential equation. Separate the variables and integrate

both sides: dy

dx =1−y

|x|

Zdy

1−y=Zdx

|x|

5. Integrating both sides gives us:

−ln |1−y|=ln |x|+C

|1−y|=e−ln |x|+C=eC·1

|x|

6. Remove the absolute value signs and simplify to find the general solution:

1−y=A

y= 1 −A

7. Therefore, the general solution to the differential equation is y= 1 −A

x, where Ais an

arbitrary constant.

17. Question 17: Solve the following initial value problem: y′+y=sin(2x),y(0) = 1.

Ans. Step 1. Find the integrating factor:

Multiply both sides of the differential equation by the integrating factor ex:

exy′+exy=exsin(2x)

Step 2. Apply the product rule:

Rewrite the left side as the derivative of the product of exand y:

(exy)′=exsin(2x)

Step 3. Integrate both sides:

Integrate both sides of the equation with respect to x:

Z(exy)′dx =Zexsin(2x)dx

exy=−1

2excos(2x) + C

Step 4. Solve for y:

Divide both sides by exto obtain the solution for y:

y=−1

2cos(2x) + Ce−x

Step 5. Use the initial condition to find C:

Substitute x= 0 and y= 1 into the solution:

1 = −1

2cos(0) + C

1 = −1

2+C

C=3

Step 6. Final solution:

Substitute C=3

2back into the solution to obtain the final answer:

y=−1

2cos(2x) + 3

2e−x

18. Question: Find the general solution of the following first-order differential equation:

dx +y=e−x

Ans. Step-by-step solution: 1. This is a first-order linear differential equation of the form

dx +P(x)y=Q(x), where P(x) = 1 and Q(x) = e−x.

2. To solve this type of differential equation, we first look for an integrating factor µ(x)which

makes the equation exact. The integrating factor is given by µ(x) = e∫P(x)dx.

3. Substitute P(x) = 1 into the formula for the integrating factor:

µ(x) = e∫1dx =ex

4. Multiply the given differential equation by the integrating factor µ(x):

exdy

dx +exy=e−xex

exdy

dx +exy= 1

5. Notice that the left side of the equation can be written in the form of the product rule:

dx(exy) = 1

6. Integrate both sides with respect to x:

dx(exy)dx =Z1dx

exy=x+C

7. Solve for y:

y=e−x(x+C)

8. Therefore, the general solution to the differential equation is:

y=xe−x+Ce−x

where C is an arbitrary constant.

19. Question:

Consider the differential equation dy

dx = 2y+e3x. Find the general solution to this differential

equation.

Ans. Step-by-step solution:

1. First, we need to write the differential equation in standard form, which means isolating

dy and dx on one side and grouping terms involving yon the other side:

dx = 2y+e3x

dx −2y=e3x

2. This is a linear first-order differential equation that can be solved using an integrating

factor. The integrating factor, denoted by I(x), is given by:

I(x) = e∫−2dx =e−2x

3. Multiply the integrating factor with the differential equation, remembering to apply the

product rule for derivatives:

e−2xdy

dx −2e−2xy=e−2xe3x

dx(e−2xy) = ex

4. Integrate both sides with respect to x:

dx(e−2xy)dx =Zexdx

e−2xy=ex+C

5. Solve for yby multiplying by e2x:

y=e3x+Ce2x

6. Therefore, the general solution to the differential equation dy

dx = 2y+e3xis y=e3x+Ce2x,

where Cis the constant of integration.

20. Question 20: Solve the following first-order linear ordinary differential equation:

y′= 2xy +x2cos(x)

Ans. Step-by-step solution:

1. First, rewrite the given differential equation in the standard form y′+P(x)y=Q(x). In

this case, we have:

y′−2xy =x2cos(x)

2. Next, find the integrating factor µ(x), which is defined as µ(x) = e∫P(x)dx. In this case,

P(x) = −2x, so:

µ(x) = e∫−2x dx =e−x2

3. Multiply the integrating factor µ(x)with the entire differential equation:

e−x2y′−2xe−x2y=x2e−x2cos(x)

4. Recognize that the left-hand side is now the derivative of the product y·e−x2. So, rewrite

the equation as:

dx(y·e−x2) = x2e−x2cos(x)

5. Integrate both sides with respect to xto get the general solution:

y·e−x2=Zx2e−x2cos(x)dx +C

6. The integral on the right-hand side may require integration by parts. Let u=x2and

dv =e−x2cos(x)dx. Then, du = 2x dx and v=−1

2e−x2sin(x).

