MATH 334 - DIFFERENTIAL
EQUATIONS - Initial value problems
Question Bank - Set 3
Liberty University
Question 1
Question
Consider the initial value problem:
dy
dx = 2x−3y, y(0) = 1
Determine the particular solution to the initial value problem.
Solution
To solve this initial value problem, we can use the method of integrating factors.
The integrating factor is given by µ(x) = eR−3dx =e−3x.
Step 1: Multiply both sides of the differential equation by the integrating
factor µ(x).
e−3xdy
dx −3e−3xy= 2xe−3x
Step 2: Recognize the left-hand side as the result of the product rule:
d
dx(ye−3x) = 2xe−3x
Step 3: Integrate both sides with respect to x:
Zd
dx(ye−3x)dx =Z2xe−3xdx
ye−3x=Z2xe−3xdx +C
Step 4: Solve the integral on the right-hand side:
ye−3x=e−3x(−2
3x−2
9) + C
y=−2
3x−2
9+Ce3x
Step 5: Use the initial condition y(0) = 1 to find the value of C:
1 = −2
9+C
C= 1 + 2
9=11
9
Step 6: Substitute C=11
9back into the particular solution:
y=−2
3x−2
9+11
9e3x
Therefore, the particular solution to the initial value problem is y=−2
3x−
2
9+11
9e3x.
Question 2
Question
Solve the initial value problem:
y′+ 2y= 4e−2x, y(0) = 3
Solution
Step 1: First, we find the integrating factor: Let P(x) = 2, then the integrating
factor is given by:
µ(x) = eRP(x)dx =eR2dx =e2x
Step 2: Next, we multiply the given equation by the integrating factor µ(x):
e2xy′+ 2e2xy= 4e2xe−2x
e2xy′+ 2e2xy= 4
Step 3: Rewrite the left side of the equation as the derivative of the product
y(x)e2x:
(e2xy)′= 4
Step 4: Integrate both sides with respect to x:
Z(e2xy)′dx =Z4dx
2
e2xy= 4x+C
Step 5: Apply the initial condition y(0) = 3 to find the value of the constant
C:
e2·0·3=4·0 + C
3 = C
Step 6: Write the final solution by substituting the value of Cback into the
equation:
e2xy= 4x+ 3
y=4x+ 3
e2x=4x+ 3
e2x
Question 3
Question
Solve the initial value problem: y′= 2y−4 with the initial condition y(0) = 3.
Solution
Step 1: First, we solve the differential equation y′= 2y−4. This is a first-
order linear differential equation. To solve it, we rewrite it in the standard form
y′−2y=−4.
Step 2: To solve the homogeneous part of the equation, we find the integrat-
ing factor. The integrating factor is given by eR−2dx =e−2x.
Step 3: Multiply the integrating factor by the given equation to get e−2xy′−
2e−2xy=−4e−2x.
Step 4: Rewriting the left side as the derivative of e−2xy, we have d
dx (e−2xy) =
−4e−2x.
Step 5: Integrating both sides with respect to x gives e−2xy= 2e−2x+C,
where Cis the constant of integration.
Step 6: Solving for y, we have y= 2 + Ce2x.
Step 7: To find the particular solution, we use the initial condition y(0) = 3.
Substitute x= 0 and y= 3 into the equation y= 2 + Ce2xto find the value of
C.
Step 8: We have 3 = 2 + C, which implies C= 1.
Step 9: Substituting C= 1 back into the equation y= 2 + Ce2xgives the
particular solution y= 2 + e2x.
Therefore, the solution to the initial value problem y′= 2y−4 with y(0) = 3
is y= 2 + e2x.
3
Question 4
Question
Solve the initial value problem:
y′+ 2xy = 2x, y(0) = 1
Solution
Step 1: First, we find the integrating factor by calculating eR2xdx.
Z2xdx =x2
Therefore, the integrating factor is ex2.
Step 2: Multiply both sides of the differential equation by the integrating
factor and rewrite the left-hand side as a product rule:
ex2y′+ 2xex2y= 2xex2
Step 3: Recognize that the left-hand side is now the derivative of a product:
(ex2y)′= 2xex2
Step 4: Integrate both sides with respect to x:
Z(ex2y)′dx =Z2xex2dx
ex2y=Z2xex2dx +C
Step 5: Solve the integral on the right-hand side:
ex2y=ex2+C
Step 6: Solve for y by dividing by ex2:
y= 1 + Ce−x2
Step 7: Use the initial condition y(0) = 1 to find the value of the constant
C:
1 = 1 + C·e0
C= 0
Step 8: Substitute the value of C back into the equation to obtain the
particular solution:
y= 1
Therefore, the solution to the initial value problem is y= 1.
