MATH 334 - DIFFERENTIAL
EQUATIONS - Ordinary differential
equations
Question Bank - Set 1
Liberty University
Question 1
Question
Find the general solution to the differential equation:
y′′ −4y′+ 4y= 0
Solution
To solve the differential equation, we will first find the characteristic equation
by assuming y=ert.
Step 1: Write down the characteristic equation using y=ert:
r2−4r+ 4 = 0
Step 2: Solve the characteristic equation to find the roots:
(r−2)2= 0 =⇒r= 2
Step 3: Write down the general solution using the roots of the characteristic
equation:
y(t) = c1e2t+c2te2t
Therefore, the general solution to the differential equation y′′ −4y′+ 4y= 0
is y(t) = c1e2t+c2te2t.
Question 2
Question
Consider the first-order ordinary differential equation:
dy
dx = 2xy
Solve the differential equation using separation of variables.
Solution
Step 1: Separate the variables by moving all terms involving yto one side and
all terms involving xto the other side.
dy
y= 2x dx
Step 2: Integrate both sides with respect to their respective variables.
Zdy
y=Z2x dx
Step 3: Perform the integrations.
ln |y|=x2+C
where Cis the constant of integration.
Step 4: Exponentiate both sides to solve for y.
y=ex2+C=Cex2
where Cis a constant.
Therefore, the general solution to the given differential equation is y=Cex2.
Question 3
Question
Solve the following initial value problem:
dy
dx =1
y, y(0) = 2
2
Solution
Step 1: Rewrite the differential equation
dy
dx =1
y
as
y dy =dx
Step 2: Integrate both sides
Zy dy =Zdx
1
2y2=x+C
where Cis the constant of integration.
Step 3: Apply the initial condition y(0) = 2
1
2·22= 0 + C
C= 2
Step 4: Substitute C= 2 back into the equation
1
2y2=x+ 2
Step 5: Solve for y
y2= 2x+ 4
y=±√2x+ 4
Step 6: Check that y(0) = 2
y(0) = ±p2(0) + 4 = ±√4 = ±2
Therefore, the solution to the initial value problem is
y=√2x+ 4
Question 4
Question
Consider the initial value problem:
y′=2
xy+x2, y(1) = 3.
Find the explicit solution to the initial value problem.
3
Solution
Step 1: First, we write the given initial value problem in the standard form of
a first-order linear differential equation y′+P(x)y=Q(x):
y′−2
xy=x2.
Step 2: Next, we find the integrating factor I(x) = eR−2
xdx =e−2 ln |x|=1
x2.
Step 3: We multiply both sides of the equation by the integrating factor:
1
x2y′−2
x3y=x2·1
x2.
Step 4: Simplifying, we get:
d
dx 1
x2y= 1.
Step 5: Integrate both sides with respect to x:
Zd
dx 1
x2ydx =Z1dx.
Step 6: This gives us: 1
x2y=x+C,
where Cis the constant of integration.
Step 7: Applying the initial condition y(1) = 3, we find C:
1
12·3 = 1 + C=⇒C= 2.
Step 8: Substituting C= 2 back into our equation, we obtain the explicit
solution:
y=x3+ 2x2.
Therefore, the explicit solution to the initial value problem is y=x3+ 2x2.
Question 5
Question
Solve the initial value problem:
y′=1
1 + x+y2, y(0) = 1
4
Solution
Step 1: We first rewrite the differential equation in a more recognizable form
by multiplying both sides by (1 + x+y2):
(1 + x+y2)dy =dx
Step 2: Next, we integrate both sides:
Z(1 + x+y2)dy =Zdx
Z1dy +Zx dy +Zy2dy =Zdx
y+1
2xy +1
3y3=x+C1
Step 3: Now, we use the initial condition y(0) = 1 to find the value of the
constant C1:
1 + 1
2(0)(1) + 1
3(1)3= 0 + C1
1 + 0 + 1
3=C1
C1=4
3
Step 4: Substituting back the value of C1into the general solution, we have:
y+1
2xy +1
3y3=x+4
3
Step 5: This equation implicitly defines yas a function of x. Although it is
difficult to solve for yexplicitly in this case, we have successfully obtained the
implicit solution for the initial value problem.
Question 6
Question
Find the general solution to the differential equation: dy
dx =y+x
y−x.
Solution
Step 1: Rewrite the equation in a suitable form for separation of variables. We
rewrite the differential equation as:
dy
dx =y+x
y−x⇒(y−x)dy = (y+x)dx
5
Step 2: Separate the variables.
Z1
y−xdy =Z1
y+xdx
Step 3: Solve the integrals. Integrating both sides gives:
ln |y−x|= ln |y+x|+C1
where C1is the constant of integration.
Step 4: Use properties of logarithms to simplify. Applying the logarithmic
property ln(a)−ln(b) = ln a
btotheequation, weget : ln y−x
y+x=C1
Step 5: Remove the logarithm using exponential functions. Taking the ex-
ponential of both sides gives:
y−x
y+x=eC1
y−x
y+x=C
where C=eC1is the new constant of integration.
Step 6: Solve for y. Cross multiplying the equation gives:
y−x=C(y+x)
y−Cy =x+Cx
(1 −C)y= (1 + C)x
y=1 + C
1−Cx
Step 7: Final answer. Therefore, the general solution to the differential
equation is y=1+C
1−Cx, where Cis an arbitrary constant.
Question 7
Question
Given the differential equation (x2+y2)dx−2xy dy = 0, find the general solution.
Solution
Step 1: We first rewrite the given differential equation in standard form by
dividing throughout by x2:
(1 + (y/x)2)dx −2(y/x)dy = 0
6
Step 2: Next, we make the substitution y=vx to reduce the differential
equation to separable form:
(1 + v2)dx −2vxdx = 0
Step 3: Simplifying the equation gives:
(1 −v2)dx = 0
Step 4: Integrating both sides gives:
Z(1 −v2)dx =Z0dx
x−x3
3=C1,
where C1is the constant of integration.
Step 5: Substituting back v=y/x gives the general solution in implicit
form:
x−x3
3=C,
where Cis the constant of integration.
Therefore, the general solution to the given differential equation is x−x3
3=
C.
Question 8
Question
Find the particular solution to the differential equation: dy
dx = 3x2−4xgiven
that y(0) = 2.
Solution
Step 1: Integrate both sides of the differential equation with respect to x.
Zdy
dx dx =Z(3x2−4x)dx
Step 2: Integrate to solve for y.
y=Z(3x2−4x)dx
Step 3: Find the antiderivative of 3x2−4x.
y=Z(3x2−4x)dx
7
y=Z3x2dx −Z4x dx
y=x3−2x2+C
Step 4: Apply the initial condition y(0) = 2 to solve for the constant C.
2 = (0)3−2(0)2+C
2 = C
Step 5: Write down the particular solution.
y=x3−2x2+ 2
Question 9
Question
Consider the differential equation y′′ −4y′+ 4y= 0. Find the general solution
to this second order linear homogeneous differential equation.
Solution
1. Let’s first write down the characteristic equation for the given differential
equation, which is obtained by replacing y′′ by r2,y′by r, and yby 1 in
the differential equation:
r2−4r+ 4 = 0
2. Next, we solve the characteristic equation to find the roots:
r2−4r+ 4 = (r−2)2= 0
This gives a repeated root of r= 2.
3. As we have a repeated root, the general solution to the differential equation
is:
y(t) = (c1+c2t)e2t
where c1and c2are arbitrary constants.
Question 10
Question
Consider the ordinary differential equation (ODE) given by dy
dx =2xy
x2+y2.
(a) Determine whether this ODE is exact or not.
(b) If it is exact, find a potential function F(x, y) such that ∂F
∂x =2xy
x2+y2.
8
Solution
(a) To determine if the ODE is exact, we check if ∂M
∂y =∂N
∂x , where dy
dx =M(x, y)
and N(x, y). Let M(x, y) = 2xy
x2+y2and N(x, y) = 1. We calculate the partial
derivatives: ∂M
∂y =2x(x2−y2)
(x2+y2)2and ∂N
∂x = 0.
Since ∂M
∂y =∂N
∂x , the ODE is not exact.
(b) We can attempt to find an integrating factor µ(x, y) to make the ODE
exact. The integrating factor is given by µ=eR
∂M
∂y
−∂N
∂x
Ndx. In our case,
∂M
∂y −∂N
∂x
N=−2x(x2−y2)
(x2+y2)2. Thus,
µ=eR−2x(x2−y2)
(x2+y2)2dx =e−2 arctan(y
x).
Multiplying the ODE by this integrating factor, we get the exact ODE:
e−2 arctan(y
x)dy
dx −e−2 arctan(y
x)2xy
x2+y2= 0.
The exact ODE can now be written as the total differential of F(x, y), i.e.,
dF = 0, where F(x, y) = cis the potential function.
Therefore, the ODE is exact and the potential function F(x, y) is F(x, y) =
e2 arctan(y
x)=c.
Question 11
Question
Consider the differential equation:
dy
dx =y(1 −x2)
Find the general solution to the differential equation.
Solution
Step 1: We will separate variables by dividing both sides by y(1 −x2):
dy
y= (1 −x2)dx
Step 2: Integrate both sides:
Zdy
y=Z(1 −x2)dx
9
Step 3: Solving the integrals, we get:
ln |y|=x−x3
3+C
where Cis the constant of integration.
