MAT 142 - COLLEGE MATHEMATICS - Differential
Equations Practice Material - Set 1
1. Find the general solution to the differential equation:
y′+ 2ty =t2
Ans. Step 1. First, write the differential equation in standard linear first-order form:
y′+ 2ty =t2
Step 2. This is a linear first-order differential equation of the form y′+P(t)y=Q(t). Here,
P(t) = 2tand Q(t) = t2.
Step 3. We can multiply the integrating factor µ(t) = e∫P(t)dt to both sides. In this case,
µ(t) = e∫2tdt =et2.
Step 4. Multiply the integrating factor with the differential equation:
et2y′+ 2tet2y=t2et2
Step 5. Notice that the left side can be rewritten as the derivative of the product et2y. We
have: d
dt(et2y) = t2et2
Step 6. Integrate both sides with respect to tto get:
et2y=∫t2et2dt
Step 7. Simplify the right-hand side integral:
∫t2et2dt =1
2et2(t2−1
2) + C
Step 8. Therefore, the general solution to the differential equation is:
y=1
2t2−1
4+Ce−t2
where Cis an arbitrary constant.
2. Question: Solve the following differential equation using the method of undetermined
coefficients: y′′ −4y′+ 4y=e2x.
Ans. Step-by-step solution: 1. First, we find the complementary solution by solving the
characteristic equation r2−4r+ 4 = 0:
r2−4r+ 4 = (r−2)2= 0
This gives us a repeated root r= 2, so the complementary solution is:
yc(x) = (c1+c2x)e2x
2. Next, we guess the particular solution yp(x)for the non-homogeneous differential equation.
Since the right-hand side contains e2x, we let:
yp(x) = Ae2x
where Ais a constant to be determined. 3. Substitute yp(x)into the differential equation and
its derivatives to find the values of the undetermined coefficients. We have:
y′
p(x) = 2Ae2x
y′′
p(x) = 4Ae2x
Substitute into the differential equation:
4Ae2x−4(2Ae2x) + 4(Ae2x) = e2x
Simplify this to get:
4Ae2x−8Ae2x+ 4Ae2x=e2x
Solving, we find A=1
2. 4. Therefore, the particular solution is:
yp(x) = 1
2e2x
5. The general solution is the sum of the complementary and particular solutions:
y(x) = yc(x) + yp(x) = (c1+c2x)e2x+1
2e2x
where c1and c2are arbitrary constants determined by initial conditions.
3. Question: Solve the following initial value problem:
y′= 2xy, y(0) = 3.
Ans. Step-by-step solution: 1. The given differential equation is a first-order linear differential
equation in the form y′=f(x)·y. To solve this equation, we will use the method of separation
of variables. We write the equation as dy
dx = 2xy.
2. Separate the variables by dividing both sides by yand multiplying by dx, yielding dy
y=
2x dx.
3. Integrate both sides. Integrating the left side gives ∫1
ydy =ln |y|, and integrating the
right side gives ∫2x dx =x2+C, where Cis the constant of integration.
4. Therefore, we have ln |y|=x2+C.
5. Now, apply the initial condition y(0) = 3. Substituting this into the equation gives
ln |3|= 0 + C, so C=ln(3).
6. Substituting C=ln(3) back into the equation, we have ln |y|=x2+ln(3).
7. Exponentiate both sides to solve for y:|y|=ex2+ln(3) =ex2·eln(3) = 3ex2.
8. We eliminate the absolute value by considering two cases: y= 3ex2if y > 0, and
y=−3ex2if y < 0.
9. Since y(0) = 3 >0, we conclude that the solution is y= 3ex2.
4. Find the general solution to the following differential equation: dy
dx =1
x+y
Ans. Let’s solve the given differential equation step by step:
1. First, rewrite the equation in the standard form of a first-order linear differential equation,
which is dy
dx −y=1
x.
2. The integrating factor I(x)is given by I(x) = e∫−1dx =e−x.
3. Multiply both sides of the equation by the integrating factor and simplify: e−xdy
dx −e−xy=
e−x1
x.
4. The left side can be rewritten using the product rule for differentiation: d
dx (e−xy) = e−x1
x.
5. Integrate both sides with respect to x:∫d
dx (e−xy)dx =∫e−x1
xdx. This gives e−xy=
∫e−x1
xdx +C.
6. Solve the integral on the right side using integration by parts: Let u=1
xand dv =e−xdx.
Then, du =−1
x2dx and v=−e−x. The integral becomes −e−x1
x+∫e−x1
x2dx.
7. Integrate the second term on the right side: ∫e−x1
x2dx =−e−x1
x−∫e−x−1
x2dx. This
simplifies to −e−x1
x+∫e−x1
x2dx.
8. Substitute this back into the equation and simplify: e−xy=−e−x1
x−e−x1
x+C.
9. Simplify further and solve for yto find the general solution: y=−2−C
ex. Hence, the
general solution to the given differential equation is y=−2−C
ex, where Cis the constant of
integration.
5. Question: Solve the initial value problem y′−2y= 4e2xwith initial condition y(0) = 1.
Ans. Step-by-step solution: 1. First, we find the integrating factor e∫−2dx =e−2x. 2. Multiply
the differential equation by the integrating factor: e−2x(y′−2y) = e−2x(4e2x). This simplifies to
e−2xy′−2e−2xy= 4. 3. The left-hand side is now the derivative of the product e−2xy, which we
can integrate to get the solution: ∫(e−2xy)′dx =∫4dx. This gives e−2xy= 4x+C, where C
is the constant of integration. 4. Using the initial condition y(0) = 1, we find the value of the
constant C:e0·1 = 4 ·0 + C, which simplifies to C= 1. 5. Substituting back into the solution,
we have e−2xy= 4x+ 1. So, the general solution is y(x) = e2x(4x+ 1).
6. Find the general solution to the following differential equation:
dy
dx =2x+ 3y−1
3x−y+ 2
Ans. Step 1. Rewrite the equation in a standard form by multiplying throughout by 3x−y+2:
(3x−y+ 2)dy
dx = 2x+ 3y−1
3xdy
dx −ydy
dx + 2dy
dx = 2x+ 3y−1
Step 2. Rearrange the terms:
3xdy
dx + 2dy
dx =y+ 2x−1+3y
(3x+ 2)dy
dx = 4x+ 4y−1
Step 3. Integrate both sides with respect to x:
∫(3x+ 2) dy
dx dx =∫(4x+ 4y−1) dx
∫(3x+ 2) dy =∫(4x+ 4y−1) dx
Step 4. Perform the integrations:
3xy + 2y= 2x2+ 2xy −x+C
Step 5. Simplify the equation:
3xy + 2y= 2x2+ 2xy −x+C
x(3y−2) = 2x2−x+C
(3y−2)dx = (2x−1)dx +C
Step 6. Integrate both sides to find the general solution:
∫(3y−2) dy =∫(2x−1) dx +∫C dx
3
2y2−2y=x2−x+C1
So, the general solution to the differential equation is:
3
2y2−2y=x2−x+C1
where C1is an arbitrary constant.
7. Find the general solution to the following differential equation:
dy
dx +y=ex
Ans. To solve this differential equation, we will use the integrating factor method.
1. Find the integrating factor: The integrating factor, denoted by µ(x), is given by
µ(x) = e∫1dx =ex.
2. Multiply both sides by the integrating factor: Multiplying both sides of the differential
equation by ex, we get:
exdy
dx +exy=e2x
3. Rewrite the left side as a product of a derivative: Since d
dx (exy) = exdy
dx +exy, we
can rewrite the left side of the equation as:
d
dx(exy) = e2x
4. Integrate both sides: Integrating both sides with respect to x, we get:
∫d
dx(exy)dx =∫e2xdx
exy=1
2e2x+C
5. Solve for y:Dividing by ex, we get the general solution:
y=1
2ex+Ce−x
where Cis an arbitrary constant.
8. Question: Solve the differential equation dy
dx =1
2ex+y.
Ans. Step-by-step solution: 1. Rewrite the given differential equation as dy
dx =1
2exey. 2.
Notice that the right-hand side of the equation can be separated into functions of xand y:
dy
dx =1
2exey=1
2exey. 3. Now, separate the variables by dividing both sides of the equation by
ey, and then multiply both sides by dx to bring all xterms to the left side and yterms to the
right side: 1
eydy =1
2exdx. 4. Integrate both sides with respect to their respective variables:
∫1
eydy =∫1
2exdx. 5. Simplify the integrals to get −e−y=1
2ex+C, where Cis the constant
of integration. 6. Solve for yby multiplying through by −1to get e−y=−1
2ex−C. 7. Finally,
solve for yby taking the natural logarithm of both sides: y=−ln (−1
2ex−C), where Cis an
arbitrary constant.
9. Question: Solve the following initial value problem: y′= 2xy −x,y(0) = 1.
Ans. Step-by-step solution: 1. We start by writing the given differential equation in the standard
form of a first-order linear differential equation: y′−2xy =−x. 2. To solve this equation, we
first find the integrating factor e∫−2xdx. 3. Integrating −2xwith respect to x, we get −x2, so the
integrating factor is e−x2. 4. Multiplying the given differential equation by the integrating factor,
we have e−x2y′−2xye−x2=−xe−x2. 5. By recognizing the left side as the derivative of the
product e−x2y, we can rewrite the equation as d
dx (e−x2y) = −xe−x2. 6. Integrating both sides
with respect to x, we get e−x2y=C−1
2e−x2+D, where Cand Dare constants of integration.