4
Question 5
Question
Solve the initial value problem:
dy
dx = 3y−6x, y(0) = 4
Solution
Step 1: First, we identify the differential equation as a first-order linear equation
and rewrite it in standard form dy/dx +P(x)y=Q(x), where P(x) and Q(x)
are functions of x.
dy
dx = 3y−6x
dy
dx −3y=−6x
Step 2: Next, we identify P(x) = −3 and Q(x) = −6x. We now find the
integrating factor I(x) = eRP(x)dx.
I(x) = eR−3dx
=e−3x
Step 3: We multiply the differential equation by the integrating factor I(x).
e−3xdy
dx −3e−3xy=−6xe−3x
d
dx(e−3xy) = −6xe−3x
Step 4: Integrate both sides of the equation with respect to x.
Zd
dx(e−3xy)dx =Z−6xe−3xdx
e−3xy=Z−6xe−3xdx
Step 5: Solve the integral on the right side.
e−3xy= 3xe−3x−Z3e−3xdx
e−3xy= 3xe−3x+C
Step 6: Multiply through by e3x.
y= 3x+Ce3x
5
Step 7: Apply the initial condition y(0) = 4 to solve for C.
4 = 3(0) + C(1)
C= 4
Step 8: Substitute C= 4 back into the general solution.
y= 3x+ 4e3x
Therefore, the solution to the initial value problem is y= 3x+ 4e3x.
Question 6
Question
Consider the initial value problem:
dy
dx =y2, y(0) = 1.
Find the particular solution to the differential equation that satisfies the
initial condition.
Solution
Step 1: Separate the variables:
1
y2
dy
dx = 1
Step 2: Integrate both sides with respect to x:
Z1
y2
dy
dx dx =Z1dx
Step 3: Simplify the left-hand side and integrate:
Z1
y2dy =Z1dx
Step 4: Integrate R1
y2dy on the left-hand side:
−1
y=x+C
where Cis the constant of integration.
Step 5: Solve for yby multiplying both sides by -1:
1
y=−x−C
6
Step 6: Apply the initial condition y(0) = 1 to find the value of the constant
C:1
1=−0−C=⇒C=−1
Step 7: Substitute the value of Cback into the equation:
1
y=−x+ 1
Step 8: Solve for y:
y=1
−x+ 1
Therefore, the particular solution to the initial value problem is y=1
1−x.
Question 7
Question
Consider the initial value problem:
dy
dx =x+y
x−y, y(0) = 1.
Find the particular solution to the initial value problem on the interval
(−∞,∞).
Solution
To solve this initial value problem, we will first write the given differential
equation in a more standard form and then proceed to separate variables to
solve it.
Step 1: Rewrite the differential equation. We rewrite the given differential
equation as:
dy
dx =x+y
x−y=⇒dy
dx =1 + y
x
1−y
x
.
Step 2: Separate variables. Separating variables in the differential equation,
we get:
1−y
x
1 + y
x
dy =dx =⇒x−y
x+ydy =dx.
Step 3: Integrate both sides. Integrating both sides of the equation, we
have: Zx−y
x+ydy =Zdx.
Step 4: Perform the integration. Integrating the left side, we get:
Zx−y
x+ydy =Zx
x+ydy −Zy
x+ydy.
7
Solving these integrals and simplifying, we obtain:
xln |x+y| − yln |x+y|=x+C,
where Cis the constant of integration.
Step 5: Apply the initial condition. Using the initial condition y(0) = 1,
we substitute x= 0 and y= 1 into the equation above:
0 ln |0+1| − 1 ln |0+1|= 0 + C.
This simplifies to:
−ln(1) = C=⇒C= 0.
Step 6: Final solution. Substitute C= 0 back into the solution to get the
particular solution:
xln |x+y| − yln |x+y|=x.
Question 8
Question
Solve the initial value problem: y′= 2y+ 3xwith y(0) = 1.
Solution
Step 1: First, we need to find the general solution to the differential equation
y′= 2y+ 3x. This is a first-order linear differential equation, which can be
written in the standard form as dy
dx −2y= 3x. The integrating factor is eR−2dx =
e−2x.
Step 2: Multiply both sides of the standard form by the integrating factor
e−2xto get: e−2xdy
dx −2e−2xy= 3xe−2x.