Step 4: Exponentiate both sides to solve for y:
|y|=ex−x3
3+C
Step 5: Rewrite the absolute value as a constant:
y=±eCex−x3
3
Step 6: Combine the constants:
y=Aex−x3
3
where A=±eC. This is the general solution to the differential equation.
Question 12
Question
Consider the initial value problem:
y′=2x+ 3y−4
3x+ 2y−1, y(0) = 2.
What is the particular solution to the differential equation above that satis-
fies the initial condition y(0) = 2?
Solution
Step 1: First, let’s rewrite the given initial value problem in the form dy/dx =
f(x, y).
y′=2x+ 3y−4
3x+ 2y−1
dy
dx =2x+ 3y−4
3x+ 2y−1
Step 2: Next, let’s check if the given differential equation is exact by com-
puting the partial derivatives of f(x, y) with respect to xand y.
∂f
∂y = 3 and ∂f
∂x = 3
10
Step 3: Since the partial derivatives are not equal, the differential equation
is not exact. We need to find an integrating factor µ(x) to make it exact. To
find µ(x), we solve the following differential equation:
dµ
dx =
∂f
∂y −∂f
∂x
fµ
dµ
dx =0
2x+3y−4
3x+2y−1
µ
dµ
dx = 0
Step 4: Integrating both sides with respect to x, we get µ(x) = C, where
Cis a constant. Since µ(x) can be any constant, we can choose µ(x) = 1 for
simplicity.
Step 5: Multiply the original differential equation by the integrating factor
µ(x) = 1:
1·dy
dx= 1 ·2x+ 3y−4
3x+ 2y−1
Step 6: Simplify the equation:
dy
dx =2x+ 3y−4
3x+ 2y−1
Step 7: Now, we need to find the solution to the differential equation. To
find the particular solution that satisfies y(0) = 2, we substitute x= 0 and
y= 2 into the equation.
dy
dx =2(0) + 3(2) −4
3(0) + 2(2) −1=6
3= 2
Therefore, the particular solution to the differential equation that satisfies
the initial condition y(0) = 2 is y(x)=2x+ 2.
Question 13
Question
Solve the initial value problem:
y′=ex+y−1, y(0) = 1.
11
Solution
Step 1: First, let’s find the integrating factor for the given differential equation.
The integrating factor is defined as µ(x) = eRp(x)dx, where p(x) is the coefficient
of yin the differential equation.
So, in this case, p(x) = 1. Thus, the integrating factor µ(x) is given by:
µ(x) = eR1dx =ex.
Step 2: Multiply both sides of the differential equation by the integrating
factor µ(x) = ex:
exy′=e2x+yex−ex.
Step 3: Recognize that the left-hand side is the derivative of the product
yexwith respect to x. Therefore, rewrite the equation as:
d
dx(yex) = e2x−ex.
Step 4: Integrate both sides with respect to x:
Zd
dx(yex)dx =Ze2x−exdx.
Step 5: Integrating, we get:
yex=1
2e2x−ex+C,
where Cis the constant of integration.
Step 6: Apply the initial condition y(0) = 1 to solve for the constant C:
1 = 1
2−1 + C⇒C= 1 + 1
2=3
2.
Step 7: Substitute the value of Cback into the expression:
yex=1
2e2x−ex+3
2.
Step 8: Finally, solve for yby dividing both sides by ex:
y=1
2ex−1 + 3
2ex.
So, the solution to the initial value problem is y=1
2ex−1 + 3
2ex.
Question 14
Question
Solve the following initial value problem:
dy
dx =2x+ 3y
3x+ 2y, y(0) = 1
12
Solution
Step 1: Rewrite the differential equation in a more convenient form.
We can rewrite the given differential equation as:
(3x+ 2y)dy = (2x+ 3y)dx
Step 2: Integrate both sides of the equation.
Integrating both sides gives:
Z(3x+ 2y)dy =Z(2x+ 3y)dx
3xy +1
2y2=x2+3
2xy +C
Step 3: Use the initial condition to solve for the constant of integration.
Substitute x= 0 and y= 1 into the equation above:
3(0)(1) + 1
2(1)2= (0)2+3
2(0)(1) + C
1
2=C
Step 4: Substitute C=1
2back into the equation.
The solution to the differential equation with the initial condition is:
3xy +1
2y2=x2+3
2xy +1
2
Step 5: Find the explicit solution.
Rearranging the equation, we get:
2y2+ 6xy −6xy −2y= 4x2
2y2−2y= 4x2
y2−y= 2x2
y2−y−2x2= 0
Step 6: Solve the quadratic equation for y.
Using the quadratic formula, we get:
y=1±√1+8x2
2
y=1±√8x2+ 1
2
Therefore, the solution to the initial value problem is:
y=1±√8x2+ 1
2
13
Question 15
Question
Find the general solution to the differential equation: dy
dx −y
x=xcos(ln(x)).
Solution
Step 1: First, we identify the form of the given differential equation, which is
linear. A linear differential equation has the general form:
dy
dx +P(x)y=Q(x)
where P(x) and Q(x) are functions of x.
Step 2: Compare the given differential equation with the general form to
find P(x) and Q(x). Here, P(x) = −1
xand Q(x) = xcos(ln(x)).
Step 3: We find the integrating factor I(x) = eRP(x)dx. In this case, I(x) =
eR−1
xdx =e−ln(x)=1
x.
Step 4: Multiply both sides of the differential equation by the integrating
factor I(x):
1
x
dy
dx −1
x2y= cos(ln(x))
Step 5: Rewrite the left side of the equation as the derivative of a product:
d
dx y
x= cos(ln(x))
Step 6: Integrate both sides with respect to x:
Zd
dx y
xdx =Zcos(ln(x))dx
Step 7: Solve the integrals to find the general solution:
y
x=Zcos(ln(x))dx +C
y
x=xsin(ln(x)) + C
Step 8: Multiply through by xto obtain the general solution:
y=x2sin(ln(x)) + Cx
Therefore, the general solution to the differential equation is y=x2sin(ln(x))+
Cx, where Cis an arbitrary constant.
14
Question 16
Question
Consider the differential equation y′+ 2xy =x. Find the general solution to
this equation.
Solution
Step 1: This is a first-order linear ordinary differential equation of the form
y′+P(x)y=Q(x), where P(x)=2xand Q(x) = x. To solve this, we first need
to find the integrating factor.
Step 2: The integrating factor, denoted by µ(x), is given by µ(x) = eRP(x)dx.
Step 3: In this case, µ(x) = eR2xdx =ex2.
Step 4: Now, we multiply the given differential equation by the integrating
factor µ(x).
Step 5: Multiplying through, we get d
dx (ex2y) = xex2.
Step 6: Integrate both sides with respect to x.
Step 7: We obtain ex2y=Rxex2dx.
Step 8: Solving the integral on the right side gives us ex2y=1
2ex2+C, where
Cis the constant of integration.
Step 9: Finally, we solve for yto find the general solution: y=1
2+Ce−x2,
where Cis an arbitrary constant.
Question 17
Question
Consider the initial value problem:
y′=x2e−y, y(0) = 1
Use Euler’s method with a step size of h= 0.5 to estimate y(1).
Solution
Step 1: Determine the number of steps required. We are integrating from x= 0
to x= 1 with a step size of h= 0.5. Therefore, the number of steps required is:
n=1−0
0.5= 2
Step 2: Initialize the values. Starting with the initial condition, y(0) = 1,
we have:
y0= 1
Step 3: Calculate the estimates for yk. Using Euler’s method, for k= 0,1,2:
yk+1 =yk+hf(xk, yk)
15
where f(x, y) = x2e−y.
For k= 0:
y1=y0+h·f(0,1) = 1 + 0.5·(02e−1) = 1 −0.5=0.5
For k= 1:
y2=y1+h·f(0.5,0.5) = 0.5+0.5·((0.5)2e−0.5)≈0.681
Step 4: Estimate y(1). The estimate for y(1) using Euler’s method with
h= 0.5 is y2≈0.681.
Question 18
Question
Consider the initial value problem:
y′+2
xy=x3, y(1) = 1.
Find the solution to the differential equation satisfying the initial condition.
Solution
Step 1: First, we find the integrating factor µ(x) by multiplying through by x2:
x2y′+ 2y=x5.
Thus, we have P(x) = 2 and Q(x) = x5. The integrating factor is given by
µ(x) = eRP(x)dx =eR2dx =e2x.
Step 2: Multiply the differential equation by the integrating factor:
x2e2xy′+ 2e2xy=x5e2x.
This simplifies to
(x2e2xy)′=x5e2x.
Step 3: Integrate both sides of the equation with respect to x:
Z(x2e(2x)y)′dx =Zx5e2xdx.
x2e2xy=1
2x5e2x−5
2Zx4e2xdx.
Step 4: Integrate Rx4e2xdx by parts with u=x4and dv =e2xdx:
u=x4, du = 4x3dx,
16
v=1
2e2x, dv =e2xdx.
This gives us 1
2x4e2x−2Zx3e2xdx.
Step 5: Integrate Rx3e2xdx by parts with u=x3and dv =e2xdx:
u=x3, du = 3x2dx,
v=1
2e2x, dv =e2xdx.
Step 6: Substitute back into the equation to find the particular solution.
Apply the initial condition y(1) = 1 to solve for the constant of integration.
Question 19
Question
Find the general solution to the differential equation: y′=1
x2+y2.
Solution
Step 1: First, separate variables by multiplying both sides of the equation by
dx and both sides of the equation by (x2+y2):
(x2+y2)dy =dx
Step 2: Integrate both sides of the equation with respect to yand xsepa-
rately:
Z(x2+y2)dy =Zdx
Step 3: Evaluate the integrals:
x2y+1
3y3=x+C
where Cis the constant of integration.