7. Applying the initial condition y(0) = 1, we find e−0·1 = C−1
2e−0+D. 8. Simplifying, we
get C−1
2+D= 1. 9. Therefore, C+D=3
2. 10. Finally, the solution to the initial value
problem is y=ex2(3
2−1
2e−x2).
10. Question: Solve the following first-order ordinary differential equation:
y′=x
y
Ans. Step-by-step solution: 1. Rewrite the differential equation in the form dy
dx =x
y. 2. Multiply
both sides by yto get y dy =x dx. 3. Integrate both sides: ∫y dy =∫x dx. 4. Integrating the
left side gives 1
2y2+C1, where C1is the constant of integration. 5. Integrating the right side
gives 1
2x2+C2, where C2is the constant of integration. 6. Therefore, the general solution is
1
2y2=1
2x2+Cwhere C=C1−C2is the constant of integration. 7. Finally, we can write the
solution in the form y=±√x2+ 2C.
11. Question: Solve the following first-order linear differential equation: dy
dx +y=e−x.
Ans. Step-by-step solution: 1. We have a first-order linear differential equation of the form
dy
dx +P(x)y=Q(x), where in this case P(x) = 1 and Q(x) = e−x. 2. To solve this equation, we
first find the integrating factor, which is given by µ(x) = e∫P(x)dx. In this case, µ(x) = e∫1dx =
ex. 3. We multiply both sides of the differential equation by the integrating factor, µ(x), to get:
exdy
dx +exy=exe−x. 4. Recognizing that the left side of the equation is the derivative of the
product of yand µ(x), we can rewrite the equation as d
dx (yex) = 1. 5. Integrating both sides
with respect to x, we get ∫d
dx (yex)dx =∫1dx. 6. Simplifying, we have yex=x+C, where
Cis the constant of integration. 7. Finally, solving for y, we get y=x
ex+Ce−x, where Cis
an arbitrary constant. Thus, the general solution to the differential equation is y=x
ex+Ce−x,
where Cis an arbitrary constant.
12. Question:
Solve the differential equation y′+ 2xy =ex2, with the initial condition y(0) = 3.
Ans. Solution:
1. First, we’ll find the integrating factor µ(x)by multiplying the entire equation by it:
µ(x) = e∫2x dx =ex2
2. Multiply the original differential equation by the integrating factor:
ex2y′+ 2xex2y=e2x2
3. Notice that the left-hand side is the derivative of the product of µ(x)and y:
d
dx(ex2y) = e2x2
4. Integrating both sides with respect to x, we get:
ex2y=∫e2x2dx +C
ex2y=√π
2erf(x) + C
5. Now, applying the initial condition y(0) = 3:
e0·3 = √π
2erf(0) + C
3 = √π
2·0 + C
C= 3
6. Therefore, the particular solution to the differential equation is:
ex2y=√π
2erf(x)+3
13. Question: Solve the initial value problem: dy
dx =1
x+ywith y(1) = 0.
Ans. Step-by-step solution: 1. Rewrite the differential equation in the form dy
dx =f(x, y)by
isolating dy on one side:
dy
dx =1
x+y
2. Rearrange the equation to separate variables:
(x+y)dy =dx
3. Integrate both sides: ∫(x+y)dy =∫dx
4. Solve the integrals: ∫x dy +∫y dy =∫dx
xy +y2
2=x+C1
5. Apply the initial condition y(1) = 0 to find the constant C1:
1·0 + 02
2= 1 + C1
0 = 1 + C1
C1=−1
6. Substitute C1=−1back into the equation:
xy +y2
2=x−1
7. To simplify further, let’s write the equation in a different form by completing the square on
the left side:
y2+ 2xy = 2x−2
y2+ 2xy +x2= 2x+x2−2
(y+x)2=x2+ 2x−2
8. Take the square root of both sides and solve for y:
y+x=±√x2+ 2x−2
y=−x±√x2+ 2x−2
Thus, the solution to the initial value problem is y=−x±√x2+ 2x−2.
14. Question: Solve the following first-order linear differential equation:
dy
dx +2y
x=x3
Ans. Step-by-step solution: 1. First, we need to write the given differential equation in the
standard form of a linear first-order ODE, which is dy
dx +P(x)y=Q(x). To do this, we divide
throughout by xto get:
dy
dx + 2y
x=x2
2. Now we can identify P(x)=2and Q(x) = x2. Next, we find the integrating factor µ(x),
which is given by:
µ(x) = e∫P(x)dx
µ(x) = e∫2dx =e2x
3. Multiplying the integrating factor by the differential equation gives us:
e2xdy
dx + 2e2xy
x=e2xx2
4. This can be rewritten as follows:
d
dx(e2xy) = x2e2x
5. Integrate both sides of the equation with respect to x:
∫d
dx(e2xy)dx =∫x2e2xdx
e2xy=1
2x2e2x−1
2∫x2e2xdx
6. Solving the integral on the right side using integration by parts will give us:
e2xy=1
2x2e2x−1
4x2e2x+1
4∫e2xdx
7. Continuing to simplify, we get:
e2xy=1
4x2e2x+1
8e2x+C
8. Finally, we solve for yby dividing by e2x:
y=1
4x2+1
8+Ce−2x
where Cis the constant of integration.
15. Question 15: Solve the following first-order linear differential equation:
dy
dx + 3y=e−4x
Ans. Step 1. Find the integrating factor µ(x)by multiplying the entire equation by it:
µ(x) = e∫3dx =e3x
Step 2. Multiply the equation by the integrating factor and rewrite it in the form of a total
derivative:
e3xdy
dx + 3e3xy=e−x
Step 3. Express the left side as a total derivative:
d
dx (e3xy)=e−4x
Step 4. Integrate both sides with respect to x:
e3xy=∫e−4xdx
Step 5. Integrate the right-hand side:
e3xy=−1
4e−4x+C
Step 6. Solve for y:
y=−1
4e−7x+Ce−3x
where Cis an arbitrary constant.
16. Question 16: Solve the following first-order linear differential equation using an integrating
factor: dy
dx +(2 + 1
x)y=x
Ans. Step-by-step solution: 1. Write the given first-order linear differential equation in the
standard form dy
dx +P(x)y=Q(x), where P(x)and Q(x)are functions of x. The given differential
equation is:
dy
dx +(2 + 1
x)y=x
Here, P(x) = 2 + 1
xand Q(x) = x.
2. Calculate the integrating factor µ(x) = e∫P(x)dx.
µ(x) = e∫(2+ 1
x)dx =e2x+ln |x|=x2e2x
3. Multiply both sides of the differential equation by the integrating factor µ(x).
x2e2xdy
dx +x2e2x(2 + 1
x)y=x3e2x
4. Rewrite the left side of the equation as a product rule of a single function.
d
dx (yx2e2x)=x3e2x
5. Integrate both sides with respect to x.
∫d(yx2e2x)=∫x3e2xdx
yx2e2x=1
2x3e2x−1
2∫x3e2xdx +C
6. Solve for yby dividing by x2e2x.
y=1
2x−1
2∫x·x2e2xdx +C
x2e2x
7. Calculate ∫x3e2xdx by using integration by parts. Let u=x3and dv =e2xdx, then
du = 3x2dx and v=1
2e2x.
1
2x3e2x−3
2∫x2e2xdx =y
Therefore, the solution to the differential equation is:
y=1
2x−1
2(1
2x3e2x−3
2∫x2e2xdx)+C
x2e2x
17. Question: Solve the differential equation y′′ + 4y=sin(2x)subject to the initial conditions
y(0) = 1 and y′(0) = 0.
Ans. Step-by-step solution:
1. First, we find the complementary solution to the homogeneous differential equation y′′ +
4y= 0. The characteristic equation is r2+ 4 = 0, which has roots r=±2i. Therefore, the
complementary solution is yc(x) = c1cos(2x) + c2sin(2x).
2. Next, we find a particular solution to the nonhomogeneous differential equation y′′ + 4y=
sin(2x). Since the right-hand side is of the form sin(2x), we guess a particular solution of the
form yp(x) = Asin(2x) + Bcos(2x).
3. We find the first and second derivatives of yp(x):y′
p(x) = 2Acos(2x)−2Bsin(2x)
y′′
p(x) = −4Asin(2x)−4Bcos(2x)
4. We substitute yp(x)and its derivatives into the differential equation to get: −4Asin(2x)−
4Bcos(2x) + 4(Asin(2x) + Bcos(2x)) = sin(2x)Simplifying, we get −4B= 1, so B=−1
4.
5. Substituting B=−1
4back into yp(x), we have yp(x) = Asin(2x)−1
4cos(2x). Now,
substitute the initial conditions into the particular solution to find A.
6. We have y(0) = 1 and y′(0) = 0: Substitute x= 0 into y(0) = Asin(0) −1
4cos(0) =
A(0) −1
4(1) = 1 This simplifies to −1
4= 1, which is not possible. Therefore, we made a mistake
in our particular solution.
7. We need to consider that the particular solution should be of the form yp(x) = Ax sin(2x)+
Bx cos(2x). Proceed with steps 3-5 with this new form to find the correct particular solution.
8. Once you have the correct particular solution, substitute the initial conditions to find the
values of the constants Aand B. Finally, add the complementary and particular solutions to get
the general solution.