Zx2e−x2cos(x)dx =−1

2x2e−x2sin(x)−Z−xe−x2sin(x)dx

7. The second integral can be integrated by parts again. Let w=xand dz =e−x2sin(x)dx.

Then, dw =dx and z=−1

2e−x2cos(x).

Z−xe−x2sin(x)dx =−1

2xe−x2cos(x)−Z−1

2e−x2cos(x)dx

8. After integrating by parts twice, substitute the results back into the solution equation and

simplify to find the general solution. Finally, solve for the constant Cusing an initial condition if

provided.

21. Question: Solve the following differential equation with initial condition: dy

dx =y2+xwith

y(0) = 1.

Ans. Solution: Let’s solve the differential equation using the method of separation of variables.

1. Write the differential equation:

dx =y2+x

2. Separate the variables:

y2+x=dx

3. Integrate both sides:

Zdy

y2+x=Zdx

4. Simplify the left side by completing the square in the denominator:

Zdy

y2+x=Zdy

(y+x

2)2−(x

2)2

5. Make a substitution u=y+x

Zdu

u2−(x

2)2=Zdx

6. The integral on the left can be simplified using partial fractions:

2Z1

u−x

2−1

u+x

2du =Zdx

7. Integrate both sides:

2ln 



u−x

u+x

2



=x+C1

8. Substitute back u=y+x

2and simplify:

2ln 



2+x

2



=x+C1

9. Solve for yby exponentiating both sides and applying the initial condition y(0) = 1:



2+x

2



=Ce2x

Applying the initial condition y(0) = 1 gives us:



2



=C⇒C=1

10. Write the final solution as two separate cases based on the absolute value:

2+x

2=±1

2e2x

Thus, the general solution to the differential equation is:

y(x) = −x+e2x, y(x) = x+e2x

22. Question: Solve the initial value problem y′= 2y−4e2xwith y(0) = 3.

Ans. Solution: 1. First, we notice that the given differential equation is a linear first-order

ODE. To solve it, we’ll start by finding the integrating factor µ(x)given by µ(x) = e∫2dx =e2x.

2. Next, we multiply both sides of the differential equation by the integrating factor µ(x):

e2xy′= 2e2xy−4

3. By the product rule for differentiation, the left side simplifies to:

dx(e2xy) = 2e2xy−4

4. Integrating both sides of the equation with respect to x, we get

dx(e2xy)dx =Z(2e2xy−4)dx

e2xy=Z(2e2xy−4)dx +C1

5. Solving the integral on the right side, we have

e2xy=e2xy−4x+C1

6. Divide by e2xto solve for y:

y=y−4xe−2x+C1e−2x

7. Now we use the initial condition y(0) = 3 to find the value of C1:

3 = 3 −0 + C1e0

3 = 3 + C1

8. Therefore, C1= 0, and the solution to the initial value problem is

y=y−4xe−2x

Thus, the solution to the initial value problem y′= 2y−4e2xwith y(0) = 3 is y=y−4xe−2x.

23. Question 23: Solve the following first-order ordinary differential equation: dy

dx + 2y= 4x,

where y(0) = 3.

Ans. Solution: 1. Write the differential equation in standard form:

dx + 2y= 4x

2. This is a first-order linear ordinary differential equation. To solve it, we need to find the

integrating factor. The integrating factor µis given by:

µ(x) = e∫2dx =e2x

3. Multiply both sides of the differential equation by the integrating factor µ(x) = e2x:

e2xdy

dx + 2e2xy= 4xe2x

4. Rewrite the left side as the derivative of the product of yand the integrating factor:

dx(ye2x) = 4xe2x

5. Integrate both sides with respect to x:

dx(ye2x)dx =Z4xe2xdx

ye2x=Z4xe2xdx

6. Solve the integral on the right side using integration by parts: Let u= 4xand dv =e2xdx

Then, du = 4 dx and v=1

2e2x

7. Apply the integration by parts formula:

Zudv =uv −Zvdu

Z4xe2xdx = 4xe2x−Z1

2e2x·4dx

= 4xe2x−2e2x+C

8. Substitute this back into the previous equation:

ye2x= 4xe2x−2e2x+C

9. Now, solve for y:

y= 4x−2 + Ce−2x

10. Using the initial condition y(0) = 3, we can find the value of C:

3 = 0 −2 + Ce0

C= 5

11. Finally, the solution to the differential equation is:

y= 4x−2+5e−2x

24. Question 24: Solve the following differential equation using the method of undetermined

coeﬀicients:

y′′ + 4y= 2e2x.

Ans. Let’s assume the particular solution has the form yp=Ae2x, where Ais a constant to be

determined.

1. Find the first and second derivatives of yp:

yp=Ae2x, y′

p= 2Ae2x, y′′

p= 4Ae2x.

2. Plug ypand its derivatives back into the differential equation:

4Ae2x+ 4Ae2x= 2e2x

8Ae2x= 2e2x

8A= 2

A=1

3. Therefore, the particular solution is yp=14e2x.

4. The complementary function ycfor the given differential equation is:

y′′ + 4y= 0.

The characteristic equation is r2+ 4 = 0, which gives r=±2i. Thus, the complementary

function is

yc=c1cos(2x) + c2sin(2x),

where c1and c2are arbitrary constants.

5. The general solution is the sum of the particular and complementary solutions:

y(x) = yc+yp=c1cos(2x) + c2sin(2x) + 1

4e2x,

where c1and c2are arbitrary constants to be determined based on initial conditions, if provided.

25. Question 25: Solve the following differential equation using the method of undetermined

coeﬀicients:

y′′ −2y′+y=ex

Ans. Step 1. Write down the characteristic equation by considering the homogenous part of

the differential equation:

r2−2r+ 1 = 0

This gives us a repeated root at r= 1.

Step 2. Write down the general solution to the homogenous part of the differential equation:

yh(x) = c1ex+c2xex

Step 3. Assume a particular solution of the form yp(x) = Aexfor the non-homogenous part

of the differential equation.

Step 4. Substitute yp(x)and its derivatives into the differential equation. Solve for A:

y′′

p−2y′

p+yp=ex

Aex−2Aex+Aex=ex

−Aex=ex

A=−1

Step 5. Write down the particular solution:

yp(x) = −ex

Step 6. The general solution to the non-homogeneous differential equation is the sum of the

homogeneous and particular solutions:

y(x) = yh(x) + yp(x) = c1ex+c2xex−ex

26. Question 26: Solve the following first-order linear differential equation:

(y−x)dy

dx +y=x

Ans. Step 1. Rewrite the differential equation in standard form.

We are given the differential equation:

(y−x)dy

dx +y=x

To put it in standard form, we divide through by (y−x):

dx +y

y−x=x

y−x

Step 2. Identify the integrating factor.

The integrating factor is given by:

(x) = e∫P(x)dx

where P(x)is the coeﬀicient of yin the standard form. In this case, P(x) = 1

y−x, so

ZP(x)dx =Z1

y−xdx =ln |y−x|

Therefore, the integrating factor is (x) = eln |y−x|=|y−x|.

Step 3. Multiply the standard form by the integrating factor.

Multiplying both sides of the standard form by the integrating factor (x) = |y−x|, we get:

|y−x|dy

dx +y

y−x|y−x|=x

y−x|y−x|

Simplifying, we have:

|y−x|dy

dx +y=x

Step 4. Integrate both sides.

Integrating the equation with respect to x, we get:

Z|y−x|dy

dxdx +Zydx =Zxdx

Z|y−x|dy +yx =1

2x2+C

Step 5. Solve for y.

At this point, it is diﬀicult to solve for yexplicitly due to the absolute value term. Depending

on the initial conditions, we can consider cases where y > x and y < x and solve for yaccordingly.

Thus, the solution to the differential equation is given implicitly by:

Z|y−x|dy +yx =1

2x2+C

27. Find the general solution to the differential equation: dy

dx =ycos(x).