Step 4: Rearrange the equation to solve for y:
y3+ 3x2y= 3x+ 3C
Step 5: The general solution to the given differential equation is:
y3+ 3x2y= 3x+ 3C
17
Question 20
Question
Consider the ordinary differential equation:
dy
dx =x2y+e2x
Find the particular solution given that y(0) = 1.
Solution
Step 1: Rewrite the given differential equation in the form dy = (x2y+e2x)dx.
Step 2: Separate the variables by dividing both sides by yand multiplying by
dx:dy
y= (x2+e2x)dx
Step 3: Integrate both sides:
Zdy
y=Z(x2+e2x)dx
Step 4: Simplify the integrals:
ln |y|=1
3x3+1
2e2x+C
Step 5: Exponentiate both sides to solve for y:
y=e(1
3x3+1
2e2x+C)
Step 6: Use the initial condition y(0) = 1 to find the value of the constant C:
1 = e(1
2+C)
Step 7: Solve for C:
C=−1
2
Step 8: Substitute the value of Cback into the particular solution:
y=e(1
3x3+1
2e2x−1
2)
Question 21
Question
Solve the following first-order ordinary differential equation using an integrating
factor:
(2x+ 1)y′−3y=x3
18
Solution
Step 1: Rewrite the equation in the form y′+P(x)y=Q(x).
(2x+ 1)y′−3y=x3
y′−3
2x+ 1y=x3
2x+ 1
Step 2: Identify P(x) and Q(x).
P(x) = −3
2x+ 1 and Q(x) = x3
2x+ 1
Step 3: Calculate the integrating factor µ(x) = eRP(x)dx.
µ(x) = eR−3
2x+1 dx
=e−3 ln |2x+1|
=eln(2x+1)−3
= (2x+ 1)−3
Step 4: Multiply the ODE by the integrating factor µ(x).
(2x+ 1)−3y′−3(2x+ 1)−4y=x3(2x+ 1)−3
d
dx[(2x+ 1)−3y] = x3(2x+ 1)−3
Step 5: Integrate both sides with respect to x.
Zd
dx[(2x+ 1)−3y]dx =Zx3(2x+ 1)−3dx
(2x+ 1)−3y=Zx3(2x+ 1)−3dx +C
Step 6: Solve the integral on the right side.
Zx3(2x+ 1)−3dx =Zx3
(2x+ 1)3dx
=Zx3
(2x+ 1)2(2x+ 1)dx
=Zu
2u2du (let u= 2x+ 1)
=1
2Z1
udu
=1
2ln |2x+ 1|+C
19
Step 7: Substitute back and solve for y.
(2x+ 1)−3y=1
2ln |2x+ 1|+C
y=(2x+ 1)3
2ln |2x+ 1|+C(2x+ 1)3
Therefore, the solution to the differential equation is y=(2x+1)3
2ln |2x+1|+
C(2x+ 1)3, where Cis an arbitrary constant.
Question 22
Question
Solve the following first-order ordinary differential equation:
dy
dx =3y−5x
2x+ 6y
Solution
Step 1: Rearrange the differential equation in the form dy
dx =P(x) + Q(x)y.
Step 2: Solve for P(x) and Q(x).
Given differential equation:
dy
dx =3y−5x
2x+ 6y
Separate variables:
(2x+ 6y)dy = (3y−5x)dx
Expand and rearrange:
2xdy + 6ydy = 3ydx −5xdx
Now, write the equation in the form dy
dx =P(x) + Q(x)y:
dy
dx =3y−5x
2x+ 6y=3
2−5
2x+ 6yy
So, P(x) = 3
2and Q(x) = −5
2x+6y.
Step 3: Solve the differential equation using an integrating factor.
The integrating factor I(x) is defined as:
I(x) = eRP(x)dx
Since P(x) = 3
2, then:
I(x) = eR3
2dx =e3
2x
20
Multiply the integrating factor with the original differential equation:
e3
2xdy
dx =e3
2x3
2−5
2x+ 6yy
Step 4: Integrate both sides of the equation.
Ze3
2xdy
dxdx =Ze3
2x3
2−5
2x+ 6yydx
e3
2xy=3
2Ze3
2xdx −5Zye 3
2xdx
Integrate using substitution method and solve for the constant of integration.
The final solution will be:
y=f(x)
Question 23
Question
Consider the differential equation:
dy
dx + 2xy = 4x
Find the general solution of the differential equation.
Solution
Step 1: First, we need to rearrange the equation to the standard form of a
first-order linear ODE: dy
dx + 2xy = 4x
Subtracting 2xy from both sides gives:
dy
dx = 4x−2xy
Step 2: We notice that this is a first-order linear differential equation. To
solve this, we will use an integrating factor. The integrating factor I(x) is given
by:
I(x) = eR2x dx =ex2
Step 3: Multiply both sides of the equation by the integrating factor:
ex2dy
dx =ex2(4x−2xy)
This simplifies to: d
dx(ex2y) = 4xex2
21
Step 4: Integrate both sides with respect to x:
Zd
dx(ex2y)dx =Z4xex2dx
ex2y=Z4xex2dx
Step 5: Solve the integral on the right side:
ex2y= 2ex2+C
where Cis the constant of integration.
Step 6: Finally, divide by ex2to find the general solution:
y= 2 + Ce−x2
So, the general solution of the differential equation is y= 2 + Ce−x2, where C
is an arbitrary constant.
Question 24
Question
Find the general solution of the following ordinary differential equation:
y′′ + 4y′+ 4y=e−2x.
Solution
Step 1: First, find the complementary solution by solving the homogeneous
equation y′′ + 4y′+ 4y= 0. The characteristic equation is r2+ 4r+ 4 = 0,
which has a repeated root at r=−2. Therefore, the complementary solution is
yc(x) = c1e−2x+c2xe−2x.
Step 2: Next, we will find a particular solution to the non-homogeneous
equation. Assume a particular solution of the form yp(x) = Ae−2x, where Ais
a constant to be determined. Plug this into the original differential equation:
−4Ae−2x+ 4Ae−2x+ 4Ae−2x=e−2x.
Simplify this to get 4Ae−2x=e−2x, which implies A=1
4. Therefore, the
particular solution is yp(x) = 1
4e−2x.
Step 3: The general solution is the sum of the complementary and particular
solutions. Thus, the general solution of the differential equation is
y(x) = yc(x) + yp(x) = c1e−2x+c2xe−2x+1
4e−2x,
where c1and c2are arbitrary constants.
22
Question 25
Question
Consider the first-order ordinary differential equation y′=ex+y. Solve the
initial value problem when y(0) = 1.
Solution
Step 1: First, we will rewrite the given differential equation in a separable form:
dy
dx =ex+y
Step 2: Next, we separate the variables by moving all terms involving yto
the left side and all terms involving xto the right side:
dy
dx −y=ex
Step 3: Now, we multiply through by the integrating factor e−xto make the
left side a perfect derivative:
e−xdy
dx −e−xy= 1
Step 4: Recognizing the left side as the derivative of ye−x, we rewrite the
equation as: d
dx(ye−x)=1
Step 5: Integrating both sides with respect to x, we have:
Zd
dx(ye−x)dx =Z1dx
ye−x=x+C
Step 6: Applying the initial condition y(0) = 1, we find the value of the
constant C:
1 = 0 + C=⇒C= 1
Step 7: Therefore, the solution to the initial value problem y′=ex+ywith
y(0) = 1 is:
ye−x=x+ 1
Step 8: Finally, we can write the solution explicitly as:
y=ex(x+ 1)
23
Question 26
Question
Consider the initial value problem:
dy
dx =−2xy2, y(0) = 1.
Find the particular solution to the differential equation.
Solution
Step 1: Let’s separate variables and solve for y:
dy
dx =−2xy2
dy
y2=−2x dx
Z1
y2dy =Z−2x dx
Zy−2dy =−x2+C1
−1
y=−x2+C1
y=1
x2−C1
Step 2: Next, we use the initial condition y(0) = 1 to find the particular
solution:
1 = 1
02−C1
1 = 1
−C1
C1=−1
Therefore, the particular solution to the differential equation is y=1
x2+1 .
Question 27
Question
Solve the initial value problem:
y′=2x+y−1
x−y, y(0) = 2
24
Solution
Step 1: First, let’s rewrite the given differential equation in a more convenient
form. We have:
y′=2x+y−1
x−y
which can be rewritten as:
y′(x−y)=2x+y−1
Step 2: Next, we expand the left-hand side and group like terms:
xy′−y= 2x+y−1
Step 3: Now, we rearrange the equation:
xy′−y−y= 2x+ 1
xy′−2y= 2x+ 1
Step 4: This is a linear first-order ODE. We need to solve it by finding an
integrating factor. The integrating factor is given by:
µ(x) = eR−2dx =e−2x
Step 5: Multiplying both sides of the ODE by the integrating factor, we get:
e−2xxy′−2e−2xy= 2e−2xx+e−2x
Step 6: The left-hand side can be simplified as the derivative of a product:
(e−2xy)′= 2e−2xx+e−2x
Step 7: Integrating both sides with respect to x gives:
Z(e−2xy)dx =Z(2e−2xx+e−2x)dx
e−2xy=−e−2x(x+ 1) + C
Step 8: Solving for y, we have:
y=−x−1 + Ce2x
Step 9: Using the initial condition y(0) = 2, we can find C:
2 = −0−1 + C
C= 3
Step 10: Therefore, the solution to the initial value problem is:
y=−x−1+3e2x
25
Question 28
Question
Solve the differential equation y′′ −4y′+ 4y=e2x, given that y(0) = 2 and
y′(0) = 1.