9. The final solution will satisfy the given initial conditions y(0) = 1 and y′(0) = 0.
18. Question: Solve the initial value problem y′−3y=e2xwith y(0) = 1.
Ans. Solution: 1. First, we find the integrating factor by solving the auxiliary equation
m−3=0, which gives m= 3. The integrating factor is then µ(x) = e∫3dx =e3x. 2. Multiply
both sides of the differential equation by the integrating factor: e3xy′−3e3xy=e5x. 3. Recognize
the left side as the derivative of the product of e3xand y:d
dx (e3xy) = e5x. 4. Integrate both
sides with respect to x:∫d
dx (e3xy)dx =∫e5xdx. 5. This gives e3xy=1
5e5x+C, where Cis
the constant of integration. 6. Solve for yby dividing by e3x:y=1
5e2x+Ce−3x. 7. Use the
initial condition y(0) = 1 to find C:1 = 1
5·1 + C·1. 8. Solve for Cto get C=4
5. 9. Therefore,
the solution to the initial value problem is y=1
5e2x+4
5e−3x.
19. Question: Solve the following differential equation using the method of integrating factors:
dy
dx +y=e−x.
Ans. Step-by-step solution: 1. We can see that the given differential equation is in the
form of first-order linear differential equation, which can be written as dy
dx +P(x)y=Q(x),
where P(x)=1and Q(x) = e−x. 2. To solve this equation, we first find the integrating
factor, denoted by I(x), which is given by I(x) = e∫P(x)dx. 3. Since P(x)=1, we have
I(x) = e∫1dx =ex. 4. We then multiply both sides of the original equation by the integrating
factor exto get exdy
dx +exy=exe−x. 5. Simplifying the above equation gives d
dx (exy) = 1. 6.
Integrating both sides with respect to xgives exy=∫1dx =x+C, where Cis the constant of
integration. 7. Therefore, the general solution to the differential equation is y=e−x(x+C).
20. Question: Find the general solution to the differential equation
y′′ −5y′+ 6y= 0
Ans. Step-by-step solution: 1. Write the characteristic equation by substituting y=ert into
the differential equation:
r2−5r+ 6 = 0
2. Factor the quadratic equation to solve for r:
(r−2)(r−3) = 0
r= 2,3
3. The general solution is then given by:
y(t) = c1e2t+c2e3t
where c1and c2are arbitrary constants.
21. Question: Solve the following differential equation using an integrating factor: dy
dx +y=x.
Ans. Step-by-step solution: 1. The given differential equation is in standard linear form
dy
dx +P(x)y=Q(x), where P(x)=1and Q(x) = x. 2. To solve this differential equation,
we first determine the integrating factor µ(x)defined by µ(x) = e∫P(x)dx. 3. In this case,
µ(x) = e∫1dx =ex. 4. We multiply both sides of the differential equation by the integrating
factor µ(x)to obtain: exdy
dx +exy=exx. 5. Notice that the left-hand side can be rewritten
using the product rule: d
dx (exy) = exdy
dx +exy. 6. Therefore, the differential equation becomes:
d
dx (exy) = exx. 7. We integrate both sides with respect to xto get: exy=∫exxdx. 8.
Integrating the right-hand side by parts (repeated integration by parts may be needed), we
find: ∫exxdx =ex(x−1) + C, where Cis the constant of integration. 9. Thus, we have
exy=ex(x−1) + C. 10. Finally, dividing by exgives us the general solution: y=x−1 + Ce−x,
where Cis an arbitrary constant.
22. Question 22: Solve the following initial value problem:
dy
dx + 2y= 3x2
Given that y(0) = 1.
Ans. Step-by-step solution:
1. First, we identify the integrating factor µ(x) = e∫2dx =e2x.
2. Multiply the given differential equation by the integrating factor:
e2xdy
dx + 2e2xy= 3x2e2x
3. Notice that the left side can be rewritten as the derivative of e2xy:
d
dx(e2xy) = 3x2e2x
4. Integrate both sides with respect to x:
∫d
dx(e2xy)dx =∫3x2e2xdx
5. Simplifying the integrals gives:
e2xy=∫3x2e2xdx =e2x(3x2−6x+ 6) + C
6. Solve for yby dividing both sides by e2x:
y= 3x2−6x+6+Ce−2x
7. Apply the initial condition y(0) = 1 to find the value of the constant C:
1 = 0 −0+6+C
C=−5
8. Therefore, the solution to the initial value problem is:
y= 3x2−6x+ 6 −5e−2x
23. Consider the following differential equation:
dy
dx =2x+y
x
Solve the differential equation above using the method of integrating factors.
Ans. Solution: 1. Notice that the given differential equation is not in standard form for
integrating factors. We can rewrite it as:
dy
dx −y
x= 2
2. Now, we identify the integrating factor, denoted by µ(x), as:
µ(x) = e∫−1
xdx =e−ln |x|=1
|x|
3. Multiplying both sides of the differential equation by the integrating factor, we get:
1
|x|
dy
dx −1
|x|
y
x= 2 ·1
|x|
4. Rewrite the left side as the derivative of the product of the integrating factor and the
function y:
d
dx (y
|x|)=2
|x|
5. Integrate both sides with respect to xto solve for y:
∫d
dx (y
|x|)dx =∫2
|x|dx
y
|x|= 2 ln |x|+C
6. Multiply through by |x|to find the general solution:
y= 2xln |x|+C|x|
Thus, the general solution to the given differential equation is y= 2xln |x|+C|x|, where C
is an arbitrary constant.
24. **Question 24:** Solve the following differential equation using the method of undetermined
coefficients:
y′′ −4y= 2e3x+ 5 cos(2x).
Ans. **Step-by-step solution:** 1. First, we find the complementary solution ycby solving the
characteristic equation r2−4 = 0:
r2−4 = 0 =⇒r=±2.
Therefore, the complementary solution is yc=c1e2x+c2e−2x, where c1and c2are arbitrary
constants.
2. Next, we find the particular solution ypby assuming it has the form Ae3x+Bcos(2x).
Since e3xand cos(2x)are already a part of the complementary solution, we need to multiply
these functions by xto prevent overlap. Thus, we try yp=Axe3x+Bx cos(2x).
3. We differentiate yptwice to find y′′
p:
y′
p= (3A+ 3Ax)e3x−Bx sin(2x),
y′′
p= (9A+ 3Ae3x)−2Bsin(2x)−2Bx cos(2x).
4. Substituting y′′
pback into the original differential equation, we have:
(9A+ 3Ae3x)−2Bsin(2x)−2Bx cos(2x)−4(Axe3x+Bx cos(2x)) = 2e3x+ 5 cos(2x).
5. Equating the coefficients of like terms, we get the following system of equations:
9A−4A= 2 =⇒A=2
5,
3A−2B= 0 =⇒B=3
10.
6. Therefore, the particular solution is:
yp=2
5xe3x+3
10xcos(2x).
7. The general solution is the sum of the complementary and particular solutions:
y=yc+yp=c1e2x+c2e−2x+2
5xe3x+3
10xcos(2x).
Therefore, the solution to the differential equation is y=c1e2x+c2e−2x+2
5xe3x+3
10xcos(2x).
25. Question 25: Solve the following initial value problem using the method of undetermined
coefficients:
y′′ + 5y′+ 6y= 3e−2t
y(0) = 2, y′(0) = −3
Ans. Step 1. Firstly, we find the complementary solution by solving the homogeneous equation
y′′ + 5y′+ 6y= 0.
The characteristic equation is r2+ 5r+ 6 = 0, which can be factored as (r+ 2)(r+ 3) = 0.
Thus, the roots are r1=−2and r2=−3.
Therefore, the complementary solution is yc(t) = c1e−2t+c2e−3t.
Step 2. Next, we find a particular solution for the non-homogeneous equation. Since the
right-hand side is in the form of 3e−2t, we assume a particular solution of the form yp(t) = Ae−2t.
Substitute yp(t)into the differential equation and solve for A:
y′′
p+ 5y′
p+ 6yp= 3e−2t
(−4A−10A+ 6A)e−2t= 3e−2t
−8A= 3
A=−3
8
Thus, the particular solution is yp(t) = −3
8e−2t.
Step 3. The general solution is given by the sum of the complementary and particular
solutions:
y(t) = yc(t) + yp(t) = c1e−2t+c2e−3t−3
8e−2t
Step 4. Finally, we apply the initial conditions to determine the values of c1and c2.
From y(0) = 2:
2 = c1+c2−3
8
From y′(0) = −3:
−3 = −2c1−3c2+3
2
Solving these two equations simultaneously, we find c1=25
8and c2=−41
8.
Therefore, the solution to the initial value problem is:
y(t) = 25
8e−2t−41
8e−3t−3
8e−2t
y(t) = 22
8e−2t−41
8e−3t
26. Question: Find the general solution to the following first-order ordinary differential equation:
dy
dx =e2xy+ex
Ans. Step-by-step solution: 1. We have the first-order linear differential equation in the form
dy
dx +P(x)y=Q(x). Here P(x) = e2xand Q(x) = ex. To solve this equation, we first find the
integrating factor, denoted by I(x):
I(x) = e∫P(x)dx
I(x) = e∫e2xdx
I(x) = e2x
2. Multiply the obtained integrating factor with the given differential equation:
e2xdy
dx +e2xP(x)y=e2xQ(x)
d
dx(e2xy) = e3x
3. Integrate both sides with respect to x:
∫d
dx(e2xy)dx =∫e3xdx
e2xy=1
3e3x+C
4. Solve for yto find the general solution:
y=1
3ex+Ce−2x
where Cis an arbitrary constant. Therefore, the general solution to the differential equation is
y=1
3ex+Ce−2x.
27. Question: Solve the initial value problem:
y′=1 + y
x, y(1) = 2.
Ans. Step-by-step solution: 1. Rewrite the given differential equation in the form y′=f(x)g(y):
y′=1 + y
x.