Ans. The general solution to the differential equation dy

dx =ycos(x)is y=kesin(x), where kis

an arbitrary constant.

1. Separation of Variables: Start by separating variables by dividing both sides of the

equation by y:

dx =ycos(x)=⇒dy

y=cos(x)dx.

2. Integration: Integrate both sides:

ydy =Zcos(x)dx =⇒ln |y|=sin(x) + C,

where Cis the constant of integration.

3. Solve for y: Exponentiate both sides to solve for y:

|y|=esin(x)+C=eCesin(x).

4. Include the Arbitrary Constant: Let k=±eCto include the arbitrary constant. Thus,

the general solution is:

y=kesin(x),

where kis an arbitrary constant.

28. Find the general solution to the differential equation: y′′ −4y′+ 4y=e2x.

Ans. Step 1. First, we find the complementary solution to the homogeneous equation

y′′ −4y′+ 4y= 0. Let y=emx, so y′=memx and y′′ =m2emx. Substituting into the

differential equation, we get:

m2emx −4memx + 4emx = 0

m2−4m+ 4 = 0

(m−2)2= 0

m= 2

So the complementary solution is yc=Ae2x+Bxe2x, where Aand Bare constants.

Step 2. Next, we find a particular solution to the non-homogeneous equation y′′ −4y′+4y=

e2x. Since the right-hand side is in the form of emx and m= 2 is already a solution to the

characteristic equation, we try a particular solution of the form yp=x2e2x. Calculate y′

pand y′′

and substitute into the differential equation:

y′′

p−4y′

p+ 4yp= 4e2x−8xe2x+ 4x2e2x

Step 3. Since this expression already appears in the non-homogeneous part, we modify the

particular solution to yp=Ax2e2x. Calculate y′

pand y′′

p, and then substitute into the differential

equation:

y′′

p−4y′

p+ 4yp= 4Ae2x−8Axe2x+ 4Ax2e2x=e2x

Equating coeﬀicients of the exponential term on both sides, we get 4A= 1, so A=1

Step 4. Therefore, the particular solution is yp=1

4x2e2x.

Step 5. The general solution to the differential equation is the sum of the complementary

solution and the particular solution:

y=yc+yp=Ae2x+Bxe2x+1

4x2e2x

29. Find the general solution to the differential equation:

y′′ −4y′+ 4y=e2x+x

Ans. To find the general solution to the given differential equation, we first solve the homoge-

neous equation y′′ −4y′+ 4y= 0 to find the complementary function, and then find a particular

solution for the non-homogeneous equation y′′ −4y′+ 4y=e2x+x.

Step 1. Solve the homogeneous equation: y′′ −4y′+ 4y= 0

Let’s assume the solution has the form y=erx. Substituting this into the homogeneous

equation gives:

r2erx −4rerx + 4erx = 0

Factor out erx:

erx(r2−4r+ 4) = 0

Solving the characteristic equation r2−4r+ 4 = 0 gives a repeated root at r= 2.

Therefore, the complementary function is yc= (c1+c2x)e2x.

Step 2. Find a particular solution: yp=Ax2e2x+Bx +C

Substitute ypand its derivatives back into the non-homogeneous equation to solve for A,B,

and C:

y′′

p−4y′

p+ 4yp=e2x+x

(4Ae2x+2Axe2x+B−8Axe2x−4Ae2x+2Bx+4Ax2e2x+Bx+C)−4(2Ae2x+Be2x+2Ax)+4(Ax2e2x+Bx+C) = e2x+x

Simplify the equation and solve for A, B, and C.