Solution
To solve the given second order linear homogeneous differential equation, we first
find the complementary solution by solving the characteristic equation r2−4r+
4 = 0.
Step 1: Find the Characteristic Equation The characteristic equation
is r2−4r+4 = 0, which factors to (r−2)2= 0. Thus, the characteristic equation
has a repeated root r= 2.
Step 2: Find the Complementary Solution The complementary solu-
tion is of the form yc(x) = (Ax +B)e2x, where Aand Bare constants to be
determined.
Step 3: Find the Particular Solution To find the particular solution,
we use the method of undetermined coefficients. Let’s assume the particular
solution has the form yp(x) = Ce2x.
Step 4: Substitute into the Differential Equation Substitute yp(x) into
the differential equation to find C. We have (2·2Ce2x−4·2Ce2x+4Ce2x) = e2x.
Simplifying, we get 4Ce2x−8Ce2x+ 4Ce2x=e2x, which simplifies to 0 = e2x,
a contradiction.
Thus, we need to modify our assumed particular solution. Let’s try yp(x) =
Axe2x.
Step 5: Find the Correct Particular Solution Substitute yp(x) into the
differential equation to find A. We have (2 ·2Axe2x−4·2Ae2x+ 4Axe2x) = e2x.
Simplifying, we get 4Axe2x−8Ae2x+4Axe2x=e2x, which simplifies to 4A= 1.
Therefore, A=1
4.
Step 6: Write the General Solution The general solution is given by
the sum of the complementary and particular solutions:
y(x) = yc(x) + yp(x) = (Ax +B)e2x+1
4xe2x
Step 7: Apply Initial Conditions Using the initial conditions y(0) = 2
and y′(0) = 1, we can find the values of Aand B.
Step 8: Solve for Constants Substitute the initial conditions into the
general solution and solve for Aand B:
y(0) = (0 + B)e0+ 0 = B= 2
y′(0) = A+1
4= 1 =⇒A=3
4
26
Step 9: Final Solution Therefore, the solution to the differential equation
y′′ −4y′+ 4y=e2xsubject to the initial conditions is:
y(x) = 3
4x+ 2e2x+1
4xe2x
Question 29
Question
Consider the differential equation: y′′ −2y′+y=e2x. Find the general solution
to this second-order linear homogeneous differential equation.
Solution
To find the general solution to the differential equation y′′ −2y′+y=e2x,
we first need to find the complementary solution by solving the characteristic
equation r2−2r+ 1 = 0.
Step 1: Find the roots of the characteristic equation The character-
istic equation is r2−2r+ 1 = (r−1)2= 0, which has a repeated root r= 1.
Step 2: Write down the complementary solution Since the character-
istic equation has a repeated root, the complementary solution is of the form
yc(x) = c1erx +c2xerx, where r= 1.
Therefore, the complementary solution is yc(x) = c1ex+c2xex.
Step 3: Find a particular solution Let’s assume the particular solution
has the form yp(x) = Ax2e2x.
Step 4: Calculate the derivatives We have y′
p(x) = (2Ax +A2)e2xand
y′′
p(x) = (4A+ 4Ax)e2x.
Step 5: Substitute into the differential equation Substitute yp(x),
y′
p(x), and y′′
p(x) into the differential equation y′′ −2y′+y=e2x: (4A+4Ax)e2x−
2(2Ax +A)e2x+Ax2e2x=e2x.
Step 6: Solve for A Simplify the equation and solve for the coefficient A:
12A+ 12Ax −4Ax −2A+Ax2= 1. This simplifies to Ax2+ 8Ax −A−2 = 0.
Comparing coefficients, we find that A=−1
2.
Step 7: Write down the particular solution Now that we have A, we
can write down the particular solution as yp(x) = −1
2x2e2x.
Step 8: Write down the general solution The general solution to the
differential equation is the sum of the complementary solution and the particular
solution: y(x) = yc(x) + yp(x) = c1ex+c2xex−1
2x2e2x, where c1and c2are
arbitrary constants.
Question 30
Question
Consider the following ordinary differential equation (ODE):
27
y′′ −2y′+ 2y=ex
Find the general solution to the ODE above.
Solution
Step 1: First, find the characteristic equation by setting the left-hand side of
the ODE equal to 0:
r2−2r+ 2 = 0
Step 2: Solve the characteristic equation to find the roots r1and r2:
r=2±p(−2)2−4(1)(2)
2(1) = 1 ±i
Step 3: Since the roots are complex, the general solution to the homogeneous
ODE is:
yh(x) = erx(c1cos(rx) + c2sin(rx))
where r= 1 and c1, c2are arbitrary constants.
Step 4: Next, find a particular solution to the non-homogeneous ODE:
yp(x) = Aex
Differentiate yp(x) twice and substitute into the original ODE to solve for A.
Step 5: Substituting yp(x) and its derivatives back into the ODE, we get:
ex−2Aex+ 2Aex=ex
Solving this equation, we find A= 1.
Step 6: The general solution to the non-homogeneous ODE is the sum of
the homogeneous and particular solutions:
y(x) = yh(x) + yp(x) = ex(c1cos(x) + c2sin(x) + 1)
where c1, c2are arbitrary constants.
Question 31
Question
Solve the following initial value problem:
dy
dx = 2x−1, y(0) = 3
28
Solution
Step 1: First, we solve the differential equation by integrating both sides with
respect to x.
Zdy
dx dx =Z(2x−1) dx
Step 2: This gives us:
y=x2−x+C, where Cis the constant of integration.
Step 3: Next, we use the initial condition y(0) = 3 to find the value of C.
3 = 02−0 + C
C= 3
Step 4: Therefore, the particular solution to the differential equation is:
y=x2−x+ 3
Question 32
Question
Find the general solution to the differential equation: dy
dx =−2xy.
Solution
Step 1: We start by separating variables. Rearranging the equation, we have:
dy
dx =−2xy
1
ydy =−2xdx
Step 2: Integrate both sides with respect to their respective variables:
Z1
ydy =Z−2xdx
ln |y|=−x2+C
Step 3: Exponentiate both sides to get rid of the natural logarithm:
|y|=e−x2+C
|y|=e−x2eC
|y|=Ce−x2
Step 4: Since Ccan be positive or negative, we can write the general solution
as:
y=±Ce−x2
29
Question 33
Question
Solve the following initial value problem:
y′= (3x2+ 2y) sin(x), y(0) = 1.
Solution
Step 1: First, rewrite the differential equation in a more recognizable form.
Separating variables, we have:
dy
dx = (3x2+ 2y) sin(x).
Step 2: Now, divide both sides by sin(x):
1
sin(x)
dy
dx = 3x2+ 2y.
Step 3: Rearrange the equation to isolate terms involving y:
dy
dx −2y= 3x2sin(x).
Step 4: This is now a first-order linear differential equation. The integrating
factor is given by e−2x.
Step 5: Multiply both sides of the equation by the integrating factor:
e−2xdy
dx −2e−2xy= 3x2sin(x)e−2x.
Step 6: The left-hand side can be rewritten as the derivative of the product
e−2xy:d
dx(e−2xy)=3x2sin(x)e−2x.
Step 7: Integrate both sides with respect to x:
e−2xy=Z3x2sin(x)e−2xdx +C,
where Cis the constant of integration.
Step 8: Solve the integral on the right-hand side. This can be done by parts
or a combination of repeated integration by parts and simplification.
Step 9: Once the integral is evaluated, solve for yby multiplying both sides
by e2xand simplifying.
Step 10: Finally, apply the initial condition y(0) = 1 to find the value of the
constant C.
30
Question 34
Question
Solve the following first-order ordinary differential equation:
dy
dx = (xy +y)√x
Solution
Step 1: Write the given equation in the form dy
dx +P(x)y=Q(x). Step 2:
Determine the integrating factor. Step 3: Use the integrating factor to solve
the differential equation.
Step 1: Rewrite the given equation in the form dy
dx +P(x)y=Q(x).
We have: dy
dx = (xy +y)√x
Divide through by √x:
1
√x
dy
dx = (xy +y)
Rewrite in the required form:
dy
dx +1
xy=y
So, P(x) = 1
xand Q(x) = y.
Step 2: Determine the integrating factor.
The integrating factor is given by:
I(x) = eRP(x)dx =eR1
xdx =eln(x)=x
Step 3: Solve the differential equation.
Multiply the entire equation by the integrating factor x:
xdy
dx +y=xy
This is now an exact differential equation.
Integrate both sides with respect to x:
Zxdy
dxdx +Zydx =Zxydx
xy +C=x2
2+D
Therefore, the general solution to the differential equation is:
y=x2
2+Cx +D
where Cand Dare constants.
31
Question 35
Question
Solve the initial value problem:
dy
dx =y
x+x2
y, y(1) = 2
Solution
Step 1: Rewrite the given differential equation as separable variables by moving
all terms involving yto one side and all terms involving xto the other side.
dy
y=dx
x+x2
ydy
Step 2: Integrate both sides with respect to their respective variables.
Z1
ydy =Z1
xdx +Zx2dy
Step 3: Solve the integrals separately.
ln |y|= ln |x|+x3
3+C1
where C1is the constant of integration.
Step 4: Exponentiate both sides to eliminate the natural logarithms.
|y|=Cxex3
3
where C=±eC1.