2. Separate the variables by rewriting the equation as:
dy
dx =1 + y
x.
3. Rearrange the equation to get all terms involving yon one side and terms involving xon
the other: dy
1 + y=dx
x.
4. Integrate both sides with respect to their respective variables:
∫1
1 + ydy =∫1
xdx.
5. Evaluate the integrals:
ln |1 + y|=ln |x|+C,
where Cis the constant of integration.
6. Solve for yby exponentiating both sides:
1 + y=±x·eC.
7. Now replace Cwith C=ln |C|:
1 + y=±x·eln |C|=±x·C.
8. Apply the initial condition y(1) = 2 to find the specific value of the constant C:
1 + 2 = ±1·C⇒C= 3.
9. Substitute C= 3 back into the equation:
1 + y=±x·3.
10. Solve for y:
y=±3x−1.
Thus, the solution to the initial value problem is y= 3x−1.
28. Question: Solve the following differential equation:
y′+2y
x=x2
Ans. Step-by-step solution:
1. We start by recognizing that this is a first-order linear differential equation in the form
y′+P(x)y=Q(x), where P(x) = 2
xand Q(x) = x2.
2. To solve this equation, we first find the integrating factor µ(x)given by:
µ(x) = e∫P(x)dx
In this case, µ(x) = e∫2
xdx =e2ln |x|=eln(x2)=x2.
3. We multiply the entire differential equation by µ(x) = x2:
x2y′+ 2xy =x4
4. The left side of the equation can be rewritten using the product rule for differentiation:
(x2y)′=x4
5. Integrating both sides, we get:
x2y=x5
5+C
6. Finally, solving for y gives:
y=x3
5+C
x2
Therefore, the general solution to the given differential equation is y=x3
5+C
x2where Cis
an arbitrary constant.
29. Find the general solution to the differential equation: dy
dx =y2
Ans. To solve the differential equation dy
dx =y2, we can separate variables and integrate both
sides.
1. Separate variables:
dy
y2=dx
2. Integrate both sides: ∫1
y2dy =∫dx
∫y−2dy =∫dx
−y−1=x+C
−1
y=x+C
3. To find the general solution, solve for y:
y=1
−x−C
y=−1
x+C
Therefore, the general solution to the differential equation dy
dx =y2is y=−1
x+C, where Cis
the constant of integration.
30. Question: Solve the following differential equation:
y′=xy +ex
Ans. Step-by-step solution:
1. This is a first-order linear ordinary differential equation of the form y′+P(x)y=Q(x).
2. To solve this type of equation, we first find the integrating factor, which is given by µ(x) =
e∫P(x)dx. 3. In this case, P(x) = x, so ∫P(x)dx =∫xdx =1
2x2. 4. Therefore, the integrating
factor is µ(x) = e1
2x2. 5. Multiply both sides of the differential equation by the integrating factor:
e1
2x2y′+xe1
2x2y=ex+1
2x2
6. Recognize that the left side is the derivative of the product of the integrating factor and y:
d
dx(e1
2x2y) = ex+1
2x2
7. Integrate both sides with respect to x:
∫d
dx(e1
2x2y)dx =∫ex+1
2x2dx
e1
2x2y=∫ex+1
2x2dx
8. The integral on the right side can be simplified by completing the square in the exponent:
ex+1
2x2=exe1
2x2=exe1
2(x2+2x+1−1) =exe1
2(x+1)2
9. Now we can evaluate the integral on the right side:
∫ex+1
2x2dx =∫exe1
2(x+1)2dx
This integral involves using Gaussian integral techniques. 10. After integrating, we get:
dy
dx =e1
2x2+Ce−1
2(x+1)2
Here, Cis the constant of integration. 11. Finally, solve for yby integrating both sides:
y=∫(e1
2x2+Ce−1
2(x+1)2)dx
This integral must now be evaluated to obtain the final solution.
31. Question: Solve the following initial value problem: y′=y2+1
x,y(1) = 1.
Ans. Step-by-step solution:
1. Rewrite the differential equation as dy
dx =y2+1
x.
2. Separate the variables by writing dy
y2+1
x
=dx.
3. Integrate both sides:
∫dy
y2+1
x
=∫dx
4. The integral on the left side can be simplified using partial fractions. We can write
1
y2+1
x
=Ax+By+C
y2x+1 .
5. Multiplying both sides by y2x+ 1 gives 1 = (Ax +By +C)(y2x+ 1).
6. Expanding and collecting like terms, we find 1 = Axy2+By3x+Ay +By +C.
7. Equating coefficients of like terms, we get the following system of equations:
Ay = 0
A+B= 0
C= 1
8. Solving the system of equations gives A= 0,B= 0, and C= 1.
9. Therefore, we can rewrite the integral as ∫dy
y2+1
x
=∫dx
1, which simplifies to ∫dy
y2+1
x
=
x+K, where Kis the constant of integration.
10. Now, we can solve the integral on the left side by making the substitution u=y√x:
∫dy
y2+1
x
=∫du
u2+ 1 =arctan(u) + C=arctan(y√x)+C
11. Therefore, we have arctan(y√x) + C=x+K.
12. Substitute the initial condition y(1) = 1 into the equation to find the value of C:
arctan(1·√1)+C= 1 + K
π
4+C= 1 + K
13. Thus, the solution to the initial value problem is y=√tan(x+π
4).
32. Question: Solve the differential equation y′+y=sin(x).
Ans. Step-by-step solution:
1. The given differential equation is a linear first-order differential equation of the form
y′+P(x)y=Q(x). In this case, P(x) = 1 and Q(x) = sin(x).
2. First, we find the integrating factor µ(x) = e∫P(x)dx. In this case, µ(x) = e∫1dx =ex.
3. Multiply the differential equation by the integrating factor: exy′+exy=exsin(x).
4. Rewrite the left-hand side as the derivative of the product y(x)ex:d
dx (y(x)ex) = exsin(x).
5. Integrate both sides with respect to x:y(x)ex=∫exsin(x)dx.
6. Integrate the right-hand side using integration by parts: Let u=sin(x)and dv =exdx.
Then du =cos(x)dx and v=ex.
7. Using the integration by parts formula ∫udv =uv −∫vdu, we have ∫exsin(x)dx =
exsin(x)−∫excos(x)dx.
8. Integrate ∫excos(x)dx by parts again, or recognize that it is a standard integral excos(x)−
∫exsin(x)dx.
9. Substitute back into the equation y(x)ex=exsin(x)−(excos(x)−∫exsin(x)dx).
10. Simplify and solve for y(x):y(x) = sin(x)−cos(x) + Ce−x, where Cis the constant of
integration.
33. Question: Solve the following differential equation:
(x3+y3)dx −3xy2dy = 0
Ans. Step-by-step solution: 1. We start by rearranging the terms in the given differential
equation:
x3dx +y3dx −3xy2dy = 0
2. Now we notice that the given DE is exact, as the coefficients of dx and dy are partial
derivatives of the same function. 3. To solve an exact differential equation, we find a function
f(x, y)such that its partial derivatives are equal to the coefficients of dx and dy. 4. Integrating
x3with respect to xgives us 1
4x4+g(y), where g(y)is a function of yonly. 5. Taking the partial
derivative of 1
4x4+g(y)with respect to ygives g′(y), which is our next term in the function
f(x, y). 6. Integrating y3with respect to ygives us 1
4y4. 7. Comparing with the original given
differential equation, we see that we need to have g(y) = 0 in order to have an exact differential
equation. 8. Therefore, the solution to the differential equation is given by 1
4x4+1
4y4=C,
where Cis the constant of integration.
34. Question: Find the general solution of the following differential equation:
y′′ −3y′+ 2y=ex
Ans. Solution: Let’s first find the complementary function by solving the homogeneous
equation y′′ −3y′+ 2y= 0.
1. We assume the solutions in the form y=erx. Substitute y=erx into the homogeneous
equation:
r2erx −3rerx + 2erx = 0
2. Divide the equation by erx:
r2−3r+ 2 = 0
3. Solve the quadratic equation r2−3r+ 2 = 0 to find the roots:
(r−1)(r−2) = 0 =⇒r= 1,2
The complementary function is then:
yc=c1ex+c2e2x
where c1and c2are arbitrary constants.
4. Next, we determine a particular solution for the non-homogeneous equation using the
method of undetermined coefficients. Since exis a solution component of the complemen-
tary function, we try a particular solution of the form yp=Axex.
5. Substitute y=Axexinto the non-homogeneous equation:
(Ax + 2A)−3(A+ 2A) + 2(Ax) = ex
6. Simplify the equation:
Ax −3Ax + 2Ax =ex
Ax =ex
So, A= 1.
7. Therefore, the particular solution is yp=xex.
8. The general solution is the sum of the complementary function and the particular solution:
y=yc+yp=c1ex+c2e2x+xex
where c1and c2are arbitrary constants.
Thus, the general solution of the differential equation y′′ −3y′+2y=exis y=c1ex+c2e2x+xex.
35. Question: Solve the following second-order differential equation using the method of
undetermined coefficients:
y′′ −2y′+y=ex
Ans. Solution:
1. First, solve the associated homogeneous equation y′′ −2y′+y= 0. The characteristic
equation is r2−2r+ 1 = 0, which factors as (r−1)2= 0. This gives us a repeated root
r1=r2= 1. So the solution to the homogeneous equation is yh= (c1+c2x)ex.