Step 3. Combine the complementary function and the particular solution to get the general

solution:

y=yc+yp= (c1+c2x)e2x+Ax2e2x+Bx +C

30. Question: Solve the following differential equation using the method of undetermined

coeﬀicients:

y′′ −4y′+ 4y= 6e2x

Ans. Solution: 1. First, find the complementary solution by solving the characteristic equation:

r2−4r+ 4 = 0

2. The characteristic equation has a repeated root at r= 2, so the complementary solution

is:

yc(x) = (c1+c2x)e2x

3. Next, guess a particular solution of the form:

yp(x) = Ae2x

4. Find the first and second derivatives:

y′

p(x) = 2Ae2x

y′′

p(x) = 4Ae2x

5. Substituting these into the differential equation:

4Ae2x−4(2Ae2x) + 4(Ae2x) = 6e2x

6. Simplifying:

4Ae2x−8Ae2x+ 4Ae2x= 6e2x

0 = 6e2x

7. This gives us 0=6e2x, which is not true. So, our assumption for the particular solution

was incorrect.

8. Since the right-hand side of the differential equation is 6e2x, we need to try a new form

for the particular solution. Since e2xappears in the complementary solution, we multiply by x,

i.e., yp(x) = Axe2x.

9. Find the first and second derivatives:

y′

p(x) = Ae2x+ 2Axe2x

y′′

p(x) = 2Ae2x+ 4Axe2x+ 2Ae2x

10. Substitute these into the differential equation:

2Ae2x+ 4Axe2x+ 2Ae2x−4(Ae2x+ 2Axe2x) + 4(Axe2x) = 6e2x

11. Simplifying:

2Ae2x+ 4Axe2x+ 2Ae2x−4Ae2x−8Axe2x+ 4Ae2x= 6e2x

12. Combine like terms: 4Ae2x= 6e2x

13. Solve for A:4A= 6 A=3

14. Therefore, the particular solution is yp(x) = 3

2xe2x.

15. The general solution is the sum of the complementary and particular solutions:

y(x) = yc(x) + yp(x)

y(x) = (c1+c2x)e2x+3

2xe2x

31. Question: Solve the following differential equation: y′′ −5y′+ 6y=e2x.

Ans. Step-by-step solution: 1. Start by finding the complementary solution. Assume a solution

of the form yc=erx. Substituting this into the differential equation, we get the characteristic

equation r2−5r+ 6 = 0. Solving this quadratic equation, we find the roots r1= 2 and r2= 3.

Therefore, the complementary solution is yc=C1e2x+C2e3x. 2. Now, we need to find the

particular solution ypfor the non-homogeneous term e2x. Since this term is of the form emx

where mis the same as one of the roots of the characteristic equation, we multiply our assumed

particular solution by xto avoid overlap. So, let’s assume yp=Axe2x. 3. Now, substitute yp

back into the differential equation and solve for Aby comparing coeﬀicients. We get:

(4A−10A+ 6A)x2e2x=e2x

A=1

4. Therefore, the particular solution is yp=1

6xe2x. 5. The general solution is the sum of the

complementary and particular solutions:

y=yc+yp=C1e2x+C2e3x+1

6xe2x

where C1and C2are constants determined by initial conditions.

32. Question 32: Solve the following first-order differential equation using an integrating factor:

y′−2y= 4x

Ans. Step 1. Write the given first-order differential equation in standard form: y′−2y= 4x.

Step 2. Identify the integrating factor µ(x), which is given by µ(x) = e∫−2dx.

Step 3. Compute the integrating factor:

µ(x) = e∫−2dx

=e−2x

Step 4. Multiply both sides of the differential equation by the integrating factor µ(x):

e−2xy′−2e−2xy= 4xe−2x

Step 5. Rewrite the equation with the left side as the derivative of the product of yand

µ(x):

dx(e−2xy) = 4xe−2x

Step 6. Integrate both sides with respect to x:

dx(e−2xy)dx =Z4xe−2xdx

e−2xy=Z4xe−2xdx

Step 7. Solve the integral on the right side:

Z4xe−2xdx =−2xe−2x+Z2e−2xdx

=−2xe−2x−e−2x

Step 8. Substitute back into the equation:

e−2xy=−2xe−2x−e−2x

Step 9. Solve for y:

y=−2x−1

Step 10. Therefore, the solution to the differential equation y′−2y= 4xis y=−2x−1.