Step 5: Apply the initial condition y(1) = 2 to find the value of C.
|2|=C(1)e13
3=⇒2 = Ce1
3=⇒C= 2e−1
3
Step 6: Plug in the value of Cback into the general solution.
y= 2e−1
3xex3
3
Therefore, the solution to the initial value problem is:
y= 2e−1
3xex3
3
32
Question 2
Question
Consider the first-order ordinary differential equation:
dy
dx = 2xy
Solve the differential equation using separation of variables.
Solution
Step 1: Separate the variables by moving all terms involving yto one side and
all terms involving xto the other side.
dy
y= 2x dx
Step 2: Integrate both sides with respect to their respective variables.
Zdy
y=Z2x dx
Step 3: Perform the integrations.
ln |y|=x2+C
where Cis the constant of integration.
Step 4: Exponentiate both sides to solve for y.
y=ex2+C=Cex2
where Cis a constant.
Therefore, the general solution to the given differential equation is y=Cex2.
Question 3
Question
Solve the following initial value problem:
dy
dx =1
y, y(0) = 2
2
Solution
Step 1: Rewrite the differential equation
dy
dx =1
y
as
y dy =dx
Step 2: Integrate both sides
Zy dy =Zdx
1
2y2=x+C
where Cis the constant of integration.
Step 3: Apply the initial condition y(0) = 2
1
2·22= 0 + C
C= 2
Step 4: Substitute C= 2 back into the equation
1
2y2=x+ 2
Step 5: Solve for y
y2= 2x+ 4
y=±√2x+ 4
Step 6: Check that y(0) = 2
y(0) = ±p2(0) + 4 = ±√4 = ±2
Therefore, the solution to the initial value problem is
y=√2x+ 4
Question 4
Question
Consider the initial value problem:
y′=2
xy+x2, y(1) = 3.
Find the explicit solution to the initial value problem.
3
Solution
Step 1: First, we write the given initial value problem in the standard form of
a first-order linear differential equation y′+P(x)y=Q(x):
y′−2
xy=x2.
Step 2: Next, we find the integrating factor I(x) = eR−2
xdx =e−2 ln |x|=1
x2.
Step 3: We multiply both sides of the equation by the integrating factor:
1
x2y′−2
x3y=x2·1
x2.
Step 4: Simplifying, we get:
d
dx 1
x2y= 1.
Step 5: Integrate both sides with respect to x:
Zd
dx 1
x2ydx =Z1dx.
Step 6: This gives us: 1
x2y=x+C,
where Cis the constant of integration.
Step 7: Applying the initial condition y(1) = 3, we find C:
1
12·3 = 1 + C=⇒C= 2.
Step 8: Substituting C= 2 back into our equation, we obtain the explicit
solution:
y=x3+ 2x2.
Therefore, the explicit solution to the initial value problem is y=x3+ 2x2.
Question 5
Question
Solve the initial value problem:
y′=1
1 + x+y2, y(0) = 1
4
Solution
Step 1: We first rewrite the differential equation in a more recognizable form
by multiplying both sides by (1 + x+y2):
(1 + x+y2)dy =dx
Step 2: Next, we integrate both sides:
Z(1 + x+y2)dy =Zdx
Z1dy +Zx dy +Zy2dy =Zdx
y+1
2xy +1
3y3=x+C1
Step 3: Now, we use the initial condition y(0) = 1 to find the value of the
constant C1:
1 + 1
2(0)(1) + 1
3(1)3= 0 + C1
1 + 0 + 1
3=C1
C1=4
3
Step 4: Substituting back the value of C1into the general solution, we have:
y+1
2xy +1
3y3=x+4
3
Step 5: This equation implicitly defines yas a function of x. Although it is
difficult to solve for yexplicitly in this case, we have successfully obtained the
implicit solution for the initial value problem.
Question 6
Question
Find the general solution to the differential equation: dy
dx =y+x
y−x.
Solution
Step 1: Rewrite the equation in a suitable form for separation of variables. We
rewrite the differential equation as:
dy
dx =y+x
y−x⇒(y−x)dy = (y+x)dx
5
Step 2: Separate the variables.
Z1
y−xdy =Z1
y+xdx
Step 3: Solve the integrals. Integrating both sides gives:
ln |y−x|= ln |y+x|+C1
where C1is the constant of integration.
Step 4: Use properties of logarithms to simplify. Applying the logarithmic
property ln(a)−ln(b) = ln a
btotheequation, weget : ln y−x
y+x=C1
Step 5: Remove the logarithm using exponential functions. Taking the ex-
ponential of both sides gives:
y−x
y+x=eC1
y−x
y+x=C
where C=eC1is the new constant of integration.
Step 6: Solve for y. Cross multiplying the equation gives:
y−x=C(y+x)
y−Cy =x+Cx
(1 −C)y= (1 + C)x
y=1 + C
1−Cx
Step 7: Final answer. Therefore, the general solution to the differential
equation is y=1+C
1−Cx, where Cis an arbitrary constant.
Question 7
Question
Given the differential equation (x2+y2)dx−2xy dy = 0, find the general solution.
Solution
Step 1: We first rewrite the given differential equation in standard form by
dividing throughout by x2:
(1 + (y/x)2)dx −2(y/x)dy = 0
6
Step 2: Next, we make the substitution y=vx to reduce the differential
equation to separable form:
(1 + v2)dx −2vxdx = 0
Step 3: Simplifying the equation gives:
(1 −v2)dx = 0
Step 4: Integrating both sides gives:
Z(1 −v2)dx =Z0dx
x−x3
3=C1,
where C1is the constant of integration.
Step 5: Substituting back v=y/x gives the general solution in implicit
form:
x−x3
3=C,
where Cis the constant of integration.
Therefore, the general solution to the given differential equation is x−x3
3=
C.
Question 8
Question
Find the particular solution to the differential equation: dy
dx = 3x2−4xgiven
that y(0) = 2.
Solution
Step 1: Integrate both sides of the differential equation with respect to x.
Zdy
dx dx =Z(3x2−4x)dx
Step 2: Integrate to solve for y.
y=Z(3x2−4x)dx
Step 3: Find the antiderivative of 3x2−4x.
y=Z(3x2−4x)dx
7
y=Z3x2dx −Z4x dx
y=x3−2x2+C
Step 4: Apply the initial condition y(0) = 2 to solve for the constant C.
2 = (0)3−2(0)2+C
2 = C
Step 5: Write down the particular solution.
y=x3−2x2+ 2
Question 9
Question
Consider the differential equation y′′ −4y′+ 4y= 0. Find the general solution
to this second order linear homogeneous differential equation.
Solution
1. Let’s first write down the characteristic equation for the given differential
equation, which is obtained by replacing y′′ by r2,y′by r, and yby 1 in
the differential equation:
r2−4r+ 4 = 0
2. Next, we solve the characteristic equation to find the roots:
r2−4r+ 4 = (r−2)2= 0
This gives a repeated root of r= 2.
3. As we have a repeated root, the general solution to the differential equation
is:
y(t) = (c1+c2t)e2t
where c1and c2are arbitrary constants.
Question 10
Question
Consider the ordinary differential equation (ODE) given by dy
dx =2xy
x2+y2.
(a) Determine whether this ODE is exact or not.
(b) If it is exact, find a potential function F(x, y) such that ∂F
∂x =2xy
x2+y2.
8
Solution
(a) To determine if the ODE is exact, we check if ∂M
∂y =∂N
∂x , where dy
dx =M(x, y)
and N(x, y). Let M(x, y) = 2xy
x2+y2and N(x, y) = 1. We calculate the partial
derivatives: ∂M
∂y =2x(x2−y2)
(x2+y2)2and ∂N
∂x = 0.
Since ∂M
∂y =∂N
∂x , the ODE is not exact.
(b) We can attempt to find an integrating factor µ(x, y) to make the ODE
exact. The integrating factor is given by µ=eR
∂M
∂y
−∂N
∂x
Ndx. In our case,
∂M
∂y −∂N
∂x
N=−2x(x2−y2)
(x2+y2)2. Thus,
µ=eR−2x(x2−y2)
(x2+y2)2dx =e−2 arctan(y
x).
Multiplying the ODE by this integrating factor, we get the exact ODE:
e−2 arctan(y
x)dy
dx −e−2 arctan(y
x)2xy
x2+y2= 0.
The exact ODE can now be written as the total differential of F(x, y), i.e.,
dF = 0, where F(x, y) = cis the potential function.
Therefore, the ODE is exact and the potential function F(x, y) is F(x, y) =
e2 arctan(y
x)=c.
Question 11
Question
Consider the differential equation:
dy
dx =y(1 −x2)
Find the general solution to the differential equation.
Solution
Step 1: We will separate variables by dividing both sides by y(1 −x2):
dy
y= (1 −x2)dx
Step 2: Integrate both sides:
Zdy
y=Z(1 −x2)dx
9
Step 3: Solving the integrals, we get:
ln |y|=x−x3
3+C
where Cis the constant of integration.
Step 4: Exponentiate both sides to solve for y:
|y|=ex−x3
3+C
Step 5: Rewrite the absolute value as a constant:
y=±eCex−x3
3
Step 6: Combine the constants:
y=Aex−x3
3
where A=±eC. This is the general solution to the differential equation.
Question 12
Question
Consider the initial value problem:
y′=2x+ 3y−4
3x+ 2y−1, y(0) = 2.
What is the particular solution to the differential equation above that satis-
fies the initial condition y(0) = 2?