2. Now, we consider the particular solution for the given non-homogeneous equation y′′ −
2y′+y=ex. Since the right-hand side is ex, we assume the particular solution has the form
yp=Axex. Substitute ypinto the differential equation:
y′′
p−2y′
p+yp=ex
Ans. Step-by-step solution: 1. First, we find the complementary solution by solving the
characteristic equation r2−4r+ 4 = 0:
r2−4r+ 4 = (r−2)2= 0
This gives us a repeated root r= 2, so the complementary solution is:
yc(x) = (c1+c2x)e2x
2. Next, we guess the particular solution yp(x)for the non-homogeneous differential equation.
Since the right-hand side contains e2x, we let:
yp(x) = Ae2x
where Ais a constant to be determined. 3. Substitute yp(x)into the differential equation and
its derivatives to find the values of the undetermined coefficients. We have:
y′
p(x) = 2Ae2x
y′′
p(x) = 4Ae2x
Substitute into the differential equation:
4Ae2x−4(2Ae2x) + 4(Ae2x) = e2x
Simplify this to get:
4Ae2x−8Ae2x+ 4Ae2x=e2x
Solving, we find A=1
2. 4. Therefore, the particular solution is:
yp(x) = 1
2e2x
5. The general solution is the sum of the complementary and particular solutions:
y(x) = yc(x) + yp(x) = (c1+c2x)e2x+1
2e2x
where c1and c2are arbitrary constants determined by initial conditions.
3. Question: Solve the following initial value problem:
y′= 2xy, y(0) = 3.
Ans. Step-by-step solution: 1. The given differential equation is a first-order linear differential
equation in the form y′=f(x)·y. To solve this equation, we will use the method of separation
of variables. We write the equation as dy
dx = 2xy.
2. Separate the variables by dividing both sides by yand multiplying by dx, yielding dy
y=
2x dx.
3. Integrate both sides. Integrating the left side gives ∫1
ydy =ln |y|, and integrating the
right side gives ∫2x dx =x2+C, where Cis the constant of integration.
4. Therefore, we have ln |y|=x2+C.
5. Now, apply the initial condition y(0) = 3. Substituting this into the equation gives
ln |3|= 0 + C, so C=ln(3).
6. Substituting C=ln(3) back into the equation, we have ln |y|=x2+ln(3).
7. Exponentiate both sides to solve for y:|y|=ex2+ln(3) =ex2·eln(3) = 3ex2.
8. We eliminate the absolute value by considering two cases: y= 3ex2if y > 0, and
y=−3ex2if y < 0.
9. Since y(0) = 3 >0, we conclude that the solution is y= 3ex2.
4. Find the general solution to the following differential equation: dy
dx =1
x+y
Ans. Let’s solve the given differential equation step by step:
1. First, rewrite the equation in the standard form of a first-order linear differential equation,
which is dy
dx −y=1
x.
2. The integrating factor I(x)is given by I(x) = e∫−1dx =e−x.
3. Multiply both sides of the equation by the integrating factor and simplify: e−xdy
dx −e−xy=
e−x1
x.
4. The left side can be rewritten using the product rule for differentiation: d
dx (e−xy) = e−x1
x.
5. Integrate both sides with respect to x:∫d
dx (e−xy)dx =∫e−x1
xdx. This gives e−xy=
∫e−x1
xdx +C.
6. Solve the integral on the right side using integration by parts: Let u=1
xand dv =e−xdx.
Then, du =−1
x2dx and v=−e−x. The integral becomes −e−x1
x+∫e−x1
x2dx.
7. Integrate the second term on the right side: ∫e−x1
x2dx =−e−x1
x−∫e−x−1
x2dx. This
simplifies to −e−x1
x+∫e−x1
x2dx.
8. Substitute this back into the equation and simplify: e−xy=−e−x1
x−e−x1
x+C.
9. Simplify further and solve for yto find the general solution: y=−2−C
ex. Hence, the
general solution to the given differential equation is y=−2−C
ex, where Cis the constant of
integration.
5. Question: Solve the initial value problem y′−2y= 4e2xwith initial condition y(0) = 1.
Ans. Step-by-step solution: 1. First, we find the integrating factor e∫−2dx =e−2x. 2. Multiply
the differential equation by the integrating factor: e−2x(y′−2y) = e−2x(4e2x). This simplifies to
e−2xy′−2e−2xy= 4. 3. The left-hand side is now the derivative of the product e−2xy, which we
can integrate to get the solution: ∫(e−2xy)′dx =∫4dx. This gives e−2xy= 4x+C, where C
is the constant of integration. 4. Using the initial condition y(0) = 1, we find the value of the
constant C:e0·1 = 4 ·0 + C, which simplifies to C= 1. 5. Substituting back into the solution,
we have e−2xy= 4x+ 1. So, the general solution is y(x) = e2x(4x+ 1).
6. Find the general solution to the following differential equation:
dy
dx =2x+ 3y−1
3x−y+ 2
Ans. Step 1. Rewrite the equation in a standard form by multiplying throughout by 3x−y+2:
(3x−y+ 2)dy
dx = 2x+ 3y−1
3xdy
dx −ydy
dx + 2dy
dx = 2x+ 3y−1
Step 2. Rearrange the terms:
3xdy
dx + 2dy
dx =y+ 2x−1+3y
(3x+ 2)dy
dx = 4x+ 4y−1
Step 3. Integrate both sides with respect to x:
∫(3x+ 2) dy
dx dx =∫(4x+ 4y−1) dx
∫(3x+ 2) dy =∫(4x+ 4y−1) dx
Step 4. Perform the integrations:
3xy + 2y= 2x2+ 2xy −x+C
Step 5. Simplify the equation:
3xy + 2y= 2x2+ 2xy −x+C
x(3y−2) = 2x2−x+C
(3y−2)dx = (2x−1)dx +C
Step 6. Integrate both sides to find the general solution:
∫(3y−2) dy =∫(2x−1) dx +∫C dx
3
2y2−2y=x2−x+C1
So, the general solution to the differential equation is:
3
2y2−2y=x2−x+C1
where C1is an arbitrary constant.
7. Find the general solution to the following differential equation:
dy
dx +y=ex
Ans. To solve this differential equation, we will use the integrating factor method.
1. Find the integrating factor: The integrating factor, denoted by µ(x), is given by
µ(x) = e∫1dx =ex.
2. Multiply both sides by the integrating factor: Multiplying both sides of the differential
equation by ex, we get:
exdy
dx +exy=e2x
3. Rewrite the left side as a product of a derivative: Since d
dx (exy) = exdy
dx +exy, we
can rewrite the left side of the equation as:
d
dx(exy) = e2x
4. Integrate both sides: Integrating both sides with respect to x, we get:
∫d
dx(exy)dx =∫e2xdx
exy=1
2e2x+C
5. Solve for y:Dividing by ex, we get the general solution:
y=1
2ex+Ce−x
where Cis an arbitrary constant.
8. Question: Solve the differential equation dy
dx =1
2ex+y.
Ans. Step-by-step solution: 1. Rewrite the given differential equation as dy
dx =1
2exey. 2.
Notice that the right-hand side of the equation can be separated into functions of xand y:
dy
dx =1
2exey=1
2exey. 3. Now, separate the variables by dividing both sides of the equation by
ey, and then multiply both sides by dx to bring all xterms to the left side and yterms to the
right side: 1
eydy =1
2exdx. 4. Integrate both sides with respect to their respective variables:
∫1
eydy =∫1
2exdx. 5. Simplify the integrals to get −e−y=1
2ex+C, where Cis the constant
of integration. 6. Solve for yby multiplying through by −1to get e−y=−1
2ex−C. 7. Finally,
solve for yby taking the natural logarithm of both sides: y=−ln (−1
2ex−C), where Cis an
arbitrary constant.
9. Question: Solve the following initial value problem: y′= 2xy −x,y(0) = 1.
Ans. Step-by-step solution: 1. We start by writing the given differential equation in the standard
form of a first-order linear differential equation: y′−2xy =−x. 2. To solve this equation, we
first find the integrating factor e∫−2xdx. 3. Integrating −2xwith respect to x, we get −x2, so the
integrating factor is e−x2. 4. Multiplying the given differential equation by the integrating factor,
we have e−x2y′−2xye−x2=−xe−x2. 5. By recognizing the left side as the derivative of the
product e−x2y, we can rewrite the equation as d
dx (e−x2y) = −xe−x2. 6. Integrating both sides
with respect to x, we get e−x2y=C−1
2e−x2+D, where Cand Dare constants of integration.
7. Applying the initial condition y(0) = 1, we find e−0·1 = C−1
2e−0+D. 8. Simplifying, we
get C−1
2+D= 1. 9. Therefore, C+D=3
2. 10. Finally, the solution to the initial value
problem is y=ex2(3
2−1
2e−x2).
10. Question: Solve the following first-order ordinary differential equation:
y′=x
y
Ans. Step-by-step solution: 1. Rewrite the differential equation in the form dy
dx =x
y. 2. Multiply
both sides by yto get y dy =x dx. 3. Integrate both sides: ∫y dy =∫x dx. 4. Integrating the
left side gives 1
2y2+C1, where C1is the constant of integration. 5. Integrating the right side
gives 1
2x2+C2, where C2is the constant of integration. 6. Therefore, the general solution is
1
2y2=1
2x2+Cwhere C=C1−C2is the constant of integration. 7. Finally, we can write the
solution in the form y=±√x2+ 2C.
11. Question: Solve the following first-order linear differential equation: dy
dx +y=e−x.