33. Question: Solve the following differential equation using the method of undetermined

coeﬀicients:

y′′ −4y′+ 4y=x2e2x

Ans. Step-by-step solution: 1. The characteristic equation for the homogeneous part of

the differential equation is r2−4r+ 4 = 0, which has a repeated root r= 2. Therefore,

the complementary solution is yc=c1e2x+c2xe2x. 2. To find a particular solution for the

non-homogeneous part, guess a solution of the form yp= (Ax2+Bx +C)e2x, where A,B,

and Care constants to be determined. 3. Calculate the first and second derivatives of yp:y′

(2Ax+B)e2x+(Ax2+Bx+C)(2e2x)y′′

p= (2A)e2x+2Axe2x+(2Ax+B)e2x+2(Ax2+Bx+C)e2x

4. Substitute ypand its derivatives back into the original differential equation:

(2A)e2x+2Axe2x+(2Ax+B)e2x+2(Ax2+Bx+C)e2x−4[(2Ax+B)e2x+(Ax2+Bx+C)(2e2x)]+4(Ax2+Bx+C)e2x=x2e2x

5. Simplify the equation and match coeﬀicients of like terms:

(−2A+ 2A)xe2x+ (−2A−4B+ 4B)xe2x+ (2A+B−4B+ 4C)e2x=x2e2x

0xe2x+ 0xe2x+ (2A+B−4B+ 4C)e2x=x2e2x

6. Equate coeﬀicients:

2A+B−4B+ 4C= 0

B−4B=−2A

3B=−2A

B=−2

7. Substitute B=−2

3Ainto the first equation:

2A−2

3A−4(−2

3A)+4C= 0

2A−2

3A+8

3A+ 4C= 0

2A+2

3A+ 4C= 0

6A+ 2A+ 12C= 0

8A+ 12C= 0

A=−3

8. Now we have A=−3

2Cand B=−2

3A. Substitute these back into the original guess for yp

to find the particular solution yp.

34. Question: Solve the following first-order linear ordinary differential equation:

dx + 2xy =ex2

Ans. Step-by-step solution: 1. We begin by identifying the integrating factor for the given

first-order linear ODE. The integrating factor is given by µ(x) = e∫2x dx =ex2.

2. We multiply the given ODE by the integrating factor µ(x)to obtain:

ex2dy

dx + 2xex2y=e2x2

3. Notice that the left-hand side of the equation now looks like the derivative of a product.

By the product rule, we have:

dx(ex2y) = e2x2

4. Integrating both sides with respect to x, we get:

dx(ex2y)dx =Ze2x2dx

5. Solving the integrals, we have:

ex2y=1

2e2x2+C

where Cis the constant of integration.

6. Finally, solving for y, we get:

y=1

2ex2+Ce−x2

where Cis the constant of integration.

35. Question:

Solve the initial value problem:

dx =1

x2y2, y(1) = 2.

Ans. Step-by-step solution:

1. Separate the variables to solve the differential equation. We have:

y2dy =1

x2dx.

2. Integrate both sides:

Zy2dy =Z1

x2dx.

3. This gives us:

3=−1

x+C,

where C is the constant of integration.

4. To find the value of the constant, we use the initial condition y(1) = 2:

Consider the differential equation y′+2xy =e−x2. Find the general solution to the differential

equation.

Ans. Step-by-step solution:

1. First, let’s rewrite the differential equation in the standard form of first-order linear ODE:

y′+P(x)y=Q(x). In this case, P(x) = 2xand Q(x) = e−x2.

2. To find the integrating factor, we calculate e∫P(x)dx. In this case, e∫2x dx =ex2.

3. Multiply the integrating factor by the original ODE to get:

ex2y′+ 2xex2y=ex2e−x2

Simplify this to:

(ex2y)′= 1

4. Integrate both sides with respect to xto get:

ex2y=x+C

where Cis the constant of integration.

5. Finally, divide by ex2to solve for y:

y=x

ex2+C

ex2

which simplifies to:

y=xe−x2+Ce−x2

Thus, the general solution to the differential equation is y=xe−x2+Ce−x2, where Cis an

arbitrary constant.

3. Question: Solve the following first-order linear differential equation:

y′+ 2xy = 4x2,

where y(0) = 3.