Step 3: Rewrite the left-hand side using the product rule: d
dx (e−2xy) =
3xe−2x.
Step 4: Integrate both sides with respect to x:Rd
dx (e−2xy)dx =R3xe−2xdx.
This gives e−2xy=−3
2xe−2x+C, where Cis the constant of integration.
Step 5: Solve for yby multiplying by e2x:y=−3
2x+Ce2x.
Step 6: Apply the initial condition y(0) = 1 to find the value of C: 1 =
−3
2(0) + Ce2(0). So, C= 1.
Step 7: Substitute C= 1 back into the general solution to obtain the par-
ticular solution: y=−3
2x+e2x.
Therefore, the solution to the initial value problem y′= 2y+3xwith y(0) = 1
is y=−3
2x+e2x.
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Question 9
Question
Solve the initial value problem:
y′+y=e−2x, y(0) = 3
Solution
Step 1: First, let’s find the integrating factor µ(x) by multiplying both sides of
the differential equation by it:
µ(x) = eR1dx =ex
Step 2: Now, multiply the differential equation by the integrating factor:
exy′+exy=e−x·e−2x
Step 3: Notice that the left side can be rewritten as the derivative of the
product exy:
(exy)′=e−3x
Step 4: Integrate both sides with respect to x:
Z(exy)′dx =Ze−3xdx
Step 5: Solve the integrals:
exy=−1
3e−3x+C
Step 6: Next, solve for yby dividing by ex:
y=−1
3e−4x+Ce−x
Step 7: Use the initial condition y(0) = 3 to find the value of C:
3 = −1
3+C⇒C=10
3
Step 8: Finally, the solution to the initial value problem is:
y=−1
3e−4x+10
3e−x
9
Question 10
Question
Consider the initial value problem
dy
dx =x2+y, y(0) = 1.
Find the particular solution to the initial value problem using the method of
integrating factors.
Solution
Step 1: First, we rewrite the given differential equation in the form y′+P(x)y=
Q(x) where P(x) and Q(x) are functions of x.
dy
dx =x2+y
dy
dx −y=x2.
Step 2: Next, we identify P(x) and Q(x):
P(x) = −1, Q(x) = x2.
Step 3: To find the integrating factor I(x), we use the formula I(x) =
eRP(x)dx.
I(x) = eR−1dx
=e−x.
Step 4: Multiply the integrating factor by the differential equation.
e−xdy
dx −e−xy=x2e−x
d
dx(e−xy) = x2e−x.
Step 5: Integrate both sides with respect to x.
Zd
dx(e−xy)dx =Zx2e−xdx
e−xy= (−x2−2x−2)e−x+C,
where Cis the constant of integration.
Step 6: Solve for y.
y=−x2−2x−2 + Cex.
10
Step 7: To find the particular solution, we substitute x= 0 and y= 1 into
the equation.
1 = −02−2(0) −2 + Ce0
1 = −2 + C
C= 3.
Therefore, the particular solution to the initial value problem is y=−x2−
2x−2+3ex.
Question 11
Question
Consider the initial value problem:
dy
dx =x2+y2
xy , y(1) = 2
Use the method of separation of variables to find the particular solution to
this initial value problem.
Solution
Step 1: Rewrite the differential equation by separating variables:
dy
dx =x2+y2
xy =⇒y
y2dy =x
x2dx
Step 2: Integrate both sides with respect to their respective variables:
Z1
ydy =Z1
xdx
Step 3: Solve the integrals:
ln |y|= ln |x|+C
where Cis the constant of integration.
Step 4: Exponentiate both sides to remove the natural logarithm:
y=eln |x|+C=eln |x|·eC=Celn |x|=C|x|
Step 5: Apply the initial condition y(1) = 2 to find the value of the constant
C:
2 = C·1 =⇒C= 2
Step 6: Substitute C= 2 back into the general solution to find the particular
solution:
y= 2|x|
Therefore, the particular solution to the initial value problem is y= 2|x|.
11
Question 12
Question
Solve the initial value problem:
dy
dx + 2y= 4x, y(0) = 3
Solution
Step 1: First, we identify the integrating factor, which is given by eR2dx =e2x.