Ans. Step-by-step solution: 1. We have a first-order linear differential equation of the form
dy
dx +P(x)y=Q(x), where in this case P(x) = 1 and Q(x) = e−x. 2. To solve this equation, we
first find the integrating factor, which is given by µ(x) = e∫P(x)dx. In this case, µ(x) = e∫1dx =
ex. 3. We multiply both sides of the differential equation by the integrating factor, µ(x), to get:
exdy
dx +exy=exe−x. 4. Recognizing that the left side of the equation is the derivative of the
product of yand µ(x), we can rewrite the equation as d
dx (yex) = 1. 5. Integrating both sides
with respect to x, we get ∫d
dx (yex)dx =∫1dx. 6. Simplifying, we have yex=x+C, where
Cis the constant of integration. 7. Finally, solving for y, we get y=x
ex+Ce−x, where Cis
an arbitrary constant. Thus, the general solution to the differential equation is y=x
ex+Ce−x,
where Cis an arbitrary constant.
12. Question:
Solve the differential equation y′+ 2xy =ex2, with the initial condition y(0) = 3.
Ans. Solution:
1. First, we’ll find the integrating factor µ(x)by multiplying the entire equation by it:
µ(x) = e∫2x dx =ex2
2. Multiply the original differential equation by the integrating factor:
ex2y′+ 2xex2y=e2x2
3. Notice that the left-hand side is the derivative of the product of µ(x)and y:
d
dx(ex2y) = e2x2
4. Integrating both sides with respect to x, we get:
ex2y=∫e2x2dx +C
ex2y=√π
2erf(x) + C
5. Now, applying the initial condition y(0) = 3:
e0·3 = √π
2erf(0) + C
3 = √π
2·0 + C
C= 3
6. Therefore, the particular solution to the differential equation is:
ex2y=√π
2erf(x)+3
13. Question: Solve the initial value problem: dy
dx =1
x+ywith y(1) = 0.
Ans. Step-by-step solution: 1. Rewrite the differential equation in the form dy
dx =f(x, y)by
isolating dy on one side:
dy
dx =1
x+y
2. Rearrange the equation to separate variables:
(x+y)dy =dx
3. Integrate both sides: ∫(x+y)dy =∫dx
4. Solve the integrals: ∫x dy +∫y dy =∫dx
xy +y2
2=x+C1
5. Apply the initial condition y(1) = 0 to find the constant C1:
1·0 + 02
2= 1 + C1
0 = 1 + C1
C1=−1
6. Substitute C1=−1back into the equation:
xy +y2
2=x−1
7. To simplify further, let’s write the equation in a different form by completing the square on
the left side:
y2+ 2xy = 2x−2
y2+ 2xy +x2= 2x+x2−2
(y+x)2=x2+ 2x−2
8. Take the square root of both sides and solve for y:
y+x=±√x2+ 2x−2
y=−x±√x2+ 2x−2
Thus, the solution to the initial value problem is y=−x±√x2+ 2x−2.
14. Question: Solve the following first-order linear differential equation:
dy
dx +2y
x=x3
Ans. Step-by-step solution: 1. First, we need to write the given differential equation in the
standard form of a linear first-order ODE, which is dy
dx +P(x)y=Q(x). To do this, we divide
throughout by xto get:
dy
dx + 2y
x=x2
2. Now we can identify P(x)=2and Q(x) = x2. Next, we find the integrating factor µ(x),
which is given by:
µ(x) = e∫P(x)dx
µ(x) = e∫2dx =e2x
3. Multiplying the integrating factor by the differential equation gives us:
e2xdy
dx + 2e2xy
x=e2xx2
4. This can be rewritten as follows:
d
dx(e2xy) = x2e2x
5. Integrate both sides of the equation with respect to x:
∫d
dx(e2xy)dx =∫x2e2xdx
e2xy=1
2x2e2x−1
2∫x2e2xdx
6. Solving the integral on the right side using integration by parts will give us:
e2xy=1
2x2e2x−1
4x2e2x+1
4∫e2xdx
7. Continuing to simplify, we get:
e2xy=1
4x2e2x+1
8e2x+C
8. Finally, we solve for yby dividing by e2x:
y=1
4x2+1
8+Ce−2x
where Cis the constant of integration.
15. Question 15: Solve the following first-order linear differential equation:
dy
dx + 3y=e−4x
Ans. Step 1. Find the integrating factor µ(x)by multiplying the entire equation by it:
µ(x) = e∫3dx =e3x
Step 2. Multiply the equation by the integrating factor and rewrite it in the form of a total
derivative:
e3xdy
dx + 3e3xy=e−x
Step 3. Express the left side as a total derivative:
d
dx (e3xy)=e−4x
Step 4. Integrate both sides with respect to x:
e3xy=∫e−4xdx
Step 5. Integrate the right-hand side:
e3xy=−1
4e−4x+C
Step 6. Solve for y:
y=−1
4e−7x+Ce−3x
where Cis an arbitrary constant.
16. Question 16: Solve the following first-order linear differential equation using an integrating
factor: dy
dx +(2 + 1
x)y=x
Ans. Step-by-step solution: 1. Write the given first-order linear differential equation in the
standard form dy
dx +P(x)y=Q(x), where P(x)and Q(x)are functions of x. The given differential
equation is:
dy
dx +(2 + 1
x)y=x
Here, P(x) = 2 + 1
xand Q(x) = x.
2. Calculate the integrating factor µ(x) = e∫P(x)dx.
µ(x) = e∫(2+ 1
x)dx =e2x+ln |x|=x2e2x
3. Multiply both sides of the differential equation by the integrating factor µ(x).
x2e2xdy
dx +x2e2x(2 + 1
x)y=x3e2x
4. Rewrite the left side of the equation as a product rule of a single function.
d
dx (yx2e2x)=x3e2x
5. Integrate both sides with respect to x.
∫d(yx2e2x)=∫x3e2xdx
yx2e2x=1
2x3e2x−1
2∫x3e2xdx +C
6. Solve for yby dividing by x2e2x.
y=1
2x−1
2∫x·x2e2xdx +C
x2e2x
7. Calculate ∫x3e2xdx by using integration by parts. Let u=x3and dv =e2xdx, then
du = 3x2dx and v=1
2e2x.
1
2x3e2x−3
2∫x2e2xdx =y
Therefore, the solution to the differential equation is:
y=1
2x−1
2(1
2x3e2x−3
2∫x2e2xdx)+C
x2e2x
17. Question: Solve the differential equation y′′ + 4y=sin(2x)subject to the initial conditions
y(0) = 1 and y′(0) = 0.
Ans. Step-by-step solution:
1. First, we find the complementary solution to the homogeneous differential equation y′′ +
4y= 0. The characteristic equation is r2+ 4 = 0, which has roots r=±2i. Therefore, the
complementary solution is yc(x) = c1cos(2x) + c2sin(2x).
2. Next, we find a particular solution to the nonhomogeneous differential equation y′′ + 4y=
sin(2x). Since the right-hand side is of the form sin(2x), we guess a particular solution of the
form yp(x) = Asin(2x) + Bcos(2x).
3. We find the first and second derivatives of yp(x):y′
p(x) = 2Acos(2x)−2Bsin(2x)
y′′
p(x) = −4Asin(2x)−4Bcos(2x)
4. We substitute yp(x)and its derivatives into the differential equation to get: −4Asin(2x)−
4Bcos(2x) + 4(Asin(2x) + Bcos(2x)) = sin(2x)Simplifying, we get −4B= 1, so B=−1
4.
5. Substituting B=−1
4back into yp(x), we have yp(x) = Asin(2x)−1
4cos(2x). Now,
substitute the initial conditions into the particular solution to find A.
6. We have y(0) = 1 and y′(0) = 0: Substitute x= 0 into y(0) = Asin(0) −1
4cos(0) =
A(0) −1
4(1) = 1 This simplifies to −1
4= 1, which is not possible. Therefore, we made a mistake
in our particular solution.
7. We need to consider that the particular solution should be of the form yp(x) = Ax sin(2x)+
Bx cos(2x). Proceed with steps 3-5 with this new form to find the correct particular solution.
8. Once you have the correct particular solution, substitute the initial conditions to find the
values of the constants Aand B. Finally, add the complementary and particular solutions to get
the general solution.
9. The final solution will satisfy the given initial conditions y(0) = 1 and y′(0) = 0.
18. Question: Solve the initial value problem y′−3y=e2xwith y(0) = 1.
Ans. Solution: 1. First, we find the integrating factor by solving the auxiliary equation
m−3=0, which gives m= 3. The integrating factor is then µ(x) = e∫3dx =e3x. 2. Multiply
both sides of the differential equation by the integrating factor: e3xy′−3e3xy=e5x. 3. Recognize
the left side as the derivative of the product of e3xand y:d
dx (e3xy) = e5x. 4. Integrate both
sides with respect to x:∫d
dx (e3xy)dx =∫e5xdx. 5. This gives e3xy=1
5e5x+C, where Cis
the constant of integration. 6. Solve for yby dividing by e3x:y=1
5e2x+Ce−3x. 7. Use the
initial condition y(0) = 1 to find C:1 = 1
5·1 + C·1. 8. Solve for Cto get C=4
5. 9. Therefore,
the solution to the initial value problem is y=1
5e2x+4
5e−3x.
19. Question: Solve the following differential equation using the method of integrating factors:
dy
dx +y=e−x.