Solution
Step 1: First, let’s rewrite the given initial value problem in the form dy/dx =
f(x, y).
y′=2x+ 3y−4
3x+ 2y−1
dy
dx =2x+ 3y−4
3x+ 2y−1
Step 2: Next, let’s check if the given differential equation is exact by com-
puting the partial derivatives of f(x, y) with respect to xand y.
∂f
∂y = 3 and ∂f
∂x = 3
10
Step 3: Since the partial derivatives are not equal, the differential equation
is not exact. We need to find an integrating factor µ(x) to make it exact. To
find µ(x), we solve the following differential equation:
dµ
dx =
∂f
∂y −∂f
∂x
fµ
dµ
dx =0
2x+3y−4
3x+2y−1
µ
dµ
dx = 0
Step 4: Integrating both sides with respect to x, we get µ(x) = C, where
Cis a constant. Since µ(x) can be any constant, we can choose µ(x) = 1 for
simplicity.
Step 5: Multiply the original differential equation by the integrating factor
µ(x) = 1:
1·dy
dx= 1 ·2x+ 3y−4
3x+ 2y−1
Step 6: Simplify the equation:
dy
dx =2x+ 3y−4
3x+ 2y−1
Step 7: Now, we need to find the solution to the differential equation. To
find the particular solution that satisfies y(0) = 2, we substitute x= 0 and
y= 2 into the equation.
dy
dx =2(0) + 3(2) −4
3(0) + 2(2) −1=6
3= 2
Therefore, the particular solution to the differential equation that satisfies
the initial condition y(0) = 2 is y(x)=2x+ 2.
Question 13
Question
Solve the initial value problem:
y′=ex+y−1, y(0) = 1.
11
Solution
Step 1: First, let’s find the integrating factor for the given differential equation.
The integrating factor is defined as µ(x) = eRp(x)dx, where p(x) is the coefficient
of yin the differential equation.
So, in this case, p(x) = 1. Thus, the integrating factor µ(x) is given by:
µ(x) = eR1dx =ex.
Step 2: Multiply both sides of the differential equation by the integrating
factor µ(x) = ex:
exy′=e2x+yex−ex.
Step 3: Recognize that the left-hand side is the derivative of the product
yexwith respect to x. Therefore, rewrite the equation as:
d
dx(yex) = e2x−ex.
Step 4: Integrate both sides with respect to x:
Zd
dx(yex)dx =Ze2x−exdx.
Step 5: Integrating, we get:
yex=1
2e2x−ex+C,
where Cis the constant of integration.
Step 6: Apply the initial condition y(0) = 1 to solve for the constant C:
1 = 1
2−1 + C⇒C= 1 + 1
2=3
2.
Step 7: Substitute the value of Cback into the expression:
yex=1
2e2x−ex+3
2.
Step 8: Finally, solve for yby dividing both sides by ex:
y=1
2ex−1 + 3
2ex.
So, the solution to the initial value problem is y=1
2ex−1 + 3
2ex.
Question 14
Question
Solve the following initial value problem:
dy
dx =2x+ 3y
3x+ 2y, y(0) = 1
12
Solution
Step 1: Rewrite the differential equation in a more convenient form.
We can rewrite the given differential equation as:
(3x+ 2y)dy = (2x+ 3y)dx
Step 2: Integrate both sides of the equation.
Integrating both sides gives:
Z(3x+ 2y)dy =Z(2x+ 3y)dx
3xy +1
2y2=x2+3
2xy +C
Step 3: Use the initial condition to solve for the constant of integration.
Substitute x= 0 and y= 1 into the equation above:
3(0)(1) + 1
2(1)2= (0)2+3
2(0)(1) + C
1
2=C
Step 4: Substitute C=1
2back into the equation.
The solution to the differential equation with the initial condition is:
3xy +1
2y2=x2+3
2xy +1
2
Step 5: Find the explicit solution.
Rearranging the equation, we get:
2y2+ 6xy −6xy −2y= 4x2
2y2−2y= 4x2
y2−y= 2x2
y2−y−2x2= 0
Step 6: Solve the quadratic equation for y.
Using the quadratic formula, we get:
y=1±√1+8x2
2
y=1±√8x2+ 1
2
Therefore, the solution to the initial value problem is:
y=1±√8x2+ 1
2
13
Question 15
Question
Find the general solution to the differential equation: dy
dx −y
x=xcos(ln(x)).
Solution
Step 1: First, we identify the form of the given differential equation, which is
linear. A linear differential equation has the general form:
dy
dx +P(x)y=Q(x)
where P(x) and Q(x) are functions of x.
Step 2: Compare the given differential equation with the general form to
find P(x) and Q(x). Here, P(x) = −1
xand Q(x) = xcos(ln(x)).
Step 3: We find the integrating factor I(x) = eRP(x)dx. In this case, I(x) =
eR−1
xdx =e−ln(x)=1
x.
Step 4: Multiply both sides of the differential equation by the integrating
factor I(x):
1
x
dy
dx −1
x2y= cos(ln(x))
Step 5: Rewrite the left side of the equation as the derivative of a product:
d
dx y
x= cos(ln(x))
Step 6: Integrate both sides with respect to x:
Zd
dx y
xdx =Zcos(ln(x))dx
Step 7: Solve the integrals to find the general solution:
y
x=Zcos(ln(x))dx +C
y
x=xsin(ln(x)) + C
Step 8: Multiply through by xto obtain the general solution:
y=x2sin(ln(x)) + Cx
Therefore, the general solution to the differential equation is y=x2sin(ln(x))+
Cx, where Cis an arbitrary constant.
14
Question 16
Question
Consider the differential equation y′+ 2xy =x. Find the general solution to
this equation.
Solution
Step 1: This is a first-order linear ordinary differential equation of the form
y′+P(x)y=Q(x), where P(x)=2xand Q(x) = x. To solve this, we first need
to find the integrating factor.
Step 2: The integrating factor, denoted by µ(x), is given by µ(x) = eRP(x)dx.
Step 3: In this case, µ(x) = eR2xdx =ex2.
Step 4: Now, we multiply the given differential equation by the integrating
factor µ(x).
Step 5: Multiplying through, we get d
dx (ex2y) = xex2.
Step 6: Integrate both sides with respect to x.
Step 7: We obtain ex2y=Rxex2dx.
Step 8: Solving the integral on the right side gives us ex2y=1
2ex2+C, where
Cis the constant of integration.
Step 9: Finally, we solve for yto find the general solution: y=1
2+Ce−x2,
where Cis an arbitrary constant.
Question 17
Question
Consider the initial value problem:
y′=x2e−y, y(0) = 1
Use Euler’s method with a step size of h= 0.5 to estimate y(1).
Solution
Step 1: Determine the number of steps required. We are integrating from x= 0
to x= 1 with a step size of h= 0.5. Therefore, the number of steps required is:
n=1−0
0.5= 2
Step 2: Initialize the values. Starting with the initial condition, y(0) = 1,
we have:
y0= 1
Step 3: Calculate the estimates for yk. Using Euler’s method, for k= 0,1,2:
yk+1 =yk+hf(xk, yk)
15
where f(x, y) = x2e−y.
For k= 0:
y1=y0+h·f(0,1) = 1 + 0.5·(02e−1) = 1 −0.5=0.5
For k= 1:
y2=y1+h·f(0.5,0.5) = 0.5+0.5·((0.5)2e−0.5)≈0.681
Step 4: Estimate y(1). The estimate for y(1) using Euler’s method with
h= 0.5 is y2≈0.681.
Question 18
Question
Consider the initial value problem:
y′+2
xy=x3, y(1) = 1.
Find the solution to the differential equation satisfying the initial condition.
Solution
Step 1: First, we find the integrating factor µ(x) by multiplying through by x2:
x2y′+ 2y=x5.
Thus, we have P(x) = 2 and Q(x) = x5. The integrating factor is given by
µ(x) = eRP(x)dx =eR2dx =e2x.
Step 2: Multiply the differential equation by the integrating factor:
x2e2xy′+ 2e2xy=x5e2x.
This simplifies to
(x2e2xy)′=x5e2x.
Step 3: Integrate both sides of the equation with respect to x:
Z(x2e(2x)y)′dx =Zx5e2xdx.
x2e2xy=1
2x5e2x−5
2Zx4e2xdx.
Step 4: Integrate Rx4e2xdx by parts with u=x4and dv =e2xdx:
u=x4, du = 4x3dx,
16
v=1
2e2x, dv =e2xdx.
This gives us 1
2x4e2x−2Zx3e2xdx.
Step 5: Integrate Rx3e2xdx by parts with u=x3and dv =e2xdx:
u=x3, du = 3x2dx,
v=1
2e2x, dv =e2xdx.
Step 6: Substitute back into the equation to find the particular solution.
Apply the initial condition y(1) = 1 to solve for the constant of integration.
Question 19
Question
Find the general solution to the differential equation: y′=1
x2+y2.
Solution
Step 1: First, separate variables by multiplying both sides of the equation by
dx and both sides of the equation by (x2+y2):
(x2+y2)dy =dx
Step 2: Integrate both sides of the equation with respect to yand xsepa-
rately:
Z(x2+y2)dy =Zdx
Step 3: Evaluate the integrals:
x2y+1
3y3=x+C
where Cis the constant of integration.
Step 4: Rearrange the equation to solve for y:
y3+ 3x2y= 3x+ 3C
Step 5: The general solution to the given differential equation is:
y3+ 3x2y= 3x+ 3C
17
Question 20
Question
Consider the ordinary differential equation:
dy
dx =x2y+e2x
Find the particular solution given that y(0) = 1.