Step 2: Multiply the integrating factor by the differential equation:
e2xdy
dx + 2e2xy= 4xe2x
Step 3: Recognize that the left-hand side is the derivative of the product of
yand the integrating factor:
d
dx(ye2x) = 4xe2x
Step 4: Integrate with respect to x:
Zd
dx(ye2x)dx =Z4xe2xdx
Step 5: Solve the integrals:
ye2x=Z4xe2xdx
ye2x= 2xe2x−Z2e2xdx
ye2x= 2xe2x−e2x+C1
Step 6: Solve for y:
y= 2x−1 + C1e−2x
Step 7: Apply the initial condition y(0) = 3 to determine the value of C1:
3 = 2(0) −1 + C1
C1= 4
Step 8: The solution to the initial value problem is:
y= 2x−1+4e−2x
12
Question 13
Question
Consider the initial value problem
dy
dx = 2x(3 −y2), y(0) = 1.
Find the particular solution to this initial value problem.
Solution
Step 1: First, let’s separate variables in the given differential equation.
dy
3−y2= 2x dx.
Step 2: Next, we integrate both sides of the equation.
Z1
3−y2dy = 2 Zx dx.
Step 3: To integrate the left side, we use partial fraction decomposition. Let
Aand Bbe constants such that
1
3−y2=A
√3 + y+B
√3−y.
Step 4: Multiplying both sides by 3 −y2gives
1 = A(√3−y) + B(√3 + y).
Step 5: Set y=√3 to find Aand set y=−√3 to find B. Solving the system
of equations gives A=−1
2√3and B=1
2√3.
Step 6: Now, we can rewrite the integral as
−1
2√3Z1
√3 + ydy +1
2√3Z1
√3−ydy = 2 Zx dx.
Step 7: After integrating both sides, we get
−1
2√3ln |√3 + y|+1
2√3ln |√3−y|=x2+C,
where Cis the constant of integration.
Step 8: Simplifying the left side and using the initial condition y(0) = 1, we
find C= 0.
Step 9: Thus, the particular solution to the initial value problem is
−1
2√3ln |√3 + y|+1
2√3ln |√3−y|=x2.
13
Question 14
Question
Solve the initial value problem:
dy
dx =2y−x
x+y, y(0) = 1.
Solution
Step 1: Rewrite the differential equation in a more standard form by dividing
both sides by x+y:
dy
dx =2y−x
x+y=⇒dy
dx =2y/x −1
1 + y/x .
Step 2: Let u=y/x, then y=ux and dy
dx =u+xdu
dx . Substituting this into
the differential equation gives:
u+xdu
dx =2u−1
1 + u=⇒xdu
dx =u−1
1 + u.
Step 3: Separate variables and integrate both sides:
Z1 + u
u−1du =Z1
xdx.
Step 4: Solve the integral on the left side:
Z1 + u
u−1du =Z(u−1) + 2
u−1du =Z1 + 2
u−1du =u+ 2 ln |u−1|+C.
Step 5: Substitute back u=y/x and solve for y:
y
x+ 2 ln |y/x −1|= ln |x|+C.
Step 6: Use the initial condition y(0) = 1 to find the value of C:
1
0+ 2 ln |1−1|= ln |0|+C=⇒0 = −∞ +C=⇒C=∞.
Step 7: The particular solution is then:
y
x+ 2 ln |y/x −1|= ln |x|+∞.
Therefore, the solution to the initial value problem is y
x+ 2 ln |y/x −1|=
ln |x|+∞.
14
Question 15
Question
Consider the initial value problem:
dy
dx =2x
y, y(0) = 1
Solution
Step 1: Rewrite the initial value problem as a separable first-order differential
equation.
y dy = 2x dx
Step 2: Integrate both sides of the equation.
Zy dy =Z2x dx
Step 3: Solve the integrals.
1
2y2=x2+C
where Cis the constant of integration.
Step 4: Use the initial condition y(0) = 1 to find the value of the constant
C.1
2·12= 02+C=⇒C=1
2
Step 5: Substitute the value of Cback into the equation.
1
2y2=x2+1
2
Step 6: Solve for yin terms of x.
y2= 2x2+ 1 ⇒y=±p2x2+ 1
Therefore, the solution to the initial value problem is y=±√2x2+ 1, but
since y(0) = 1, the solution is y=√2x2+ 1.
Question 16
Question
Solve the initial value problem:
dy
dx = sin(2x) + 1
2y, y(0) = 1
15
Solution
Step 1: Rewrite the initial value problem:
dy
dx = sin(2x) + 1
2ywith y(0) = 1
Step 2: This is a first-order linear differential equation in the form dy
dx +
p(x)y=q(x), where p(x) = 1
2and q(x) = sin(2x).
