MAT 142 - COLLEGE MATHEMATICS - Differential
Equations Practice Material - Set 3
1. Question:
Consider the following differential equation:
dy
dx =x2+y2
Find the particular solution to the initial value problem given that y(0) = 1.
Ans. Step-by-step solution:
1. Write the differential equation as:
dy
dx =x2+y2
2. Separate the variables by moving all terms involving yto one side and all terms involving
xto the other side:
dy
dx −y2=x2
3. This is a first order nonlinear ordinary differential equation that we can attempt to solve
using the method of integrating factors. To do this, we first identify the integrating factor µ(x)
by multiplying through by it:
µ(x)dy
dx −µ(x)y2=x2µ(x)
4. Now, we choose µ(x)such that the left side becomes the derivative with respect to xof
µ(x)y:
d
dx[µ(x)y] = x2µ(x)
5. So, in order to identify µ(x), we solve the differential equation:
µ′(x) = x2µ(x)
6. This is a separable differential equation that can be solved by integrating both sides:
dµ
µ=x2dx
Zdµ
µ=Zx2dx
ln |µ|=x3
3+C
µ=ex3
3+C=Cex3
3
Since µ(x)cannot be zero, we can set C = 1.
7. Multiply the original differential equation by the integrating factor:
ex3
3dy
dx −ex3
3y2=x2ex3
3
8. Notice that the left side of the equation can be simplified using the product rule:
d
dx[ex3
3y] = x2ex3
3
9. Integrate both sides with respect to xto solve for y:
ex3
3y=Zx2ex3
3dx +C
y=e−x3
3Zx2ex3
3dx +Ce−x3
3
10. The integral on the right side can be computed by substitution. Given that y(0) = 1, we
can solve for C.
Thus, the particular solution will be the solution to the initial value problem.
2. Question:
Solve the initial value problem: y′= 2xy2with y(0) = 1.
Ans. Step-by-step solution:
1. First, we separate the variables in the differential equation. We can rewrite the differential
equation as dy
dx = 2xy2. 2. Next, we separate the variables by dividing both sides by y2and
multiplying both sides by dx:dy
y2= 2xdx. 3. Integrate both sides with respect to their respective
variables. For the left side, we have R1
y2dy =Ry−2dy =−y−1+C1, where C1is the constant
of integration. 4. For the right side, we have R2xdx =x2+C2, where C2is the constant of
integration. 5. Putting these together, we have −y−1=x2+Cwhere C=C2−C1. 6. To solve
for y, we can solve for y:y=1
−x2−C=−1
x2+C. 7. Now, we use the initial condition y(0) = 1 to
find the value of C. Substituting y= 1 and x= 0 into y=−1
x2+C, we get 1 = −1
C, which implies
that C=−1. 8. Finally, the solution to the initial value problem is y=−1
x2−1.
3. Question: Solve the following first-order linear differential equation:
dy
dx + 2y= 4e−2x
Ans. Step-by-step solution: 1. We first identify the integrating factor, which is given by the
exponential of the integral of the coefficient of y:
Integrating factor, µ(x) = e∫2dx =e2x
2. Multiply both sides of the differential equation by the integrating factor:
e2xdy
dx + 2e2xy= 4e2xe−2x
This simplifies to:
e2xdy
dx + 2e2xy= 4
3. Notice that the left side is now a derivative of the product ye2x. Rewrite the equation
accordingly:
d
dx ye2x= 4
4. Integrate both sides with respect to x:
Zd
dx ye2xdx =Z4dx
This gives us:
ye2x= 4x+C
5. Solve for y:
y=4x+C
e2x=4x
e2x+C
e2x
where Cis an arbitrary constant.
4. Question: Solve the following differential equation: dy
dx +y=e−x.
Ans. Step-by-step solution: 1. First, we need to identify the integrating factor. The integrating
factor is given by I(x) = e∫1dx =ex.
2. Multiply the given differential equation by the integrating factor:
exdy
dx +exy=exe−x
exdy
dx +exy= 1
3. Notice that the left side can be rewritten as the derivative of the product of the integrating
factor and yusing the product rule:
d
dx(exy) = 1
4. Integrate both sides with respect to x:
Zd
dx(exy)dx =Z1dx
5. This simplifies to:
exy=x+C
where Cis the constant of integration.
6. Finally, solve for yto obtain the general solution:
y=e−x(x+C)
where Cis an arbitrary constant.
5. Question: Solve the following first-order linear differential equation: dy
dx + 2xy =x.
Ans. Step-by-step solution: 1. Write the given differential equation in standard form, which is
dy
dx +p(x)y=q(x). 2. In this case, p(x)=2xand q(x) = x. 3. To solve this equation, we can
use an integrating factor given by e∫p(x)dx. 4. Calculate the integrating factor as e∫2xdx =ex2. 5.
Multiply both sides of the differential equation by the integrating factor: ex2dy
dx + 2xex2y=xex2.
6. Recognize the left side as the derivative of the integrating factor times y:d
dx (ex2y) = xex2.
7. Integrate both sides with respect to x:Rd
dx (ex2y)dx =Rxex2dx. 8. This simplifies to
ex2y=1
2ex2+C, where Cis the constant of integration. 9. Solve for yto get y=1
2+Ce−x2,
where Cis an arbitrary constant. Thus, the solution to the differential equation is y=1
2+Ce−x2.
6. Question:
Find the general solution to the differential equation y′′ + 4y= 0.
Ans. Solution:
1. Homogeneous Differential Equation:
The given differential equation is y′′ + 4y= 0. Let’s first write this in standard form:
y′′ + 4y= 0.
2. Characteristic Equation:
The characteristic equation for the given differential equation is obtained by assuming the
solution is of the form y=ert, where ris a constant to be determined. Substituting this into
the differential equation, we get:
r2ert + 4ert = 0.
Dividing both sides by ert, we have:
r2+ 4 = 0.
3. Finding the roots of the characteristic equation:
Solving the characteristic equation r2+ 4 = 0, we get r=±2i.
4. General Solution:
Since the roots are complex (r=±2i), the general solution to the differential equation is
given by:
y(t) = Acos(2t) + Bsin(2t),
where Aand Bare arbitrary constants determined by initial conditions if provided. This
represents the complete solution set to the differential equation y′′ + 4y= 0.
7. Question: Solve the following first-order linear differential equation:
y′+ (2x−1)y=x.
Ans. Solution: 1. The equation is in the form y′+P(x)y=Q(x), where P(x)=2x−1and
Q(x) = x. We will use an integrating factor to solve the equation.
2. The integrating factor I(x)is given by I(x) = e∫P(x)dx. Calculating the integral, we have:
Z(2x−1) dx =x2−x.
So, the integrating factor is I(x) = ex2−x.
3. Multiplying both sides of the differential equation by the integrating factor, we get:
ex2−xy′+ex2−x(2x−1)y=ex2−xx.
4. Recognizing the left side as the product rule of a product of two functions, we have:
(ex2−xy)′=ex2−xx.
5. Integrating both sides with respect to x, we get:
Z(ex2−xy)′dx =Zex2−xx dx.
6. Solving the integrals on both sides, we have:
ex2−xy=Zx·ex2−xdx.
Since the integral on the right side is not straightforward, we will make a substitution to simplify
it.
7. Let z=x2−x, then dz = (2x−1)dx. Solving for dx, we get dx =dz
2x−1. Substituting
into the integral, we have: Zx·ex2−xdx =Zz
2·ezdz.
8. The integral now becomes more manageable:
Zz
2·ezdz =1
2Zzezdz.
Using integration by parts, u=zand dv =ezdz, we get:
1
2Zzezdz =1
2(zez−Zezdz) = 1
2(zez−ez).
9. Substituting back z=x2−x, we have:
Zx·ex2−xdx =1
2((x2−x)ex2−x−ex2−x).
10. Therefore, the solution to the differential equation is:
ex2−xy=1
2((x2−x)ex2−x−ex2−x) + C,
where Cis the constant of integration.
11. Finally, dividing by ex2−x, we obtain the general solution:
y=1
2(x2−x−1) + Ce−x2+x,
where Cis an arbitrary constant.
8. Question: Solve the following first-order differential equation:
dy
dx + 2xy =x2
Ans. Step-by-step solution:
1. This is a first-order linear differential equation in standard form: dy
dx +P(x)y=Q(x),
where P(x) = 2xand Q(x) = x2.
2. To solve this equation, we first find the integrating factor, denoted by I(x) = e∫P(x)dx.
I(x) = e∫2xdx =ex2
3. Multiply both sides of the differential equation by the integrating factor.
ex2dy
dx + 2xex2y=x2ex2
4. Recognize that the left hand side can be written as the derivative of the product of the
integrating factor and yusing the product rule.
d
dx(ex2y) = x2ex2
5. Integrate both sides to solve for y.
Zd
dx(ex2y)dx =Zx2ex2dx
6. After integrating, we get:
ex2y=1
2x2ex2+C
7. Finally, solve for yby dividing by the integrating factor ex2.
y=1
2x2+Ce−x2
So, the general solution to the differential equation is y=1
2x2+Ce−x2, where Cis an
arbitrary constant.