Solution
Step 1: Rewrite the given differential equation in the form dy = (x2y+e2x)dx.
Step 2: Separate the variables by dividing both sides by yand multiplying by
dx:dy
y= (x2+e2x)dx
Step 3: Integrate both sides:
Zdy
y=Z(x2+e2x)dx
Step 4: Simplify the integrals:
ln |y|=1
3x3+1
2e2x+C
Step 5: Exponentiate both sides to solve for y:
y=e(1
3x3+1
2e2x+C)
Step 6: Use the initial condition y(0) = 1 to find the value of the constant C:
1 = e(1
2+C)
Step 7: Solve for C:
C=−1
2
Step 8: Substitute the value of Cback into the particular solution:
y=e(1
3x3+1
2e2x−1
2)
Question 21
Question
Solve the following first-order ordinary differential equation using an integrating
factor:
(2x+ 1)y′−3y=x3
18
Solution
Step 1: Rewrite the equation in the form y′+P(x)y=Q(x).
(2x+ 1)y′−3y=x3
y′−3
2x+ 1y=x3
2x+ 1
Step 2: Identify P(x) and Q(x).
P(x) = −3
2x+ 1 and Q(x) = x3
2x+ 1
Step 3: Calculate the integrating factor µ(x) = eRP(x)dx.
µ(x) = eR−3
2x+1 dx
=e−3 ln |2x+1|
=eln(2x+1)−3
= (2x+ 1)−3
Step 4: Multiply the ODE by the integrating factor µ(x).
(2x+ 1)−3y′−3(2x+ 1)−4y=x3(2x+ 1)−3
d
dx[(2x+ 1)−3y] = x3(2x+ 1)−3
Step 5: Integrate both sides with respect to x.
Zd
dx[(2x+ 1)−3y]dx =Zx3(2x+ 1)−3dx
(2x+ 1)−3y=Zx3(2x+ 1)−3dx +C
Step 6: Solve the integral on the right side.
Zx3(2x+ 1)−3dx =Zx3
(2x+ 1)3dx
=Zx3
(2x+ 1)2(2x+ 1)dx
=Zu
2u2du (let u= 2x+ 1)
=1
2Z1
udu
=1
2ln |2x+ 1|+C
19
Step 7: Substitute back and solve for y.
(2x+ 1)−3y=1
2ln |2x+ 1|+C
y=(2x+ 1)3
2ln |2x+ 1|+C(2x+ 1)3
Therefore, the solution to the differential equation is y=(2x+1)3
2ln |2x+1|+
C(2x+ 1)3, where Cis an arbitrary constant.
Question 22
Question
Solve the following first-order ordinary differential equation:
dy
dx =3y−5x
2x+ 6y
Solution
Step 1: Rearrange the differential equation in the form dy
dx =P(x) + Q(x)y.
Step 2: Solve for P(x) and Q(x).
Given differential equation:
dy
dx =3y−5x
2x+ 6y
Separate variables:
(2x+ 6y)dy = (3y−5x)dx
Expand and rearrange:
2xdy + 6ydy = 3ydx −5xdx
Now, write the equation in the form dy
dx =P(x) + Q(x)y:
dy
dx =3y−5x
2x+ 6y=3
2−5
2x+ 6yy
So, P(x) = 3
2and Q(x) = −5
2x+6y.
Step 3: Solve the differential equation using an integrating factor.
The integrating factor I(x) is defined as:
I(x) = eRP(x)dx
Since P(x) = 3
2, then:
I(x) = eR3
2dx =e3
2x
20
Multiply the integrating factor with the original differential equation:
e3
2xdy
dx =e3
2x3
2−5
2x+ 6yy
Step 4: Integrate both sides of the equation.
Ze3
2xdy
dxdx =Ze3
2x3
2−5
2x+ 6yydx
e3
2xy=3
2Ze3
2xdx −5Zye 3
2xdx
Integrate using substitution method and solve for the constant of integration.
The final solution will be:
y=f(x)
Question 23
Question
Consider the differential equation:
dy
dx + 2xy = 4x
Find the general solution of the differential equation.
Solution
Step 1: First, we need to rearrange the equation to the standard form of a
first-order linear ODE: dy
dx + 2xy = 4x
Subtracting 2xy from both sides gives:
dy
dx = 4x−2xy
Step 2: We notice that this is a first-order linear differential equation. To
solve this, we will use an integrating factor. The integrating factor I(x) is given
by:
I(x) = eR2x dx =ex2
Step 3: Multiply both sides of the equation by the integrating factor:
ex2dy
dx =ex2(4x−2xy)
This simplifies to: d
dx(ex2y) = 4xex2
21
Step 4: Integrate both sides with respect to x:
Zd
dx(ex2y)dx =Z4xex2dx
ex2y=Z4xex2dx
Step 5: Solve the integral on the right side:
ex2y= 2ex2+C
where Cis the constant of integration.
Step 6: Finally, divide by ex2to find the general solution:
y= 2 + Ce−x2
So, the general solution of the differential equation is y= 2 + Ce−x2, where C
is an arbitrary constant.
Question 24
Question
Find the general solution of the following ordinary differential equation:
y′′ + 4y′+ 4y=e−2x.
Solution
Step 1: First, find the complementary solution by solving the homogeneous
equation y′′ + 4y′+ 4y= 0. The characteristic equation is r2+ 4r+ 4 = 0,
which has a repeated root at r=−2. Therefore, the complementary solution is
yc(x) = c1e−2x+c2xe−2x.
Step 2: Next, we will find a particular solution to the non-homogeneous
equation. Assume a particular solution of the form yp(x) = Ae−2x, where Ais
a constant to be determined. Plug this into the original differential equation:
−4Ae−2x+ 4Ae−2x+ 4Ae−2x=e−2x.
Simplify this to get 4Ae−2x=e−2x, which implies A=1
4. Therefore, the
particular solution is yp(x) = 1
4e−2x.
Step 3: The general solution is the sum of the complementary and particular
solutions. Thus, the general solution of the differential equation is
y(x) = yc(x) + yp(x) = c1e−2x+c2xe−2x+1
4e−2x,
where c1and c2are arbitrary constants.
22
Question 25
Question
Consider the first-order ordinary differential equation y′=ex+y. Solve the
initial value problem when y(0) = 1.
Solution
Step 1: First, we will rewrite the given differential equation in a separable form:
dy
dx =ex+y
Step 2: Next, we separate the variables by moving all terms involving yto
the left side and all terms involving xto the right side:
dy
dx −y=ex
Step 3: Now, we multiply through by the integrating factor e−xto make the
left side a perfect derivative:
e−xdy
dx −e−xy= 1
Step 4: Recognizing the left side as the derivative of ye−x, we rewrite the
equation as: d
dx(ye−x)=1
Step 5: Integrating both sides with respect to x, we have:
Zd
dx(ye−x)dx =Z1dx
ye−x=x+C
Step 6: Applying the initial condition y(0) = 1, we find the value of the
constant C:
1 = 0 + C=⇒C= 1
Step 7: Therefore, the solution to the initial value problem y′=ex+ywith
y(0) = 1 is:
ye−x=x+ 1
Step 8: Finally, we can write the solution explicitly as:
y=ex(x+ 1)
23
Question 26
Question
Consider the initial value problem:
dy
dx =−2xy2, y(0) = 1.
Find the particular solution to the differential equation.
Solution
Step 1: Let’s separate variables and solve for y:
dy
dx =−2xy2
dy
y2=−2x dx
Z1
y2dy =Z−2x dx
Zy−2dy =−x2+C1
−1
y=−x2+C1
y=1
x2−C1
Step 2: Next, we use the initial condition y(0) = 1 to find the particular
solution:
1 = 1
02−C1
1 = 1
−C1
C1=−1
Therefore, the particular solution to the differential equation is y=1
x2+1 .
Question 27
Question
Solve the initial value problem:
y′=2x+y−1
x−y, y(0) = 2
24
Solution
Step 1: First, let’s rewrite the given differential equation in a more convenient
form. We have:
y′=2x+y−1
x−y
which can be rewritten as:
y′(x−y)=2x+y−1
Step 2: Next, we expand the left-hand side and group like terms:
xy′−y= 2x+y−1
Step 3: Now, we rearrange the equation:
xy′−y−y= 2x+ 1
xy′−2y= 2x+ 1
Step 4: This is a linear first-order ODE. We need to solve it by finding an
integrating factor. The integrating factor is given by:
µ(x) = eR−2dx =e−2x
Step 5: Multiplying both sides of the ODE by the integrating factor, we get:
e−2xxy′−2e−2xy= 2e−2xx+e−2x
Step 6: The left-hand side can be simplified as the derivative of a product:
(e−2xy)′= 2e−2xx+e−2x
Step 7: Integrating both sides with respect to x gives:
Z(e−2xy)dx =Z(2e−2xx+e−2x)dx
e−2xy=−e−2x(x+ 1) + C
Step 8: Solving for y, we have:
y=−x−1 + Ce2x
Step 9: Using the initial condition y(0) = 2, we can find C:
2 = −0−1 + C
C= 3
Step 10: Therefore, the solution to the initial value problem is:
y=−x−1+3e2x
25
Question 28
Question
Solve the differential equation y′′ −4y′+ 4y=e2x, given that y(0) = 2 and
y′(0) = 1.
Solution
To solve the given second order linear homogeneous differential equation, we first
find the complementary solution by solving the characteristic equation r2−4r+
4 = 0.