Step 3: We start by finding the integrating factor I(x) = eRp(x)dx:
I(x) = eR1
2dx =e1
2x
Step 4: Multiply both sides of the differential equation by the integrating
factor I(x):
e1
2xdy
dx +e1
2x·1
2y=e1
2xsin(2x)
Step 5: Rewrite the left-hand side as the derivative of the product I(x)y
using the product rule:
d
dx e1
2xy=e1
2xsin(2x)
Step 6: Integrate both sides with respect to x:
Zd
dx e1
2xydx =Ze1
2xsin(2x)dx
Step 7: Integrating the right-hand side with respect to x, we get:
e1
2xy=−1
2e1
2xcos(2x) + C
Step 8: Solve for yby multiplying through by e−1
2x:
y=−1
2cos(2x) + Ce−1
2x
Step 9: Use the initial condition y(0) = 1 to solve for the constant C:
1 = −1
2cos(0) + Ce−1
2·0=⇒1 = −1
2+C=⇒C=3
2
Step 10: Substitute the value of Cback into the general solution to find the
particular solution:
y=−1
2cos(2x) + 3
2e−1
2x
Therefore, the solution to the initial value problem dy
dx = sin(2x) + 1
2ywith
y(0) = 1 is y=−1
2cos(2x) + 3
2e−1
2x.
16
Question 17
Question
Solve the initial value problem:
y′= 4y−e5x, y(0) = 3
Solution
Step 1: First, we find the general solution to the given differential equation.
The standard form of the linear first-order differential equation is y′+
P(x)y=Q(x). Comparing the given equation y′= 4y−e5xwith the stan-
dard form, we have P(x) = 4 and Q(x) = e5x.
The integrating factor I(x) is given by:
I(x) = eRP(x)dx =eR4dx =e4x
Multiplying both sides of the differential equation by the integrating factor,
we get:
e4xy′+ 4e4xy=e9x
Now, we can rewrite the left side of the above equation as the derivative of
the product of e4xand yusing the product rule:
d
dx(e4xy) = e9x
Integrating both sides with respect to xgives us:
e4xy=1
9e9x+C
(Where Cis the constant of integration)
Step 2: Next, we apply the initial condition y(0) = 3 to find the value of the
constant C:
e4(0) ·3 = 1
9e9(0) +C
3 = 1
9+C
C= 3 −1
9=26
9
Therefore, the particular solution to the initial value problem is:
e4xy=1
9e9x+26
9
y=1
9e5x+26
9e−4x
17
Question 18
Question
Consider the initial value problem given by
dy
dx =x+y
x−y, y(0) = 1.
Find the particular solution to the initial value problem.
Solution
Step 1: Rewrite the initial value problem in a separable form.
dy
dx =x+y
x−y
dy
dx =x+y
x−y
(x−y)dy = (x+y)dx
Step 2: Integrate both sides to solve for y.
Z(x−y)dy =Z(x+y)dx
Zx dy −Zy dy =Zx dx +Zy dx
xy −y2
2=x2
2+y2
2+C1
where C1is the constant of integration.
Step 3: Use the initial condition y(0) = 1 to find the value of C1.
(0)(1) −(1)2
2=(0)2
2+(1)2
2+C1
−1
2=1
2+C1
−1 = 1 + 2C1
C1=−1
Step 4: Substitute C1=−1 back into the equation to find the particular
solution.
xy −y2
2=x2
2+y2
2−1
2xy −2y2=x2+y2−2
2xy −2y2=x2+y2−2
2xy +y2=x2+ 2
Therefore, the particular solution to the initial value problem is 2xy +y2=
x2+ 2.