9. Question: Solve the differential equation xy′+ 2y=x3given that y(1) = 2.
Ans. Step-by-step solution:
1. Write the differential equation in standard form: y′+2
xy=x2.
2. This is a first-order linear differential equation. To solve it, we first find the integrating
factor I(x) = e∫2
xdx =e2ln |x|=x2.
3. Multiply the differential equation by the integrating factor: x2y′+ 2x2y=x4.
4. Notice that the left side can be written as (x2y)′=x4.
5. Integrate both sides with respect to x: R(x2y)′dx =Rx4dx.
6. Solve the integrals: x2y=1
5x5+C.
7. Divide by x2to get the general solution: y=1
5x3+C
x2.
8. Apply the initial condition y(1) = 2 to find the particular solution. Substituting x= 1 and
y= 2 into the general solution, we find C= 9.
9. Therefore, the solution to the differential equation is y=1
5x3+9
x2.
10. Question:
Solve the following differential equation: dy
dx =y2cos(x), where y(0) = 1.
Ans. Step-by-step solution:
1. Separate the variables in the differential equation:
dy
y2=cos(x)dx
2. Integrate both sides with respect to their respective variables:
Z1
y2dy =Zcos(x)dx
⇒ −1
y=sin(x) + C
3. Solve for yby rearranging the equation and using the initial condition y(0) = 1:
y=−1
sin(x) + C
4. Plug in the initial condition y(0) = 1 to find the value of C:
1 = −1
sin(0) + C=−1
0 + C=−1
C
⇒C=−1
5. Substitute C=−1back into the equation to get the final solution:
y=−1
sin(x)−1
11. Question: Solve the initial value problem:
dy
dx =x
y+e−y, y(0) = 1
Ans. Solution: 1. Rewrite the initial value problem in the form of a separable differential
equation:
dy
dx =x
y+e−y
dy
dx =x
y+e−y
y dy =x dx +e−ydy
2. Integrate both sides: Zy dy =Zx dx +Ze−ydy
y2
2=x2
2−e−y+C
3. Since we are given the initial condition y(0) = 1, we can find the constant C:
1
2= 0 −e−1+C
C=1
2+e−1
4. Substitute the value of Cback into the equation:
y2
2=x2
2−e−y+1
2+e−1
5. To solve for yexplicitly, we can rewrite the equation as:
y2+ 2e−y=x2+ 2e−1+ 1
6. This equation is transcendental and cannot be solved algebraically. We can use numerical
methods to approximate the solution.
Therefore, the solution to the initial value problem is:
y2+ 2e−y=x2+ 2e−1+ 1
12. Question: Find the general solution to the following differential equation:
d2y
dx2+ 4dy
dx + 4y=e−2x
Ans. Solution: Let’s first find the complementary function by solving the corresponding
homogenous differential equation:
1. Setting the RHS to 0, we get the homogenous differential equation:
d2y
dx2+ 4dy
dx + 4y= 0
The characteristic equation is given by m2+ 4m+ 4 = 0, which simplifies to (m+ 2)2= 0.
Hence, we have a repeated root m=−2.
2. Therefore, the complementary function is given by:
yc(x) = c1e−2x+c2xe−2x
Next, let’s find the particular solution using the method of undetermined coefficients.
3. Since the RHS of the differential equation is e−2x, we assume the particular solution has
the form yp=ae−2x.
4. We substitute ypback into the differential equation to find a. Taking the derivatives and
plugging into the differential equation, we get:
4a+ 4a+ 4ae−2x=e−2x
5. Simplifying, we find that a=1
8.
6. Therefore, the particular solution is yp(x) = 1
8e−2x.
7. The general solution is then the sum of the complementary and particular solutions:
y(x) = yc(x) + yp(x) = c1e−2x+c2xe−2x+1
8e−2x
where c1and c2are constants determined by initial conditions if provided.
13. Question: Solve the differential equation y′+ 2xy =xwith the initial condition y(0) = 1.
Ans. Solution: 1. First, we rewrite the differential equation in standard form by multiplying
through by the integrating factor e∫2xdx:
e∫2xdxy′+ 2xe∫2xdxy=xe∫2xdx
ex2y′+ 2xex2y=xex2
2. Notice that the left-hand side can be simplified using the product rule for differentiation
in reverse: d
dx(ex2y) = xex2
3. Integrating both sides with respect to x, we have:
Zd
dx(ex2y)dx =Zxex2dx
ex2y=1
2ex2+C
4. Now, we solve for y:
y=1
2+Ce−x2
5. To find the value of the constant C, we use the initial condition y(0) = 1:
1 = 1
2+Ce0
C=1
2
6. Therefore, the solution to the differential equation with the initial condition is:
y=1
2+1
2e−x2
14. Question: Solve the differential equation y′′ −y′−2y= 0 with initial conditions y(0) = 2
and y′(0) = −1.
Ans. Solution: 1. Write down the characteristic equation: r2−r−2 = 0. 2. Solve the
characteristic equation to find the roots: r= 2,−1. 3. Write down the general solution of the
differential equation: y(t) = c1e2t+c2e−t. 4. Apply the initial conditions to find the values of c1
and c2. 5. From y(0) = 2, we have 2 = c1+c2. 6. From y′(0) = −1, we have −1=2c1−c2.
7. Solve the system of equations to find c1and c2. 8. From 2 = c1+c2, we get c1=1
3and
c2=5
3. 9. Therefore, the particular solution to the initial value problem is y(t) = 1
3e2t+5
3e−t.
15. Find the general solution to the differential equation:
dy
dx =y2−4
x2
Ans. Solution:
1. We can rewrite the given differential equation as:
dy
dx =(y−2)(y+ 2)
x2
2. Next, we can separate the variables:
dy
(y−2)(y+ 2) =dx
x2
3. Now, we can perform partial fraction decomposition on the left-hand side:
1
41
y−2−1
y+ 2=dx
x2
4. Integrating both sides:
1
4ln
y−2
y+ 2
=−1
x+C
5. Exponentiating both sides:
y−2
y+ 2
=e−4/x+4C
6. Simplifying further:
y−2
y+ 2 =±e−4/xe4C
7. Since e4Cis a positive constant, we can rewrite as:
y−2
y+ 2 =±Ce−4/x
8. Cross multiply to solve for y:
y−2 = ±Cye−4/x⇒y=2±2Cye−4/x
1∓Cye−4/x
9. Therefore, the general solution to the differential equation is:
y=2±2Cye−4/x
1∓Cye−4/x
16. Find the general solution of the following differential equation:
y′′ −4y′+ 4y= 0
Ans. To solve the differential equation y′′ −4y′+ 4y= 0, we can assume a solution of the form
y=ert, where ris a constant to be determined.
1. Write down the characteristic equation by substituting y=ert into the differential equa-
tion:
(r2−4r+ 4)ert = 0
2. Solve the characteristic equation r2−4r+ 4 = 0 to find the roots r:
(r−2)2= 0 =⇒r= 2
3. Since we have a repeated root, the general solution of the differential equation is:
y= (c1+c2t)e2t
where c1and c2are arbitrary constants.
17. Question: Find the particular solution to the differential equation dy
dx =e2x+ywith the
initial condition y(0) = 1.
Ans. Step-by-step solution: 1. We start by separating variables in the differential equation:
dy
dx =e2x+y
dy
dx −y=e2x
2. Next, we multiply both sides by the integrating factor e−xto simplify the equation:
e−xdy
dx −e−xy=ex
3. Rewrite the left side as the derivative of the product of e−xand y:
d
dx(e−xy) = ex
4. Integrate both sides with respect to xto solve for y:
Zd
dx(e−xy)dx =Zexdx
e−xy=Zexdx
5. Integrate the right side and solve for y:
e−xy=ex+C
y=e2x+Cex
6. Use the initial condition y(0) = 1 to find the value of the constant C:
1 = e0+C·e0
1 = 1 + C
C= 0
7. Substituting C= 0 back into the solution gives the particular solution to the differential
equation:
y=e2x
18. Question: Solve the following first-order linear differential equation:
y′+ 2xy = 3x
Ans. Step-by-step solution: 1. This is a first-order linear differential equation in the form
y′+P(x)y=Q(x), where P(x) = 2xand Q(x) = 3x. 2. To solve this equation, we first find
the integrating factor µ(x) = e∫P(x)dx. In this case, µ(x) = e∫2xdx =ex2. 3. Multiply both
sides of the differential equation by the integrating factor µ(x):
ex2y′+ 2xex2y= 3xex2
4. Notice that the left-hand side is the derivative of the product ex2ywith respect to x. Therefore,
we can rewrite the equation as:
d
dx(ex2y) = 3xex2
5. Integrate both sides with respect to x:
Zd
dx(ex2y)dx =Z3xex2dx
ex2y=3
2ex2+C
where Cis the constant of integration. 6. Finally, solve for yby dividing both sides by ex2:
y=3
2+Ce−x2
7. Therefore, the general solution to the differential equation y′+ 2xy = 3xis y=3
2+Ce−x2,
where Cis an arbitrary constant.