Ans. Step-by-step solution: 1. We can see that the given differential equation is in the
form of first-order linear differential equation, which can be written as dy
dx +P(x)y=Q(x),
where P(x)=1and Q(x) = e−x. 2. To solve this equation, we first find the integrating
factor, denoted by I(x), which is given by I(x) = e∫P(x)dx. 3. Since P(x)=1, we have
I(x) = e∫1dx =ex. 4. We then multiply both sides of the original equation by the integrating
factor exto get exdy
dx +exy=exe−x. 5. Simplifying the above equation gives d
dx (exy) = 1. 6.
Integrating both sides with respect to xgives exy=∫1dx =x+C, where Cis the constant of
integration. 7. Therefore, the general solution to the differential equation is y=e−x(x+C).
20. Question: Find the general solution to the differential equation
y′′ −5y′+ 6y= 0
Ans. Step-by-step solution: 1. Write the characteristic equation by substituting y=ert into
the differential equation:
r2−5r+ 6 = 0
2. Factor the quadratic equation to solve for r:
(r−2)(r−3) = 0
r= 2,3
3. The general solution is then given by:
y(t) = c1e2t+c2e3t
where c1and c2are arbitrary constants.
21. Question: Solve the following differential equation using an integrating factor: dy
dx +y=x.
Ans. Step-by-step solution: 1. The given differential equation is in standard linear form
dy
dx +P(x)y=Q(x), where P(x)=1and Q(x) = x. 2. To solve this differential equation,
we first determine the integrating factor µ(x)defined by µ(x) = e∫P(x)dx. 3. In this case,
µ(x) = e∫1dx =ex. 4. We multiply both sides of the differential equation by the integrating
factor µ(x)to obtain: exdy
dx +exy=exx. 5. Notice that the left-hand side can be rewritten
using the product rule: d
dx (exy) = exdy
dx +exy. 6. Therefore, the differential equation becomes:
d
dx (exy) = exx. 7. We integrate both sides with respect to xto get: exy=∫exxdx. 8.
Integrating the right-hand side by parts (repeated integration by parts may be needed), we
find: ∫exxdx =ex(x−1) + C, where Cis the constant of integration. 9. Thus, we have
exy=ex(x−1) + C. 10. Finally, dividing by exgives us the general solution: y=x−1 + Ce−x,
where Cis an arbitrary constant.
22. Question 22: Solve the following initial value problem:
dy
dx + 2y= 3x2
Given that y(0) = 1.
Ans. Step-by-step solution:
1. First, we identify the integrating factor µ(x) = e∫2dx =e2x.
2. Multiply the given differential equation by the integrating factor:
e2xdy
dx + 2e2xy= 3x2e2x
3. Notice that the left side can be rewritten as the derivative of e2xy:
d
dx(e2xy) = 3x2e2x
4. Integrate both sides with respect to x:
∫d
dx(e2xy)dx =∫3x2e2xdx
5. Simplifying the integrals gives:
e2xy=∫3x2e2xdx =e2x(3x2−6x+ 6) + C
6. Solve for yby dividing both sides by e2x:
y= 3x2−6x+6+Ce−2x
7. Apply the initial condition y(0) = 1 to find the value of the constant C:
1 = 0 −0+6+C
C=−5
8. Therefore, the solution to the initial value problem is:
y= 3x2−6x+ 6 −5e−2x
23. Consider the following differential equation:
dy
dx =2x+y
x
Solve the differential equation above using the method of integrating factors.
Ans. Solution: 1. Notice that the given differential equation is not in standard form for
integrating factors. We can rewrite it as:
dy
dx −y
x= 2
2. Now, we identify the integrating factor, denoted by µ(x), as:
µ(x) = e∫−1
xdx =e−ln |x|=1
|x|
3. Multiplying both sides of the differential equation by the integrating factor, we get:
1
|x|
dy
dx −1
|x|
y
x= 2 ·1
|x|
4. Rewrite the left side as the derivative of the product of the integrating factor and the
function y:
d
dx (y
|x|)=2
|x|
5. Integrate both sides with respect to xto solve for y:
∫d
dx (y
|x|)dx =∫2
|x|dx
y
|x|= 2 ln |x|+C
6. Multiply through by |x|to find the general solution:
y= 2xln |x|+C|x|
Thus, the general solution to the given differential equation is y= 2xln |x|+C|x|, where C
is an arbitrary constant.
24. **Question 24:** Solve the following differential equation using the method of undetermined
coefficients:
y′′ −4y= 2e3x+ 5 cos(2x).
Ans. **Step-by-step solution:** 1. First, we find the complementary solution ycby solving the
characteristic equation r2−4 = 0:
r2−4 = 0 =⇒r=±2.
Therefore, the complementary solution is yc=c1e2x+c2e−2x, where c1and c2are arbitrary
constants.
2. Next, we find the particular solution ypby assuming it has the form Ae3x+Bcos(2x).
Since e3xand cos(2x)are already a part of the complementary solution, we need to multiply
these functions by xto prevent overlap. Thus, we try yp=Axe3x+Bx cos(2x).
3. We differentiate yptwice to find y′′
p:
y′
p= (3A+ 3Ax)e3x−Bx sin(2x),
y′′
p= (9A+ 3Ae3x)−2Bsin(2x)−2Bx cos(2x).
4. Substituting y′′
pback into the original differential equation, we have:
(9A+ 3Ae3x)−2Bsin(2x)−2Bx cos(2x)−4(Axe3x+Bx cos(2x)) = 2e3x+ 5 cos(2x).
5. Equating the coefficients of like terms, we get the following system of equations:
9A−4A= 2 =⇒A=2
5,
3A−2B= 0 =⇒B=3
10.
6. Therefore, the particular solution is:
yp=2
5xe3x+3
10xcos(2x).
7. The general solution is the sum of the complementary and particular solutions:
y=yc+yp=c1e2x+c2e−2x+2
5xe3x+3
10xcos(2x).
Therefore, the solution to the differential equation is y=c1e2x+c2e−2x+2
5xe3x+3
10xcos(2x).
25. Question 25: Solve the following initial value problem using the method of undetermined
coefficients:
y′′ + 5y′+ 6y= 3e−2t
y(0) = 2, y′(0) = −3
Ans. Step 1. Firstly, we find the complementary solution by solving the homogeneous equation
y′′ + 5y′+ 6y= 0.
The characteristic equation is r2+ 5r+ 6 = 0, which can be factored as (r+ 2)(r+ 3) = 0.
Thus, the roots are r1=−2and r2=−3.
Therefore, the complementary solution is yc(t) = c1e−2t+c2e−3t.
Step 2. Next, we find a particular solution for the non-homogeneous equation. Since the
right-hand side is in the form of 3e−2t, we assume a particular solution of the form yp(t) = Ae−2t.
Substitute yp(t)into the differential equation and solve for A:
y′′
p+ 5y′
p+ 6yp= 3e−2t
(−4A−10A+ 6A)e−2t= 3e−2t
−8A= 3
A=−3
8
Thus, the particular solution is yp(t) = −3
8e−2t.
Step 3. The general solution is given by the sum of the complementary and particular
solutions:
y(t) = yc(t) + yp(t) = c1e−2t+c2e−3t−3
8e−2t
Step 4. Finally, we apply the initial conditions to determine the values of c1and c2.
From y(0) = 2:
2 = c1+c2−3
8
From y′(0) = −3:
−3 = −2c1−3c2+3
2
Solving these two equations simultaneously, we find c1=25
8and c2=−41
8.
Therefore, the solution to the initial value problem is:
y(t) = 25
8e−2t−41
8e−3t−3
8e−2t
y(t) = 22
8e−2t−41
8e−3t
26. Question: Find the general solution to the following first-order ordinary differential equation:
dy
dx =e2xy+ex
Ans. Step-by-step solution: 1. We have the first-order linear differential equation in the form
dy
dx +P(x)y=Q(x). Here P(x) = e2xand Q(x) = ex. To solve this equation, we first find the
integrating factor, denoted by I(x):
I(x) = e∫P(x)dx
I(x) = e∫e2xdx
I(x) = e2x
2. Multiply the obtained integrating factor with the given differential equation:
e2xdy
dx +e2xP(x)y=e2xQ(x)
d
dx(e2xy) = e3x
3. Integrate both sides with respect to x:
∫d
dx(e2xy)dx =∫e3xdx
e2xy=1
3e3x+C
4. Solve for yto find the general solution:
y=1
3ex+Ce−2x
where Cis an arbitrary constant. Therefore, the general solution to the differential equation is
y=1
3ex+Ce−2x.
27. Question: Solve the initial value problem:
y′=1 + y
x, y(1) = 2.
Ans. Step-by-step solution: 1. Rewrite the given differential equation in the form y′=f(x)g(y):
y′=1 + y
x.
2. Separate the variables by rewriting the equation as:
dy
dx =1 + y
x.
3. Rearrange the equation to get all terms involving yon one side and terms involving xon
the other: dy
1 + y=dx
x.
4. Integrate both sides with respect to their respective variables:
∫1
1 + ydy =∫1
xdx.
5. Evaluate the integrals:
ln |1 + y|=ln |x|+C,
where Cis the constant of integration.
6. Solve for yby exponentiating both sides:
1 + y=±x·eC.
7. Now replace Cwith C=ln |C|:
1 + y=±x·eln |C|=±x·C.
8. Apply the initial condition y(1) = 2 to find the specific value of the constant C:
1 + 2 = ±1·C⇒C= 3.
9. Substitute C= 3 back into the equation:
1 + y=±x·3.
10. Solve for y:
y=±3x−1.
Thus, the solution to the initial value problem is y= 3x−1.
28. Question: Solve the following differential equation:
y′+2y
x=x2
Ans. Step-by-step solution:
1. We start by recognizing that this is a first-order linear differential equation in the form
y′+P(x)y=Q(x), where P(x) = 2
xand Q(x) = x2.