Ans. Step-by-step solution: 1. To solve the given first-order linear differential equation, we first

need to rearrange it into the standard form y′+P(x)y=Q(x). In this case, P(x) = 2xand

Q(x)=4x2. 2. We can find the integrating factor µ(x)by computing µ(x) = e∫P(x)dx. Thus,

we have:

µ(x) = e∫2xdx =ex2.

3. Multiply both sides of the differential equation by the integrating factor µ(x):

ex2y′+ 2xex2y= 4x2ex2.

4. Notice that the left-hand side of the equation can be written as the derivative of the product

of µ(x)and y:

dx(ex2y) = 4x2ex2.

5. Integrate both sides with respect to x:

dx(ex2y)dx =Z4x2ex2dx.

6. This simplifies to:

ex2y=Z4x2ex2dx +C.

7. Solve the integral on the right-hand side:

Z4x2ex2dx = 2ex2(2x2−1) + K,

where Kis a constant of integration. 8. Substitute this back into the equation:

ex2y= 2ex2(2x2−1) + K.

9. Finally, solve for yby dividing both sides by ex2and simplifying:

y= 2(2x2−1) + K

ex2.

10. Apply the initial condition y(0) = 3 to find the value of K:

3 = 2(0) −1 + K

e0=⇒K= 4.

Therefore, the solution to the initial value problem is:

y= 2(2x2−1) + 4

ex2.

4. Question:

Solve the following differential equation: dy

dx = (x2+y2)1/2 with the initial condition y(0) = 1.

Ans. Solution:

1. Separate variables to get all yterms on one side and all xterms on the other side:

(x2+y2)1/2 =dx

2. Integrate both sides. Let’s perform a substitution where y=xtan(θ):R1

(x2+x2tan2(θ))1/2 xsec(θ)tan(θ)dθ =

Rdx

3. Simplify the integrand: Rxsec(θ)tan(θ)

(x2(1+tan2(θ)))1/2 dθ =Rdx

4. Further simplify the integrand to: Rxsec(θ)

x(1+tan2(θ))1/2 dθ =Rdx

5. After simplifying, we get: Rsec(θ)dθ =Rdx

6. Integrating both sides, we get: ln |sec(θ) + tan(θ)|=x+C

7. Substitute back x=ycot(θ)and simplify: ln |sec(θ) + tan(θ)|=ycot(θ) + C

8. Apply the initial condition y(0) = 1 to find the value of C:ln |1|= 1 ·cot(θ) + C

9. Solve for C:C= 0

10. Thus, the solution to the differential equation is: ln |sec(θ) + tan(θ)|=ycot(θ)

5. Question: Solve the differential equation (y′)2−y′′ = 0.

Ans. Solution: 1. Let v=y′, so the differential equation becomes v2−v′= 0. 2. Rewrite

the equation as v(v−1) = 0. 3. Solve for vby setting each factor to zero: v= 0 or v= 1. 4.

For v= 0, integrate vwith respect to xto find y:y=R0dx =C1, where C1is an arbitrary

constant. 5. For v= 1, integrate vwith respect to xto find y:y=R1dx =x+C2, where C2

is an arbitrary constant. 6. Therefore, the general solution to the differential equation is y=C1

or y=x+C2, where C1and C2are arbitrary constants.

6. Question: Solve the differential equation y′=x

y2.

Ans. Step-by-step solution:

1. Rewrite the given differential equation as:

y2dy

dx =x

2. Separate variables by moving all terms involving yto the left side and all terms involving

xto the right side:

y2dy =x dx

3. Integrate both sides:

Zy2dy =Zx dx

4. Evaluate the integrals:

3=x2

2+C

where Cis the constant of integration.

5. Solve for y:

y3=3x2

2+C

6. Finally, solve for yby taking the cube root of both sides:

y=3

r3x2

2+C

So the solution to the differential equation y′=x

y2is y=3

q3x2

2+C, where Cis the constant

of integration.

7. Question: Find the general solution to the differential equation y′′ −4y′+ 4y= 0.

Ans. Step 1. Write down the characteristic equation by assuming a solution of the form

y=ert. The characteristic equation is: r2−4r+ 4 = 0.

Step 2. Solve the characteristic equation to find the roots r1and r2. Factorizing the

characteristic equation gives: (r−2)2= 0. Therefore, the repeated root is r= 2.