19. Question 19: Solve the differential equation y′−2xy = 2x, where y=y(x).
Ans. Step-by-step solution: 1. First, we will rewrite the given differential equation in the
standard form of a first-order linear differential equation: y′−2xy = 2x. 2. The integrating
factor µ(x)is given by µ(x) = e∫−2x dx =e−x2. 3. Multiply both sides of the differential
equation by the integrating factor µ(x):e−x2y′−2xe−x2y= 2xe−x2. 4. Rewriting the left-hand
side as the derivative of the product e−x2y, we get d
dx (e−x2y) = 2xe−x2. 5. Integrate both
sides with respect to x:e−x2y=R2xe−x2dx. 6. Solve the integral on the right-hand side:
R2xe−x2dx =−e−x2+C, where Cis the constant of integration. 7. Therefore, we have
e−x2y=−e−x2+C. 8. Finally, solve for y:y=−1 + Cex2, where Cis an arbitrary constant.
So, the general solution to the differential equation y′−2xy = 2xis y=−1 + Cex2, where C
is an arbitrary constant.
20. Question: Solve the differential equation dy
dx =x2y
1+x2.
Ans. Step-by-step solution: 1. Rewrite the differential equation in separable form. We can
rewrite it as: 1
ydy =x2
1 + x2dx
2. Integrate both sides. Integrating the left side gives:
Z1
ydy =ln |y|+C1
3. To integrate the right side, let’s perform a substitution. Let u= 1 + x2, then du = 2x dx.
This means we can rewrite the right side as:
Zx2
1 + x2dx =1
2Z2x
1 + x2·x dx
=1
2Z1
udu =1
2ln |u|+C2=1
2ln |1 + x2|+C2
4. Putting everything together, we have:
ln |y|=1
2ln |1 + x2|+C
y=e1
2ln |1+x2|+C=eC√1 + x2
5. Finally, we can write the general solution as:
y=k√1 + x2
where k=eCis an arbitrary constant.
21. Question: Find the general solution to the differential equation:
dy
dx =2x+ 1
y2
Ans. Solution: 1. Rewrite the given differential equation in separable form by multiplying both
sides by y2:
y2dy
dx = 2x+ 1
2. Integrate both sides with respect to x:
Zy2dy =Z(2x+ 1) dx
Zy2dy = 2 Zx dx +Z1dx
y3
3=x2+x+C
3. Multiply through by 3 to get the general solution:
y3= 3x2+ 3x+ 3C
where Cis the constant of integration.
22. Question: Solve the initial value problem: y′−y2= 0,y(0) = 1.
Ans. Step-by-step solution:
1. First, we rewrite the differential equation as dy
dx −y2= 0.
2. We can separate variables by dividing both sides by y2:dy
dx =y2.
3. Let’s rewrite this as dy
y2=dx.
4. Now we integrate both sides:
Z1
y2dy =Zdx.
5. The integral on the left gives us −1
y=x+C, where Cis the constant of integration.
6. Solving for y, we get y=−1
x+C.
7. Applying the initial condition y(0) = 1, we have: 1 = −1
C⇒C=−1.
8. Therefore, the solution to the initial value problem is y=−1
x−1.
23. Find the particular solution to the differential equation y′′ −y=exsatisfying the initial
conditions y(0) = 1 and y′(0) = 0.
Ans. Solution: 1. First, we find the complementary function by solving the homogeneous
equation y′′ −y= 0. The characteristic equation is r2−1 = 0, which gives us r=±1. Thus,
the complementary function is yc(x) = c1ex+c2e−x.
2. Next, we need to find a particular solution to the non-homogeneous equation y′′ −y=ex.
Let’s guess a particular solution in the form yp(x) = Aex, where Ais a constant to be determined.
3. We substitute yp(x) = Aexinto the differential equation:
y′′
p−yp=ex
Aex−Aex=ex
0 = ex
4. Since 0=exfor all x, our assumption for the particular solution was incorrect. Let’s try
a new guess in the form yp(x) = Axex.
5. Substituting yp(x) = Axexinto the differential equation:
y′′
p−yp=ex
Aex−Axex=ex
A−Ax = 1
6. We get A= 1 and A= 0, which is a contradiction. Therefore, let’s try a new guess in
the form yp(x) = −1
2x2ex.
7. Substituting yp(x) = −1
2x2exinto the differential equation:
y′′
p−yp=ex
−x2ex−2xex−ex=ex
−x2−2x−1 = 1
8. We get −x2−2x= 2, which implies x2+ 2x+ 2 = 0 has no real roots. Therefore, the
correct particular solution is yp(x) = −1
2x2ex.
9. The general solution is then y(x) = yc(x) + yp(x) = c1ex+c2e−x−1
2x2ex.
10. Finally, applying the initial conditions, we have y(0) = 1 and y′(0) = 0:
y(0) = c1+c2= 1
y′(0) = c1−c2= 0
11. Solving the system of equations, we find c1=1
2and c2=1
2.
12. Therefore, the particular solution to the differential equation satisfying the initial condi-
tions is y(x) = 1
2ex+1
2e−x−1
2x2ex.
24. Question: Solve the differential equation dy
dx =2x+3y
3x+2y.
Ans. Solution: 1. Write the given differential equation in the form dy
dx =A(x,y)+B(x,y)dy
dx
C(x,y)+D(x,y)dy
dx
.
2. Here, A(x, y)=2x,B(x, y)=3y,C(x, y)=3x, and D(x, y)=2y. 3. Notice that
M(x, y) = A(x, y)−B(x, y)dy
dx = 2x−3yand N(x, y) = C(x, y)−D(x, y)dy
dx = 3x−2y. 4.
Calculate the partial derivatives ∂M
∂y =−3and ∂N
∂x = 3. 5. Since ∂M
∂y =∂N
∂x , the differential
equation is exact. 6. To solve an exact differential equation, let u(x, y)be a function such that
∂u
∂x =Mand ∂u
∂y =N. 7. Integrate ∂u
∂x = 2x−3yto find u(x, y). 8. Integrate ∂u
∂y = 3x−2yto
find u(x, y). 9. Equate the two expressions for u(x, y)to eliminate the arbitrary constant. 10.
The solution to the differential equation is u(x, y) = x2−3xy +y2=Constant.
25. Let y(t)be a differentiable function that satisfies the differential equation:
dy
dt = 2ty2
with initial condition y(0) = 1
3.
Determine the value of y(1).
Ans. To solve the differential equation, we can separate variables and integrate both sides.
Let’s find the solution step by step: 1. Separate variables:
dy
y2= 2tdt
2. Integrate both sides: Z1
y2dy =Z2tdt
3. Integrate the left side:
−1
y=t2+C
4. Apply the initial condition y(0) = 1
3:
−1
1/3 = 02+C
−3 = C
5. Substitute C=−3back into the equation:
−1
y=t2−3
1
y= 3 −t2
y=1
3−t2
6. Find y(1):
y(1) = 1
3−12=1
2
Therefore, the value of y(1) is 1
2.
26. Consider the first-order differential equation:
(y3−3y)dx + (x2+ 2x)dy = 0
Find the general solution to this differential equation.
Ans. To solve this differential equation, we will rearrange the equation and separate the
variables:
1. Rearranging the equation: Rewrite the given equation in the form of M(x, y)dx +
N(x, y)dy = 0:
M(x, y) = y3−3y
N(x, y) = x2+ 2x
2. Checking for exactness: Calculate the partial derivatives of Mwith respect to yand
Nwith respect to x:
∂M
∂y = 3y2−3
∂N
∂x = 2x+ 2
3. Solving for exactness: The equation is exact if ∂M
∂y =∂N
∂x . Since this condition is not
satisfied, the equation is not exact.
4. Integrating factor: To find an integrating factor, we use the formula (x, y) = e∫Nx−My
Mdx.
Calculate Nx−My:
Nx−My= (2x+ 2) −(3y2−3) = 2x+ 2 −3y2+ 3
Nx−My= 2x−3y2+ 5
The integrating factor (x, y)is e∫2x−3y2+5
y3−3ydx.
5. Multiplier calculation: Calculate the integrating factor:
(x, y) = e∫2x−3y2+5
y3−3ydx =e∫2x−3y2+5
y(y2−3) dx
6. General solution: Multiply the original equation by the integrating factor (x, y):
(y3−3y)e∫2x−3y2+5
y(y2−3) dxdx + (x2+ 2x)e∫2x−3y2+5
y(y2−3) dxdy = 0
This equation can be written in the form of an exact differential and be solved further to find the
general solution.