2. To solve this equation, we first find the integrating factor µ(x)given by:
µ(x) = e∫P(x)dx
In this case, µ(x) = e∫2
xdx =e2ln |x|=eln(x2)=x2.
3. We multiply the entire differential equation by µ(x) = x2:
x2y′+ 2xy =x4
4. The left side of the equation can be rewritten using the product rule for differentiation:
(x2y)′=x4
5. Integrating both sides, we get:
x2y=x5
5+C
6. Finally, solving for y gives:
y=x3
5+C
x2
Therefore, the general solution to the given differential equation is y=x3
5+C
x2where Cis
an arbitrary constant.
29. Find the general solution to the differential equation: dy
dx =y2
Ans. To solve the differential equation dy
dx =y2, we can separate variables and integrate both
sides.
1. Separate variables:
dy
y2=dx
2. Integrate both sides: ∫1
y2dy =∫dx
∫y−2dy =∫dx
−y−1=x+C
−1
y=x+C
3. To find the general solution, solve for y:
y=1
−x−C
y=−1
x+C
Therefore, the general solution to the differential equation dy
dx =y2is y=−1
x+C, where Cis
the constant of integration.
30. Question: Solve the following differential equation:
y′=xy +ex
Ans. Step-by-step solution:
1. This is a first-order linear ordinary differential equation of the form y′+P(x)y=Q(x).
2. To solve this type of equation, we first find the integrating factor, which is given by µ(x) =
e∫P(x)dx. 3. In this case, P(x) = x, so ∫P(x)dx =∫xdx =1
2x2. 4. Therefore, the integrating
factor is µ(x) = e1
2x2. 5. Multiply both sides of the differential equation by the integrating factor:
e1
2x2y′+xe1
2x2y=ex+1
2x2
6. Recognize that the left side is the derivative of the product of the integrating factor and y:
d
dx(e1
2x2y) = ex+1
2x2
7. Integrate both sides with respect to x:
∫d
dx(e1
2x2y)dx =∫ex+1
2x2dx
e1
2x2y=∫ex+1
2x2dx
8. The integral on the right side can be simplified by completing the square in the exponent:
ex+1
2x2=exe1
2x2=exe1
2(x2+2x+1−1) =exe1
2(x+1)2
9. Now we can evaluate the integral on the right side:
∫ex+1
2x2dx =∫exe1
2(x+1)2dx
This integral involves using Gaussian integral techniques. 10. After integrating, we get:
dy
dx =e1
2x2+Ce−1
2(x+1)2
Here, Cis the constant of integration. 11. Finally, solve for yby integrating both sides:
y=∫(e1
2x2+Ce−1
2(x+1)2)dx
This integral must now be evaluated to obtain the final solution.
31. Question: Solve the following initial value problem: y′=y2+1
x,y(1) = 1.
Ans. Step-by-step solution:
1. Rewrite the differential equation as dy
dx =y2+1
x.
2. Separate the variables by writing dy
y2+1
x
=dx.
3. Integrate both sides:
∫dy
y2+1
x
=∫dx
4. The integral on the left side can be simplified using partial fractions. We can write
1
y2+1
x
=Ax+By+C
y2x+1 .
5. Multiplying both sides by y2x+ 1 gives 1 = (Ax +By +C)(y2x+ 1).
6. Expanding and collecting like terms, we find 1 = Axy2+By3x+Ay +By +C.
7. Equating coefficients of like terms, we get the following system of equations:
Ay = 0
A+B= 0
C= 1
8. Solving the system of equations gives A= 0,B= 0, and C= 1.
9. Therefore, we can rewrite the integral as ∫dy
y2+1
x
=∫dx
1, which simplifies to ∫dy
y2+1
x
=
x+K, where Kis the constant of integration.
10. Now, we can solve the integral on the left side by making the substitution u=y√x:
∫dy
y2+1
x
=∫du
u2+ 1 =arctan(u) + C=arctan(y√x)+C
11. Therefore, we have arctan(y√x) + C=x+K.
12. Substitute the initial condition y(1) = 1 into the equation to find the value of C:
arctan(1·√1)+C= 1 + K
π
4+C= 1 + K
13. Thus, the solution to the initial value problem is y=√tan(x+π
4).
32. Question: Solve the differential equation y′+y=sin(x).
Ans. Step-by-step solution:
1. The given differential equation is a linear first-order differential equation of the form
y′+P(x)y=Q(x). In this case, P(x) = 1 and Q(x) = sin(x).
2. First, we find the integrating factor µ(x) = e∫P(x)dx. In this case, µ(x) = e∫1dx =ex.
3. Multiply the differential equation by the integrating factor: exy′+exy=exsin(x).
4. Rewrite the left-hand side as the derivative of the product y(x)ex:d
dx (y(x)ex) = exsin(x).
5. Integrate both sides with respect to x:y(x)ex=∫exsin(x)dx.
6. Integrate the right-hand side using integration by parts: Let u=sin(x)and dv =exdx.
Then du =cos(x)dx and v=ex.
7. Using the integration by parts formula ∫udv =uv −∫vdu, we have ∫exsin(x)dx =
exsin(x)−∫excos(x)dx.
8. Integrate ∫excos(x)dx by parts again, or recognize that it is a standard integral excos(x)−
∫exsin(x)dx.
9. Substitute back into the equation y(x)ex=exsin(x)−(excos(x)−∫exsin(x)dx).
10. Simplify and solve for y(x):y(x) = sin(x)−cos(x) + Ce−x, where Cis the constant of
integration.
33. Question: Solve the following differential equation:
(x3+y3)dx −3xy2dy = 0
Ans. Step-by-step solution: 1. We start by rearranging the terms in the given differential
equation:
x3dx +y3dx −3xy2dy = 0
2. Now we notice that the given DE is exact, as the coefficients of dx and dy are partial
derivatives of the same function. 3. To solve an exact differential equation, we find a function
f(x, y)such that its partial derivatives are equal to the coefficients of dx and dy. 4. Integrating
x3with respect to xgives us 1
4x4+g(y), where g(y)is a function of yonly. 5. Taking the partial
derivative of 1
4x4+g(y)with respect to ygives g′(y), which is our next term in the function
f(x, y). 6. Integrating y3with respect to ygives us 1
4y4. 7. Comparing with the original given
differential equation, we see that we need to have g(y) = 0 in order to have an exact differential
equation. 8. Therefore, the solution to the differential equation is given by 1
4x4+1
4y4=C,
where Cis the constant of integration.
34. Question: Find the general solution of the following differential equation:
y′′ −3y′+ 2y=ex
Ans. Solution: Let’s first find the complementary function by solving the homogeneous
equation y′′ −3y′+ 2y= 0.
1. We assume the solutions in the form y=erx. Substitute y=erx into the homogeneous
equation:
r2erx −3rerx + 2erx = 0
2. Divide the equation by erx:
r2−3r+ 2 = 0
3. Solve the quadratic equation r2−3r+ 2 = 0 to find the roots:
(r−1)(r−2) = 0 =⇒r= 1,2
The complementary function is then:
yc=c1ex+c2e2x
where c1and c2are arbitrary constants.
4. Next, we determine a particular solution for the non-homogeneous equation using the
method of undetermined coefficients. Since exis a solution component of the complemen-
tary function, we try a particular solution of the form yp=Axex.
5. Substitute y=Axexinto the non-homogeneous equation:
(Ax + 2A)−3(A+ 2A) + 2(Ax) = ex
6. Simplify the equation:
Ax −3Ax + 2Ax =ex
Ax =ex
So, A= 1.
7. Therefore, the particular solution is yp=xex.
8. The general solution is the sum of the complementary function and the particular solution:
y=yc+yp=c1ex+c2e2x+xex
where c1and c2are arbitrary constants.
Thus, the general solution of the differential equation y′′ −3y′+2y=exis y=c1ex+c2e2x+xex.
35. Question: Solve the following second-order differential equation using the method of
undetermined coefficients:
y′′ −2y′+y=ex
Ans. Solution:
1. First, solve the associated homogeneous equation y′′ −2y′+y= 0. The characteristic
equation is r2−2r+ 1 = 0, which factors as (r−1)2= 0. This gives us a repeated root
r1=r2= 1. So the solution to the homogeneous equation is yh= (c1+c2x)ex.
2. Now, we consider the particular solution for the given non-homogeneous equation y′′ −
2y′+y=ex. Since the right-hand side is ex, we assume the particular solution has the form
yp=Axex. Substitute ypinto the differential equation:
y′′
p−2y′
p+yp=ex
⇒(Aex)′′ −2(Aex)′+Aex=ex
⇒Aex−2Aex+Aex=ex
⇒ −Aex+Aex=ex
⇒0 = ex
This leads to a contradiction, so we must use the method of undetermined coefficients. Let’s
try a new form for ypwith Axex:
yp=Ax2ex
y′
p= (2Ax +Ax2)ex
y′′
p= (2A+ 4Ax +Ax2)ex
Substitute ypinto the differential equation:
(2A+ 4Ax +Ax2)ex−2(2Ax +Ax2)ex+Ax2ex=ex
2A+ 4Ax +Ax2−4Ax −2Ax2+Ax2= 1
2A= 1
A=1
2
So, the particular solution is yp=1
2x2ex.
3. The general solution is the sum of the homogeneous and particular solutions:
y=yh+yp= (c1+c2x)ex+1
2x2ex
Hence, the solution to the differential equation is y= (c1+1
2)ex+c2xex+1
2x2ex.