Step 1: Find the Characteristic Equation The characteristic equation
is r2−4r+4 = 0, which factors to (r−2)2= 0. Thus, the characteristic equation
has a repeated root r= 2.
Step 2: Find the Complementary Solution The complementary solu-
tion is of the form yc(x) = (Ax +B)e2x, where Aand Bare constants to be
determined.
Step 3: Find the Particular Solution To find the particular solution,
we use the method of undetermined coefficients. Let’s assume the particular
solution has the form yp(x) = Ce2x.
Step 4: Substitute into the Differential Equation Substitute yp(x) into
the differential equation to find C. We have (2·2Ce2x−4·2Ce2x+4Ce2x) = e2x.
Simplifying, we get 4Ce2x−8Ce2x+ 4Ce2x=e2x, which simplifies to 0 = e2x,
a contradiction.
Thus, we need to modify our assumed particular solution. Let’s try yp(x) =
Axe2x.
Step 5: Find the Correct Particular Solution Substitute yp(x) into the
differential equation to find A. We have (2 ·2Axe2x−4·2Ae2x+ 4Axe2x) = e2x.
Simplifying, we get 4Axe2x−8Ae2x+4Axe2x=e2x, which simplifies to 4A= 1.
Therefore, A=1
4.
Step 6: Write the General Solution The general solution is given by
the sum of the complementary and particular solutions:
y(x) = yc(x) + yp(x) = (Ax +B)e2x+1
4xe2x
Step 7: Apply Initial Conditions Using the initial conditions y(0) = 2
and y′(0) = 1, we can find the values of Aand B.
Step 8: Solve for Constants Substitute the initial conditions into the
general solution and solve for Aand B:
y(0) = (0 + B)e0+ 0 = B= 2
y′(0) = A+1
4= 1 =⇒A=3
4
26
Step 9: Final Solution Therefore, the solution to the differential equation
y′′ −4y′+ 4y=e2xsubject to the initial conditions is:
y(x) = 3
4x+ 2e2x+1
4xe2x
Question 29
Question
Consider the differential equation: y′′ −2y′+y=e2x. Find the general solution
to this second-order linear homogeneous differential equation.
Solution
To find the general solution to the differential equation y′′ −2y′+y=e2x,
we first need to find the complementary solution by solving the characteristic
equation r2−2r+ 1 = 0.
Step 1: Find the roots of the characteristic equation The character-
istic equation is r2−2r+ 1 = (r−1)2= 0, which has a repeated root r= 1.
Step 2: Write down the complementary solution Since the character-
istic equation has a repeated root, the complementary solution is of the form
yc(x) = c1erx +c2xerx, where r= 1.
Therefore, the complementary solution is yc(x) = c1ex+c2xex.
Step 3: Find a particular solution Let’s assume the particular solution
has the form yp(x) = Ax2e2x.
Step 4: Calculate the derivatives We have y′
p(x) = (2Ax +A2)e2xand
y′′
p(x) = (4A+ 4Ax)e2x.
Step 5: Substitute into the differential equation Substitute yp(x),
y′
p(x), and y′′
p(x) into the differential equation y′′ −2y′+y=e2x: (4A+4Ax)e2x−
2(2Ax +A)e2x+Ax2e2x=e2x.
Step 6: Solve for A Simplify the equation and solve for the coefficient A:
12A+ 12Ax −4Ax −2A+Ax2= 1. This simplifies to Ax2+ 8Ax −A−2 = 0.
Comparing coefficients, we find that A=−1
2.
Step 7: Write down the particular solution Now that we have A, we
can write down the particular solution as yp(x) = −1
2x2e2x.
Step 8: Write down the general solution The general solution to the
differential equation is the sum of the complementary solution and the particular
solution: y(x) = yc(x) + yp(x) = c1ex+c2xex−1
2x2e2x, where c1and c2are
arbitrary constants.
Question 30
Question
Consider the following ordinary differential equation (ODE):
27
y′′ −2y′+ 2y=ex
Find the general solution to the ODE above.
Solution
Step 1: First, find the characteristic equation by setting the left-hand side of
the ODE equal to 0:
r2−2r+ 2 = 0
Step 2: Solve the characteristic equation to find the roots r1and r2:
r=2±p(−2)2−4(1)(2)
2(1) = 1 ±i
Step 3: Since the roots are complex, the general solution to the homogeneous
ODE is:
yh(x) = erx(c1cos(rx) + c2sin(rx))
where r= 1 and c1, c2are arbitrary constants.
Step 4: Next, find a particular solution to the non-homogeneous ODE:
yp(x) = Aex
Differentiate yp(x) twice and substitute into the original ODE to solve for A.
Step 5: Substituting yp(x) and its derivatives back into the ODE, we get:
ex−2Aex+ 2Aex=ex
Solving this equation, we find A= 1.
Step 6: The general solution to the non-homogeneous ODE is the sum of
the homogeneous and particular solutions:
y(x) = yh(x) + yp(x) = ex(c1cos(x) + c2sin(x) + 1)
where c1, c2are arbitrary constants.
Question 31
Question
Solve the following initial value problem:
dy
dx = 2x−1, y(0) = 3
28
Solution
Step 1: First, we solve the differential equation by integrating both sides with
respect to x.
Zdy
dx dx =Z(2x−1) dx
Step 2: This gives us:
y=x2−x+C, where Cis the constant of integration.
Step 3: Next, we use the initial condition y(0) = 3 to find the value of C.
3 = 02−0 + C
C= 3
Step 4: Therefore, the particular solution to the differential equation is:
y=x2−x+ 3
Question 32
Question
Find the general solution to the differential equation: dy
dx =−2xy.
Solution
Step 1: We start by separating variables. Rearranging the equation, we have:
dy
dx =−2xy
1
ydy =−2xdx
Step 2: Integrate both sides with respect to their respective variables:
Z1
ydy =Z−2xdx
ln |y|=−x2+C
Step 3: Exponentiate both sides to get rid of the natural logarithm:
|y|=e−x2+C
|y|=e−x2eC
|y|=Ce−x2
Step 4: Since Ccan be positive or negative, we can write the general solution
as:
y=±Ce−x2
29
Question 33
Question
Solve the following initial value problem:
y′= (3x2+ 2y) sin(x), y(0) = 1.
Solution
Step 1: First, rewrite the differential equation in a more recognizable form.
Separating variables, we have:
dy
dx = (3x2+ 2y) sin(x).
Step 2: Now, divide both sides by sin(x):
1
sin(x)
dy
dx = 3x2+ 2y.
Step 3: Rearrange the equation to isolate terms involving y:
dy
dx −2y= 3x2sin(x).
Step 4: This is now a first-order linear differential equation. The integrating
factor is given by e−2x.
Step 5: Multiply both sides of the equation by the integrating factor:
e−2xdy
dx −2e−2xy= 3x2sin(x)e−2x.
Step 6: The left-hand side can be rewritten as the derivative of the product
e−2xy:d
dx(e−2xy)=3x2sin(x)e−2x.
Step 7: Integrate both sides with respect to x:
e−2xy=Z3x2sin(x)e−2xdx +C,
where Cis the constant of integration.
Step 8: Solve the integral on the right-hand side. This can be done by parts
or a combination of repeated integration by parts and simplification.
Step 9: Once the integral is evaluated, solve for yby multiplying both sides
by e2xand simplifying.
Step 10: Finally, apply the initial condition y(0) = 1 to find the value of the
constant C.
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Question 34
Question
Solve the following first-order ordinary differential equation:
dy
dx = (xy +y)√x
Solution
Step 1: Write the given equation in the form dy
dx +P(x)y=Q(x). Step 2:
Determine the integrating factor. Step 3: Use the integrating factor to solve
the differential equation.
Step 1: Rewrite the given equation in the form dy
dx +P(x)y=Q(x).
We have: dy
dx = (xy +y)√x
Divide through by √x:
1
√x
dy
dx = (xy +y)
Rewrite in the required form:
dy
dx +1
xy=y
So, P(x) = 1
xand Q(x) = y.
Step 2: Determine the integrating factor.
The integrating factor is given by:
I(x) = eRP(x)dx =eR1
xdx =eln(x)=x
Step 3: Solve the differential equation.
Multiply the entire equation by the integrating factor x:
xdy
dx +y=xy
This is now an exact differential equation.
Integrate both sides with respect to x:
Zxdy
dxdx +Zydx =Zxydx
xy +C=x2
2+D
Therefore, the general solution to the differential equation is:
y=x2
2+Cx +D
where Cand Dare constants.
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Question 35
Question
Solve the initial value problem:
dy
dx =y
x+x2
y, y(1) = 2
Solution
Step 1: Rewrite the given differential equation as separable variables by moving
all terms involving yto one side and all terms involving xto the other side.
dy
y=dx
x+x2
ydy
Step 2: Integrate both sides with respect to their respective variables.
Z1
ydy =Z1
xdx +Zx2dy
Step 3: Solve the integrals separately.
ln |y|= ln |x|+x3
3+C1
where C1is the constant of integration.
Step 4: Exponentiate both sides to eliminate the natural logarithms.
|y|=Cxex3
3
where C=±eC1.
Step 5: Apply the initial condition y(1) = 2 to find the value of C.
|2|=C(1)e13
3=⇒2 = Ce1
3=⇒C= 2e−1
3
Step 6: Plug in the value of Cback into the general solution.
y= 2e−1
3xex3
3
Therefore, the solution to the initial value problem is:
y= 2e−1
3xex3
3
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