27. Question 27: Solve the following initial value problem:
dy
dx = 2x+y, y(0) = 1
Ans. Solution: 1. We rearrange the given first-order linear differential equation into standard
form:
dy
dx = 2x+y=⇒dy
dx −y= 2x
2. Next, we identify the integrating factor, which is e∫−1dx =e−x.
3. We multiply the standard form of the equation by the integrating factor:
e−xdy
dx −e−xy= 2xe−x
4. Notice that the left side can be simplified using the product rule of differentiation:
d
dx(e−xy) = 2xe−x
5. Integrating both sides with respect to xgives:
e−xy=Z2xe−xdx
6. Solving the integral on the right side gives:
e−xy=−2xe−x−2e−x+C
7. Dividing by e−x, we find the general solution:
y=−2x−2 + Cex
8. Using the initial condition y(0) = 1, we can find the value of the constant C:
1 = −2(0) −2 + Ce0=⇒C= 3
9. Therefore, the particular solution to the initial value problem is:
y=−2x−2+3ex
28. Question: Solve the initial value problem y′′ −2y′+y= 0,y(0) = 1,y′(0) = 2.
Ans. Solution: 1. We first find the characteristic equation by assuming y=ert. Substituting
this into the differential equation gives us r2−2r+ 1 = 0.
2. Solving the characteristic equation, we find r= 1. Since we have a repeated root, the
general solution will be y(t) = c1et+c2tet.
3. Applying the initial conditions y(0) = 1 and y′(0) = 2, we get c1= 1 and c1+c2= 2.
4. Therefore, the solution to the initial value problem is y(t) = et+ (2 −t)et.
29. Question: Solve the following differential equation:
y′′ −2y′+y= 3 cos(x)
Ans. Solution: 1. First, we find the complementary function by solving the characteristic
equation: r2−2r+ 1 = 0. This gives us (r−1)2= 0, so r= 1 is a repeated root. Thus, the
complementary function is of the form yc(x) = (c1+c2x)ex.
2. Next, we find a particular solution for the non-homogeneous part of the equation. Since
3cos(x)is a trigonometric function, we assume the particular solution has the form yp(x) =
Acos(x) + Bsin(x).
3. We find the first and second derivatives of yp:
y′
p=−Asin(x) + Bcos(x)
y′′
p=−Acos(x)−Bsin(x)
4. Substituting the particular solution and its derivatives into the differential equation, we
have:
−Acos(x)−Bsin(x)−2(−Asin(x) + Bcos(x)) + Acos(x) + Bsin(x) = 3 cos(x)
5. Simplifying, we get:
(−2A−B+A)cos(x)+(−2B+A+B)sin(x) = 3 cos(x)
6. Equating coefficients of cos(x)and sin(x), we get the system of equations:
A−B= 0
A+B=−3
7. Solving the system of equations gives A=−3
2and B=−3
2. Therefore, the particular
solution is yp(x) = −3
2cos(x)−3
2sin(x).
8. Combining the complementary function and the particular solution, the general solution
to the differential equation is:
y(x) = (c1+c2x)ex−3
2cos(x)−3
2sin(x)
where c1and c2are constants determined by initial conditions if given.
30. Question 30: Solve the following differential equation using the method of integrating
factors:
2xy2+ (x2−2y)dx = 0
Ans. Solution: 1. We first rearrange the given differential equation into the form Mdx+N dy =
0:
2xy2dx + (x2−2y)dx = 0
2. Now, we identify Mand Nas follows:
M= 2xy2, N =x2−2y
3. The integrating factor Iis given by:
I=e∫Nx−My
Mdx
4. Calculating Nxand My:
Nx= 2x, My= 2xy
Nx−My= 2x−2xy = 2x(1 −y)
5. Substituting Nx−Myback into the formula for I:
I=e∫2x(1−y)dx =ex2(1−y)
6. Multiply the original differential equation by the integrating factor I:
2xy2ex2(1−y)dx + (x2−2y)ex2(1−y)dx = 0
7. Rewrite the equation as a total differential by recognizing it as the result of the product
rule: d
dx(x2y2ex2(1−y)) = 0
8. Integrate both sides with respect to x:
x2y2ex2(1−y)=C
9. Finally, we have the general solution to the given differential equation as:
x2y2ex2(1−y)=C
where Cis the constant of integration.
31. Question: Find the general solution to the differential equation y′′ −6y′+ 9y= 0.
Ans. Step-by-step solution: 1. Write the characteristic equation by substituting y=erx into
the differential equation.
r2−6r+ 9 = 0
2. Solve the characteristic equation to find the roots r.
(r−3)2= 0
r= 3 (double root)
3. Write down the general solution based on the roots found:
y(x) = (c1+c2x)e3x
where c1and c2are arbitrary constants.
32. Question 32: Solve the following differential equation using the method of integrating
factors:
(2x+y)dx + (x−y)dy = 0
Ans. Solution:
1. Write the given differential equation in the form: M(x, y)dx +N(x, y)dy = 0. Here,
M(x, y) = 2x+yand N(x, y) = x−y.
2. Check if the given differential equation is exact. Calculate ∂M
∂y and ∂N
∂x .
∂M
∂y = 1 =−1 = ∂N
∂x
Since ∂M
∂y =∂N
∂x , the given differential equation is not exact.
3. Find the integrating factor, denoted by µ, to make the differential equation exact. The
integrating factor is given by:
µ=e∫Nx−My
Mdx
4. Calculate Nx−My.
Nx−My=−1−1 = −2
5. Substitute Nx−Myinto the formula for the integrating factor:
µ=e∫−2
2x+ydx =e−2ln |2x+y|=1
|2x+y|2
6. Multiply the given differential equation by the integrating factor µ.
1
|2x+y|2(2x+y)dx +1
|2x+y|2(x−y)dy = 0
7. Rearrange terms and simplify.
2x+y
|2x+y|2dx +x−y
|2x+y|2dy = 0
8. Rewrite the equation in exact form and integrate. The equation now becomes exact:
dx2
|2x+y|= 0
9. Integrate with respect to xto find the general solution.
x2
|2x+y|=C
where Cis the constant of integration.
10. Solve for yto find the particular solution.
y=±px2−C(2x+y)
33. Question:
Solve the following initial value problem:
y′+ 2xy =sin(x), y(0) = 1
Ans. Solution:
1. Homogeneous Solution:
First, we find the solution to the homogeneous differential equation with no forcing term
(sin(x)):
y′+ 2xy = 0
This is a first-order linear differential equation. Let’s rewrite it in standard form y′+P(x)y= 0
where P(x) = 2x.
The integrating factor IF (x)is given by IF (x) = e∫P(x)dx =e∫2x dx =ex2.
Multiplying the differential equation by the integrating factor, we get:
ex2y′+ 2xex2y= 0
Now, using the product rule on the left side, we have:
d
dx(ex2y) = 0
Integrating both sides gives:
ex2y=C
where Cis the constant of integration. Therefore, the general solution to the homogeneous
differential equation is yh=Ce−x2.
2. Particular Solution:
Now, we need to find a particular solution to the non-homogeneous differential equation with
the forcing term sin(x).
Considering a particular solution of the form yp=Asin(x) + Bcos(x), we substitute it back
into the differential equation:
(Acos x−Bsin x)+2x(Asin x+Bcos x) = sin x
Equating coefficients of sin xand cos x:
For sin x:2xA −B= 1
For cos x:−A+ 2xB = 0
Solving this system of equations gives A=1
4and B=1
8.
Therefore, the particular solution is yp=1
4sin x+1
8cos x.
3. Overall Solution:
The overall solution is the sum of the homogeneous and particular solutions:
y=yh+yp=Ce−x2+1
4sin x+1
8cos x
4. Initial Condition:
Applying the initial condition y(0) = 1, we get:
1 = Ce−0+1
4sin 0 + 1
8cos 0
Simplifying, we find C= 1 −1
8=7
8.
5. Final Solution:
The final solution to the initial value problem is:
y=7
8e−x2+1
4sin x+1
8cos x
34. Question:
Solve the following first-order differential equation:
dy
dx =x2+y2
xy
Ans. Solution:
1. Rearrange the terms in the differential equation to separate variables:
dy
dx =x2+y2
xy =⇒y dy =x dx +y2
xdx
2. Integrate both sides of the equation:
Zy dy =Zx dx +Zy2
xdx
3. Solve the integrals:
Integrating y dy with respect to ygives 1
2y2+C1, where C1is the constant of integration.
Integrating x dx with respect to xgives 1
2x2+C2, where C2is another constant of integration.
Integrating y2
xdx involves substituting u=y2and d(u) = 2y dy:
Zy2
xdx =Z1
2u1
√udu =1
2Z√u du =1
3u3
2+C3
where C3is another constant of integration.
Thus, the equation becomes:
1
2y2+C1=1
2x2+C2+1
3y3+C3
4. Rearrange the equation to solve for y:
y2−1
3y3=x2+C
where C=C2−C1+C3.
35. Question 35: Solve the following first-order differential equation using an integrating factor:
y′+ 2y=ex.
ln |µ|=x3
3+C
µ=ex3
3+C=Cex3
3
Since µ(x)cannot be zero, we can set C = 1.
7. Multiply the original differential equation by the integrating factor:
ex3
3dy
dx −ex3
3y2=x2ex3
3
8. Notice that the left side of the equation can be simplified using the product rule:
d
dx[ex3
3y] = x2ex3
3
9. Integrate both sides with respect to xto solve for y:
ex3
3y=Zx2ex3
3dx +C
y=e−x3
3Zx2ex3
3dx +Ce−x3
3
10. The integral on the right side can be computed by substitution. Given that y(0) = 1, we
can solve for C.
Thus, the particular solution will be the solution to the initial value problem.
2. Question:
Solve the initial value problem: y′= 2xy2with y(0) = 1.
Ans. Step-by-step solution:
1. First, we separate the variables in the differential equation. We can rewrite the differential
equation as dy
dx = 2xy2. 2. Next, we separate the variables by dividing both sides by y2and
multiplying both sides by dx:dy
y2= 2xdx. 3. Integrate both sides with respect to their respective
variables. For the left side, we have R1
y2dy =Ry−2dy =−y−1+C1, where C1is the constant
of integration. 4. For the right side, we have R2xdx =x2+C2, where C2is the constant of
integration. 5. Putting these together, we have −y−1=x2+Cwhere C=C2−C1. 6. To solve
for y, we can solve for y:y=1
−x2−C=−1
x2+C. 7. Now, we use the initial condition y(0) = 1 to
find the value of C. Substituting y= 1 and x= 0 into y=−1
x2+C, we get 1 = −1
C, which implies
that C=−1. 8. Finally, the solution to the initial value problem is y=−1
x2−1.
3. Question: Solve the following first-order linear differential equation:
dy
dx + 2y= 4e−2x
Ans. Step-by-step solution: 1. We first identify the integrating factor, which is given by the
exponential of the integral of the coefficient of y:
Integrating factor, µ(x) = e∫2dx =e2x
2. Multiply both sides of the differential equation by the integrating factor:
e2xdy
dx + 2e2xy= 4e2xe−2x
This simplifies to:
e2xdy
dx + 2e2xy= 4
3. Notice that the left side is now a derivative of the product ye2x. Rewrite the equation
accordingly:
d
dx ye2x= 4
4. Integrate both sides with respect to x:
Zd
dx ye2xdx =Z4dx
This gives us:
ye2x= 4x+C
5. Solve for y:
y=4x+C
e2x=4x
e2x+C
e2x
where Cis an arbitrary constant.
4. Question: Solve the following differential equation: dy
dx +y=e−x.
Ans. Step-by-step solution: 1. First, we need to identify the integrating factor. The integrating
factor is given by I(x) = e∫1dx =ex.
2. Multiply the given differential equation by the integrating factor:
exdy
dx +exy=exe−x
exdy
dx +exy= 1
3. Notice that the left side can be rewritten as the derivative of the product of the integrating
factor and yusing the product rule:
d
dx(exy) = 1
4. Integrate both sides with respect to x:
Zd
dx(exy)dx =Z1dx
5. This simplifies to:
exy=x+C
where Cis the constant of integration.
6. Finally, solve for yto obtain the general solution:
y=e−x(x+C)
where Cis an arbitrary constant.
5. Question: Solve the following first-order linear differential equation: dy
dx + 2xy =x.
Ans. Step-by-step solution: 1. Write the given differential equation in standard form, which is
dy
dx +p(x)y=q(x). 2. In this case, p(x)=2xand q(x) = x. 3. To solve this equation, we can
use an integrating factor given by e∫p(x)dx. 4. Calculate the integrating factor as e∫2xdx =ex2. 5.
Multiply both sides of the differential equation by the integrating factor: ex2dy
dx + 2xex2y=xex2.
6. Recognize the left side as the derivative of the integrating factor times y:d
dx (ex2y) = xex2.
7. Integrate both sides with respect to x:Rd
dx (ex2y)dx =Rxex2dx. 8. This simplifies to
ex2y=1
2ex2+C, where Cis the constant of integration. 9. Solve for yto get y=1
2+Ce−x2,
where Cis an arbitrary constant. Thus, the solution to the differential equation is y=1
2+Ce−x2.
6. Question:
Find the general solution to the differential equation y′′ + 4y= 0.
Ans. Solution:
1. Homogeneous Differential Equation:
The given differential equation is y′′ + 4y= 0. Let’s first write this in standard form:
y′′ + 4y= 0.
2. Characteristic Equation:
The characteristic equation for the given differential equation is obtained by assuming the
solution is of the form y=ert, where ris a constant to be determined. Substituting this into
the differential equation, we get:
r2ert + 4ert = 0.
Dividing both sides by ert, we have:
r2+ 4 = 0.
3. Finding the roots of the characteristic equation:
Solving the characteristic equation r2+ 4 = 0, we get r=±2i.
4. General Solution:
Since the roots are complex (r=±2i), the general solution to the differential equation is
given by:
y(t) = Acos(2t) + Bsin(2t),
where Aand Bare arbitrary constants determined by initial conditions if provided. This
represents the complete solution set to the differential equation y′′ + 4y= 0.
7. Question: Solve the following first-order linear differential equation:
y′+ (2x−1)y=x.
Ans. Solution: 1. The equation is in the form y′+P(x)y=Q(x), where P(x)=2x−1and
Q(x) = x. We will use an integrating factor to solve the equation.
2. The integrating factor I(x)is given by I(x) = e∫P(x)dx. Calculating the integral, we have:
Z(2x−1) dx =x2−x.
So, the integrating factor is I(x) = ex2−x.
3. Multiplying both sides of the differential equation by the integrating factor, we get:
ex2−xy′+ex2−x(2x−1)y=ex2−xx.
4. Recognizing the left side as the product rule of a product of two functions, we have:
(ex2−xy)′=ex2−xx.
5. Integrating both sides with respect to x, we get:
Z(ex2−xy)′dx =Zex2−xx dx.
6. Solving the integrals on both sides, we have:
ex2−xy=Zx·ex2−xdx.
Since the integral on the right side is not straightforward, we will make a substitution to simplify
it.
7. Let z=x2−x, then dz = (2x−1)dx. Solving for dx, we get dx =dz
2x−1. Substituting
into the integral, we have: Zx·ex2−xdx =Zz
2·ezdz.
8. The integral now becomes more manageable:
Zz
2·ezdz =1
2Zzezdz.
Using integration by parts, u=zand dv =ezdz, we get:
1
2Zzezdz =1
2(zez−Zezdz) = 1
2(zez−ez).
9. Substituting back z=x2−x, we have:
Zx·ex2−xdx =1
2((x2−x)ex2−x−ex2−x).
10. Therefore, the solution to the differential equation is:
ex2−xy=1
2((x2−x)ex2−x−ex2−x) + C,
where Cis the constant of integration.
11. Finally, dividing by ex2−x, we obtain the general solution:
y=1
2(x2−x−1) + Ce−x2+x,
where Cis an arbitrary constant.
8. Question: Solve the following first-order differential equation:
dy
dx + 2xy =x2
Ans. Step-by-step solution:
1. This is a first-order linear differential equation in standard form: dy
dx +P(x)y=Q(x),
where P(x) = 2xand Q(x) = x2.
2. To solve this equation, we first find the integrating factor, denoted by I(x) = e∫P(x)dx.
I(x) = e∫2xdx =ex2
3. Multiply both sides of the differential equation by the integrating factor.
ex2dy
dx + 2xex2y=x2ex2
4. Recognize that the left hand side can be written as the derivative of the product of the
integrating factor and yusing the product rule.
d
dx(ex2y) = x2ex2
5. Integrate both sides to solve for y.
Zd
dx(ex2y)dx =Zx2ex2dx
6. After integrating, we get:
ex2y=1
2x2ex2+C
7. Finally, solve for yby dividing by the integrating factor ex2.
y=1
2x2+Ce−x2
So, the general solution to the differential equation is y=1
2x2+Ce−x2, where Cis an
arbitrary constant.
9. Question: Solve the differential equation xy′+ 2y=x3given that y(1) = 2.
Ans. Step-by-step solution:
1. Write the differential equation in standard form: y′+2
xy=x2.
2. This is a first-order linear differential equation. To solve it, we first find the integrating
factor I(x) = e∫2
xdx =e2ln |x|=x2.
3. Multiply the differential equation by the integrating factor: x2y′+ 2x2y=x4.
4. Notice that the left side can be written as (x2y)′=x4.
5. Integrate both sides with respect to x: R(x2y)′dx =Rx4dx.
6. Solve the integrals: x2y=1
5x5+C.
7. Divide by x2to get the general solution: y=1
5x3+C
x2.
8. Apply the initial condition y(1) = 2 to find the particular solution. Substituting x= 1 and
y= 2 into the general solution, we find C= 9.
9. Therefore, the solution to the differential equation is y=1
5x3+9
x2.
10. Question:
Solve the following differential equation: dy
dx =y2cos(x), where y(0) = 1.
Ans. Step-by-step solution:
1. Separate the variables in the differential equation:
dy
y2=cos(x)dx
2. Integrate both sides with respect to their respective variables:
Z1
y2dy =Zcos(x)dx
⇒ −1
y=sin(x) + C
3. Solve for yby rearranging the equation and using the initial condition y(0) = 1:
y=−1
sin(x) + C
4. Plug in the initial condition y(0) = 1 to find the value of C:
1 = −1
sin(0) + C=−1
0 + C=−1
C
⇒C=−1
5. Substitute C=−1back into the equation to get the final solution:
y=−1
sin(x)−1
11. Question: Solve the initial value problem:
dy
dx =x
y+e−y, y(0) = 1
Ans. Solution: 1. Rewrite the initial value problem in the form of a separable differential
equation:
dy
dx =x
y+e−y
dy
dx =x
y+e−y
y dy =x dx +e−ydy
2. Integrate both sides: Zy dy =Zx dx +Ze−ydy
y2
2=x2
2−e−y+C
3. Since we are given the initial condition y(0) = 1, we can find the constant C:
1
2= 0 −e−1+C
C=1
2+e−1
4. Substitute the value of Cback into the equation:
y2
2=x2
2−e−y+1
2+e−1
5. To solve for yexplicitly, we can rewrite the equation as:
y2+ 2e−y=x2+ 2e−1+ 1
6. This equation is transcendental and cannot be solved algebraically. We can use numerical
methods to approximate the solution.
Therefore, the solution to the initial value problem is:
y2+ 2e−y=x2+ 2e−1+ 1
12. Question: Find the general solution to the following differential equation:
d2y
dx2+ 4dy
dx + 4y=e−2x
Ans. Solution: Let’s first find the complementary function by solving the corresponding
homogenous differential equation:
1. Setting the RHS to 0, we get the homogenous differential equation:
d2y
dx2+ 4dy
dx + 4y= 0
The characteristic equation is given by m2+ 4m+ 4 = 0, which simplifies to (m+ 2)2= 0.
Hence, we have a repeated root m=−2.
2. Therefore, the complementary function is given by:
yc(x) = c1e−2x+c2xe−2x
Next, let’s find the particular solution using the method of undetermined coefficients.
3. Since the RHS of the differential equation is e−2x, we assume the particular solution has
the form yp=ae−2x.
4. We substitute ypback into the differential equation to find a. Taking the derivatives and
plugging into the differential equation, we get:
4a+ 4a+ 4ae−2x=e−2x
5. Simplifying, we find that a=1
8.
6. Therefore, the particular solution is yp(x) = 1
8e−2x.
7. The general solution is then the sum of the complementary and particular solutions:
y(x) = yc(x) + yp(x) = c1e−2x+c2xe−2x+1
8e−2x
where c1and c2are constants determined by initial conditions if provided.
13. Question: Solve the differential equation y′+ 2xy =xwith the initial condition y(0) = 1.
Ans. Solution: 1. First, we rewrite the differential equation in standard form by multiplying
through by the integrating factor e∫2xdx:
e∫2xdxy′+ 2xe∫2xdxy=xe∫2xdx
ex2y′+ 2xex2y=xex2
2. Notice that the left-hand side can be simplified using the product rule for differentiation
in reverse: d
dx(ex2y) = xex2
3. Integrating both sides with respect to x, we have:
Zd
dx(ex2y)dx =Zxex2dx
ex2y=1
2ex2+C
4. Now, we solve for y:
y=1
2+Ce−x2
5. To find the value of the constant C, we use the initial condition y(0) = 1:
1 = 1
2+Ce0
C=1
2
6. Therefore, the solution to the differential equation with the initial condition is:
y=1
2+1
2e−x2
14. Question: Solve the differential equation y′′ −y′−2y= 0 with initial conditions y(0) = 2
and y′(0) = −1.
Ans. Solution: 1. Write down the characteristic equation: r2−r−2 = 0. 2. Solve the
characteristic equation to find the roots: r= 2,−1. 3. Write down the general solution of the
differential equation: y(t) = c1e2t+c2e−t. 4. Apply the initial conditions to find the values of c1
and c2. 5. From y(0) = 2, we have 2 = c1+c2. 6. From y′(0) = −1, we have −1=2c1−c2.
7. Solve the system of equations to find c1and c2. 8. From 2 = c1+c2, we get c1=1
3and
c2=5
3. 9. Therefore, the particular solution to the initial value problem is y(t) = 1
3e2t+5
3e−t.
15. Find the general solution to the differential equation:
dy
dx =y2−4
x2
Ans. Solution:
1. We can rewrite the given differential equation as:
dy
dx =(y−2)(y+ 2)
x2
2. Next, we can separate the variables:
dy
(y−2)(y+ 2) =dx
x2
3. Now, we can perform partial fraction decomposition on the left-hand side:
1
41
y−2−1
y+ 2=dx
x2
4. Integrating both sides:
1
4ln
y−2
y+ 2
=−1
x+C
5. Exponentiating both sides:
y−2
y+ 2
=e−4/x+4C
6. Simplifying further:
y−2
y+ 2 =±e−4/xe4C
7. Since e4Cis a positive constant, we can rewrite as:
y−2
y+ 2 =±Ce−4/x
8. Cross multiply to solve for y:
y−2 = ±Cye−4/x⇒y=2±2Cye−4/x
1∓Cye−4/x
9. Therefore, the general solution to the differential equation is:
y=2±2Cye−4/x
1∓Cye−4/x
16. Find the general solution of the following differential equation:
y′′ −4y′+ 4y= 0
Ans. To solve the differential equation y′′ −4y′+ 4y= 0, we can assume a solution of the form
y=ert, where ris a constant to be determined.
1. Write down the characteristic equation by substituting y=ert into the differential equa-
tion:
(r2−4r+ 4)ert = 0
2. Solve the characteristic equation r2−4r+ 4 = 0 to find the roots r:
(r−2)2= 0 =⇒r= 2
3. Since we have a repeated root, the general solution of the differential equation is:
y= (c1+c2t)e2t
where c1and c2are arbitrary constants.
17. Question: Find the particular solution to the differential equation dy
dx =e2x+ywith the
initial condition y(0) = 1.
Ans. Step-by-step solution: 1. We start by separating variables in the differential equation:
dy
dx =e2x+y
dy
dx −y=e2x
2. Next, we multiply both sides by the integrating factor e−xto simplify the equation:
e−xdy
dx −e−xy=ex
3. Rewrite the left side as the derivative of the product of e−xand y:
d
dx(e−xy) = ex
4. Integrate both sides with respect to xto solve for y:
Zd
dx(e−xy)dx =Zexdx
e−xy=Zexdx
5. Integrate the right side and solve for y:
e−xy=ex+C
y=e2x+Cex
6. Use the initial condition y(0) = 1 to find the value of the constant C:
1 = e0+C·e0
1 = 1 + C
C= 0
7. Substituting C= 0 back into the solution gives the particular solution to the differential
equation:
y=e2x
18. Question: Solve the following first-order linear differential equation:
y′+ 2xy = 3x
Ans. Step-by-step solution: 1. This is a first-order linear differential equation in the form
y′+P(x)y=Q(x), where P(x) = 2xand Q(x) = 3x. 2. To solve this equation, we first find
the integrating factor µ(x) = e∫P(x)dx. In this case, µ(x) = e∫2xdx =ex2. 3. Multiply both
sides of the differential equation by the integrating factor µ(x):
ex2y′+ 2xex2y= 3xex2
4. Notice that the left-hand side is the derivative of the product ex2ywith respect to x. Therefore,
we can rewrite the equation as:
d
dx(ex2y) = 3xex2
5. Integrate both sides with respect to x:
Zd
dx(ex2y)dx =Z3xex2dx
ex2y=3
2ex2+C
where Cis the constant of integration. 6. Finally, solve for yby dividing both sides by ex2:
y=3
2+Ce−x2
7. Therefore, the general solution to the differential equation y′+ 2xy = 3xis y=3
2+Ce−x2,
where Cis an arbitrary constant.
19. Question 19: Solve the differential equation y′−2xy = 2x, where y=y(x).
Ans. Step-by-step solution: 1. First, we will rewrite the given differential equation in the
standard form of a first-order linear differential equation: y′−2xy = 2x. 2. The integrating
factor µ(x)is given by µ(x) = e∫−2x dx =e−x2. 3. Multiply both sides of the differential
equation by the integrating factor µ(x):e−x2y′−2xe−x2y= 2xe−x2. 4. Rewriting the left-hand
side as the derivative of the product e−x2y, we get d
dx (e−x2y) = 2xe−x2. 5. Integrate both
sides with respect to x:e−x2y=R2xe−x2dx. 6. Solve the integral on the right-hand side:
R2xe−x2dx =−e−x2+C, where Cis the constant of integration. 7. Therefore, we have
e−x2y=−e−x2+C. 8. Finally, solve for y:y=−1 + Cex2, where Cis an arbitrary constant.
So, the general solution to the differential equation y′−2xy = 2xis y=−1 + Cex2, where C
is an arbitrary constant.
20. Question: Solve the differential equation dy
dx =x2y
1+x2.
Ans. Step-by-step solution: 1. Rewrite the differential equation in separable form. We can
rewrite it as: 1
ydy =x2
1 + x2dx
2. Integrate both sides. Integrating the left side gives:
Z1
ydy =ln |y|+C1
3. To integrate the right side, let’s perform a substitution. Let u= 1 + x2, then du = 2x dx.
This means we can rewrite the right side as:
Zx2
1 + x2dx =1
2Z2x
1 + x2·x dx
=1
2Z1
udu =1
2ln |u|+C2=1
2ln |1 + x2|+C2
4. Putting everything together, we have:
ln |y|=1
2ln |1 + x2|+C
y=e1
2ln |1+x2|+C=eC√1 + x2
5. Finally, we can write the general solution as:
y=k√1 + x2
where k=eCis an arbitrary constant.
21. Question: Find the general solution to the differential equation:
dy
dx =2x+ 1
y2
Ans. Solution: 1. Rewrite the given differential equation in separable form by multiplying both
sides by y2:
y2dy
dx = 2x+ 1
2. Integrate both sides with respect to x:
Zy2dy =Z(2x+ 1) dx
Zy2dy = 2 Zx dx +Z1dx
y3
3=x2+x+C
3. Multiply through by 3 to get the general solution:
y3= 3x2+ 3x+ 3C
where Cis the constant of integration.
22. Question: Solve the initial value problem: y′−y2= 0,y(0) = 1.
Ans. Step-by-step solution:
1. First, we rewrite the differential equation as dy
dx −y2= 0.
2. We can separate variables by dividing both sides by y2:dy
dx =y2.
3. Let’s rewrite this as dy
y2=dx.
4. Now we integrate both sides:
Z1
y2dy =Zdx.
5. The integral on the left gives us −1
y=x+C, where Cis the constant of integration.
6. Solving for y, we get y=−1
x+C.
7. Applying the initial condition y(0) = 1, we have: 1 = −1
C⇒C=−1.
8. Therefore, the solution to the initial value problem is y=−1
x−1.
23. Find the particular solution to the differential equation y′′ −y=exsatisfying the initial
conditions y(0) = 1 and y′(0) = 0.
Ans. Solution: 1. First, we find the complementary function by solving the homogeneous
equation y′′ −y= 0. The characteristic equation is r2−1 = 0, which gives us r=±1. Thus,
the complementary function is yc(x) = c1ex+c2e−x.
2. Next, we need to find a particular solution to the non-homogeneous equation y′′ −y=ex.
Let’s guess a particular solution in the form yp(x) = Aex, where Ais a constant to be determined.
3. We substitute yp(x) = Aexinto the differential equation:
y′′
p−yp=ex
Aex−Aex=ex
0 = ex
4. Since 0=exfor all x, our assumption for the particular solution was incorrect. Let’s try
a new guess in the form yp(x) = Axex.
5. Substituting yp(x) = Axexinto the differential equation:
y′′
p−yp=ex
Aex−Axex=ex
A−Ax = 1
6. We get A= 1 and A= 0, which is a contradiction. Therefore, let’s try a new guess in
the form yp(x) = −1
2x2ex.
7. Substituting yp(x) = −1
2x2exinto the differential equation:
y′′
p−yp=ex
−x2ex−2xex−ex=ex
−x2−2x−1 = 1
8. We get −x2−2x= 2, which implies x2+ 2x+ 2 = 0 has no real roots. Therefore, the
correct particular solution is yp(x) = −1
2x2ex.
9. The general solution is then y(x) = yc(x) + yp(x) = c1ex+c2e−x−1
2x2ex.
10. Finally, applying the initial conditions, we have y(0) = 1 and y′(0) = 0:
y(0) = c1+c2= 1
y′(0) = c1−c2= 0
11. Solving the system of equations, we find c1=1
2and c2=1
2.
12. Therefore, the particular solution to the differential equation satisfying the initial condi-
tions is y(x) = 1
2ex+1
2e−x−1
2x2ex.
24. Question: Solve the differential equation dy
dx =2x+3y
3x+2y.
Ans. Solution: 1. Write the given differential equation in the form dy
dx =A(x,y)+B(x,y)dy
dx
C(x,y)+D(x,y)dy
dx
.
2. Here, A(x, y)=2x,B(x, y)=3y,C(x, y)=3x, and D(x, y)=2y. 3. Notice that
M(x, y) = A(x, y)−B(x, y)dy
dx = 2x−3yand N(x, y) = C(x, y)−D(x, y)dy
dx = 3x−2y. 4.
Calculate the partial derivatives ∂M
∂y =−3and ∂N
∂x = 3. 5. Since ∂M
∂y =∂N
∂x , the differential
equation is exact. 6. To solve an exact differential equation, let u(x, y)be a function such that
∂u
∂x =Mand ∂u
∂y =N. 7. Integrate ∂u
∂x = 2x−3yto find u(x, y). 8. Integrate ∂u
∂y = 3x−2yto
find u(x, y). 9. Equate the two expressions for u(x, y)to eliminate the arbitrary constant. 10.
The solution to the differential equation is u(x, y) = x2−3xy +y2=Constant.
25. Let y(t)be a differentiable function that satisfies the differential equation:
dy
dt = 2ty2
with initial condition y(0) = 1
3.
Determine the value of y(1).
Ans. To solve the differential equation, we can separate variables and integrate both sides.
Let’s find the solution step by step: 1. Separate variables:
dy
y2= 2tdt
2. Integrate both sides: Z1
y2dy =Z2tdt
3. Integrate the left side:
−1
y=t2+C
4. Apply the initial condition y(0) = 1
3:
−1
1/3 = 02+C
−3 = C
5. Substitute C=−3back into the equation:
−1
y=t2−3
1
y= 3 −t2
y=1
3−t2
6. Find y(1):
y(1) = 1
3−12=1
2
Therefore, the value of y(1) is 1
2.
26. Consider the first-order differential equation:
(y3−3y)dx + (x2+ 2x)dy = 0
Find the general solution to this differential equation.
Ans. To solve this differential equation, we will rearrange the equation and separate the
variables:
1. Rearranging the equation: Rewrite the given equation in the form of M(x, y)dx +
N(x, y)dy = 0:
M(x, y) = y3−3y
N(x, y) = x2+ 2x
2. Checking for exactness: Calculate the partial derivatives of Mwith respect to yand
Nwith respect to x:
∂M
∂y = 3y2−3
∂N
∂x = 2x+ 2
3. Solving for exactness: The equation is exact if ∂M
∂y =∂N
∂x . Since this condition is not
satisfied, the equation is not exact.
4. Integrating factor: To find an integrating factor, we use the formula (x, y) = e∫Nx−My
Mdx.
Calculate Nx−My:
Nx−My= (2x+ 2) −(3y2−3) = 2x+ 2 −3y2+ 3
Nx−My= 2x−3y2+ 5
The integrating factor (x, y)is e∫2x−3y2+5
y3−3ydx.
5. Multiplier calculation: Calculate the integrating factor:
(x, y) = e∫2x−3y2+5
y3−3ydx =e∫2x−3y2+5
y(y2−3) dx
6. General solution: Multiply the original equation by the integrating factor (x, y):
(y3−3y)e∫2x−3y2+5
y(y2−3) dxdx + (x2+ 2x)e∫2x−3y2+5
y(y2−3) dxdy = 0
This equation can be written in the form of an exact differential and be solved further to find the
general solution.
27. Question 27: Solve the following initial value problem:
dy
dx = 2x+y, y(0) = 1
Ans. Solution: 1. We rearrange the given first-order linear differential equation into standard
form:
dy
dx = 2x+y=⇒dy
dx −y= 2x
2. Next, we identify the integrating factor, which is e∫−1dx =e−x.
3. We multiply the standard form of the equation by the integrating factor:
e−xdy
dx −e−xy= 2xe−x
4. Notice that the left side can be simplified using the product rule of differentiation:
d
dx(e−xy) = 2xe−x
5. Integrating both sides with respect to xgives:
e−xy=Z2xe−xdx
6. Solving the integral on the right side gives:
e−xy=−2xe−x−2e−x+C
7. Dividing by e−x, we find the general solution:
y=−2x−2 + Cex
8. Using the initial condition y(0) = 1, we can find the value of the constant C:
1 = −2(0) −2 + Ce0=⇒C= 3
9. Therefore, the particular solution to the initial value problem is:
y=−2x−2+3ex
28. Question: Solve the initial value problem y′′ −2y′+y= 0,y(0) = 1,y′(0) = 2.
Ans. Solution: 1. We first find the characteristic equation by assuming y=ert. Substituting
this into the differential equation gives us r2−2r+ 1 = 0.
2. Solving the characteristic equation, we find r= 1. Since we have a repeated root, the
general solution will be y(t) = c1et+c2tet.
3. Applying the initial conditions y(0) = 1 and y′(0) = 2, we get c1= 1 and c1+c2= 2.
4. Therefore, the solution to the initial value problem is y(t) = et+ (2 −t)et.
29. Question: Solve the following differential equation:
y′′ −2y′+y= 3 cos(x)
Ans. Solution: 1. First, we find the complementary function by solving the characteristic
equation: r2−2r+ 1 = 0. This gives us (r−1)2= 0, so r= 1 is a repeated root. Thus, the
complementary function is of the form yc(x) = (c1+c2x)ex.
2. Next, we find a particular solution for the non-homogeneous part of the equation. Since
3cos(x)is a trigonometric function, we assume the particular solution has the form yp(x) =
Acos(x) + Bsin(x).
3. We find the first and second derivatives of yp:
y′
p=−Asin(x) + Bcos(x)
y′′
p=−Acos(x)−Bsin(x)
4. Substituting the particular solution and its derivatives into the differential equation, we
have:
−Acos(x)−Bsin(x)−2(−Asin(x) + Bcos(x)) + Acos(x) + Bsin(x) = 3 cos(x)
5. Simplifying, we get:
(−2A−B+A)cos(x)+(−2B+A+B)sin(x) = 3 cos(x)
6. Equating coefficients of cos(x)and sin(x), we get the system of equations:
A−B= 0
A+B=−3
7. Solving the system of equations gives A=−3
2and B=−3
2. Therefore, the particular
solution is yp(x) = −3
2cos(x)−3
2sin(x).
8. Combining the complementary function and the particular solution, the general solution
to the differential equation is:
y(x) = (c1+c2x)ex−3
2cos(x)−3
2sin(x)
where c1and c2are constants determined by initial conditions if given.
30. Question 30: Solve the following differential equation using the method of integrating
factors:
2xy2+ (x2−2y)dx = 0
Ans. Solution: 1. We first rearrange the given differential equation into the form Mdx+N dy =
0:
2xy2dx + (x2−2y)dx = 0
2. Now, we identify Mand Nas follows:
M= 2xy2, N =x2−2y
3. The integrating factor Iis given by:
I=e∫Nx−My
Mdx
4. Calculating Nxand My:
Nx= 2x, My= 2xy
Nx−My= 2x−2xy = 2x(1 −y)
5. Substituting Nx−Myback into the formula for I:
I=e∫2x(1−y)dx =ex2(1−y)
6. Multiply the original differential equation by the integrating factor I:
2xy2ex2(1−y)dx + (x2−2y)ex2(1−y)dx = 0
7. Rewrite the equation as a total differential by recognizing it as the result of the product
rule: d
dx(x2y2ex2(1−y)) = 0
8. Integrate both sides with respect to x:
x2y2ex2(1−y)=C
9. Finally, we have the general solution to the given differential equation as:
x2y2ex2(1−y)=C
where Cis the constant of integration.
31. Question: Find the general solution to the differential equation y′′ −6y′+ 9y= 0.
Ans. Step-by-step solution: 1. Write the characteristic equation by substituting y=erx into
the differential equation.
r2−6r+ 9 = 0
2. Solve the characteristic equation to find the roots r.
(r−3)2= 0
r= 3 (double root)
3. Write down the general solution based on the roots found:
y(x) = (c1+c2x)e3x
where c1and c2are arbitrary constants.
32. Question 32: Solve the following differential equation using the method of integrating
factors:
(2x+y)dx + (x−y)dy = 0
Ans. Solution:
1. Write the given differential equation in the form: M(x, y)dx +N(x, y)dy = 0. Here,
M(x, y) = 2x+yand N(x, y) = x−y.
2. Check if the given differential equation is exact. Calculate ∂M
∂y and ∂N
∂x .
∂M
∂y = 1 =−1 = ∂N
∂x
Since ∂M
∂y =∂N
∂x , the given differential equation is not exact.
3. Find the integrating factor, denoted by µ, to make the differential equation exact. The
integrating factor is given by:
µ=e∫Nx−My
Mdx
4. Calculate Nx−My.
Nx−My=−1−1 = −2
5. Substitute Nx−Myinto the formula for the integrating factor:
µ=e∫−2
2x+ydx =e−2ln |2x+y|=1
|2x+y|2
6. Multiply the given differential equation by the integrating factor µ.
1
|2x+y|2(2x+y)dx +1
|2x+y|2(x−y)dy = 0
7. Rearrange terms and simplify.
2x+y
|2x+y|2dx +x−y
|2x+y|2dy = 0
8. Rewrite the equation in exact form and integrate. The equation now becomes exact:
dx2
|2x+y|= 0
9. Integrate with respect to xto find the general solution.
x2
|2x+y|=C
where Cis the constant of integration.
10. Solve for yto find the particular solution.
y=±px2−C(2x+y)
33. Question:
Solve the following initial value problem:
y′+ 2xy =sin(x), y(0) = 1
Ans. Solution:
1. Homogeneous Solution:
First, we find the solution to the homogeneous differential equation with no forcing term
(sin(x)):
y′+ 2xy = 0
This is a first-order linear differential equation. Let’s rewrite it in standard form y′+P(x)y= 0
where P(x) = 2x.
The integrating factor IF (x)is given by IF (x) = e∫P(x)dx =e∫2x dx =ex2.
Multiplying the differential equation by the integrating factor, we get:
ex2y′+ 2xex2y= 0
Now, using the product rule on the left side, we have:
d
dx(ex2y) = 0
Integrating both sides gives:
ex2y=C
where Cis the constant of integration. Therefore, the general solution to the homogeneous
differential equation is yh=Ce−x2.
2. Particular Solution:
Now, we need to find a particular solution to the non-homogeneous differential equation with
the forcing term sin(x).
Considering a particular solution of the form yp=Asin(x) + Bcos(x), we substitute it back
into the differential equation:
(Acos x−Bsin x)+2x(Asin x+Bcos x) = sin x
Equating coefficients of sin xand cos x:
For sin x:2xA −B= 1
For cos x:−A+ 2xB = 0
Solving this system of equations gives A=1
4and B=1
8.
Therefore, the particular solution is yp=1
4sin x+1
8cos x.
3. Overall Solution:
The overall solution is the sum of the homogeneous and particular solutions:
y=yh+yp=Ce−x2+1
4sin x+1
8cos x
4. Initial Condition:
Applying the initial condition y(0) = 1, we get:
1 = Ce−0+1
4sin 0 + 1
8cos 0
Simplifying, we find C= 1 −1
8=7
8.
5. Final Solution:
The final solution to the initial value problem is:
y=7
8e−x2+1
4sin x+1
8cos x
34. Question:
Solve the following first-order differential equation:
dy
dx =x2+y2
xy
Ans. Solution:
1. Rearrange the terms in the differential equation to separate variables:
dy
dx =x2+y2
xy =⇒y dy =x dx +y2
xdx
2. Integrate both sides of the equation:
Zy dy =Zx dx +Zy2
xdx
3. Solve the integrals:
Integrating y dy with respect to ygives 1
2y2+C1, where C1is the constant of integration.
Integrating x dx with respect to xgives 1
2x2+C2, where C2is another constant of integration.
Integrating y2
xdx involves substituting u=y2and d(u) = 2y dy:
Zy2
xdx =Z1
2u1
√udu =1
2Z√u du =1
3u3
2+C3
where C3is another constant of integration.
Thus, the equation becomes:
1
2y2+C1=1
2x2+C2+1
3y3+C3
4. Rearrange the equation to solve for y:
y2−1
3y3=x2+C
where C=C2−C1+C3.
35. Question 35: Solve the following first-order differential equation using an integrating factor:
y′+ 2y=ex.
Ans. Solution: 1. The given first-order differential equation is of the form y′+P(x)y=Q(x),
where P(x) = 2 and Q(x) = ex.
2. To solve this differential equation, we first find the integrating factor, denoted by I(x),
defined as I(x) = e∫P(x)dx.
3. In this case, P(x) = 2, so the integrating factor is calculated as I(x) = e∫2dx =e2x.
4. Multiply the given differential equation by the integrating factor e2x:
e2xy′+ 2e2xy=e3x
5. Notice that the left side can be expressed as the derivative of a product:
(e2xy)′=e3x
6. Integrate both sides with respect to xto solve for y:
Z(e2xy)dx =Ze3xdx
e2xy=1
3e3x+C
7. Solve for yby dividing by e2x:
y=1
3ex+Ce−2x
8. Therefore, the general solution to the differential equation y′+2y=exis y=1
3ex+Ce−2x,
where Cis an arbitrary constant.