18
Question 19
Question
Consider the initial value problem:
dy
dx = 2e3x−5y, y(0) = 3
Solution
Step 1: First we rewrite the differential equation as a separable equation:
dy
dx + 5y= 2e3x
Step 2: To solve the separable equation, we separate variables:
dy
dx + 5y= 2e3x=⇒dy
dx = 2e3x−5y
Step 3: We now solve the separable equation: Separating variables, we have:
dy
2e3x−5y=dx
Step 4: Next, we integrate both sides of the equation:
Z1
2e3x−5ydy =Zdx
Step 5: By substituting u= 2e3x−5y,du =−5dy, we get:
Z−1
5udu =Zdx
Step 6: Integrating both sides, we have:
−1
5ln |u|=x+C
Step 7: Substituting back u= 2e3x−5yand rearranging, we get:
ln |2e3x−5y|=−5x−5C
Step 8: Next, we solve for y:
2e3x−5y=e−5x−5C
Step 9: Applying the initial condition y(0) = 3, we can solve for C:
2e0−5(3) = e0=⇒2−15 = 1 =⇒C=−13
Step 10: Substituting this value of Cback in, we have:
2e3x−5y=e−5x−5(−13)
Step 11: Simplifying, we get the solution to the initial value problem:
2e3x−5y=e−5x+65
19
Question 20
Question
Solve the initial value problem:
dy
dx =x
y, y(0) = 1
Solution
Step 1: Rewrite the differential equation as dy
dx =x
y. Step 2: Rearrange the
terms to separate variables: y dy =x dx. Step 3: Integrate both sides of the
equation:
Zy dy =Zx dx
Step 4: Solve the integrals to get:
y2
2=x2
2+C
where Cis the constant of integration. Step 5: Find the value of the constant
Cusing the initial condition y(0) = 1:
12
2=02
2+C=⇒1
2=C
So, the equation becomes y2
2=x2
2+1
2. Step 6: Solve for yin terms of x:
y2=x2+ 1 =⇒y=±px2+ 1
Step 7: Since y(0) = 1, we choose y=√x2+ 1. Therefore, the solution to the
initial value problem is y=√x2+ 1.
Question 21
Question
Consider the initial value problem:
dy
dx = 2y+x, y(0) = 1
Find the particular solution to the differential equation that satisfies the initial
condition.
20
Solution
Step 1: First, let’s solve the differential equation dy
dx = 2y+xusing the method
of integrating factors. The integrating factor is given by eR2dx =e2x.
d
dx(e2xy) = e2x(2y+x)
e2xy=Ze2x(2y+x)dx
e2xy=Z2e2xydx +Zxe2xdx
Step 2: We can solve the integrals on the right side of the equation to get:
e2xy= 2 Ze2xydx +Zxe2xdx
e2xy= 2e2xy−Ze2xdy +1
2xe2x−1
2Ze2xdx
e2xy= 2e2xy−e2x+1
2xe2x−1
4e2x+C
Step 3: Now, applying the initial condition y(0) = 1, we get:
e0·1 = 2e0·1−e0+1
2·0·e0−1
4·e0+C
1 = 2 −1−1
4
1 = 1
4+C
C=3
4
Step 4: Substituting C=3
4back into our equation, we find:
e2xy= 2e2xy−e2x+1
2xe2x−1
4e2x+3
4
e2xy−2e2xy=−e2x+1
2xe2x−1
4e2x+3
4
−e2xy=−e2x+1
2xe2x−1
4e2x+3
4
y= 1 −1
2x+1
4+3
4e−2x
Therefore, the particular solution to the differential equation that satisfies
the initial condition y(0) = 1 is y= 1 −1
2x+1
4+3
4e−2x.
21
Question 22
Question
Consider the initial value problem:
dy
dx = 2x+ 3y, y(0) = 1.
Find the particular solution to the initial value problem.
Solution
Step 1: Rewrite the given initial value problem as:
dy
dx −3y= 2x, y(0) = 1.
Step 2: This is a first-order linear differential equation of the form dy
dx +
P(x)y=Q(x). The integrating factor is given by eRP(x)dx.
Step 3: In this case, P(x) = −3, so the integrating factor is eR−3dx =e−3x.
Step 4: Multiply both sides of the differential equation by the integrating
factor:
e−3xdy
dx −3e−3xy= 2xe−3x.
Step 5: Rewrite the equation as the derivative of a product:
d
dx(ye−3x)=2xe−3x.
Step 6: Integrate both sides with respect to x:
Zd
dx(ye−3x)dx =Z2xe−3xdx.
Step 7: This simplifies to:
ye−3x=−2
3xe−3x−2
9e−3x+C,
where Cis the constant of integration.
Step 8: Applying the initial condition y(0) = 1, we have:
1=0−2
9+C⇒C=11
9.
Step 9: Substitute C=11
9back into the equation to find the particular
solution:
ye−3x=−2
3xe−3x−2
9e−3x+11
9.
Thus, the particular solution to the initial value problem is y=−2
3x−2
9+
11
9e3x.
22
Question 23
Question
Solve the initial value problem:
y′= 3y−6, y(0) = 2
Solution
Step 1: First, let’s solve the differential equation y′= 3y−6 by separating
variables: dy
dt = 3y−6
dy
3y−6=dt
Step 2: Integrate both sides with respect to yand t:
Z1
3y−6dy =Zdt
1
3ln |3y−6|=t+C1
ln |3y−6|= 3t+C2
|3y−6|=e3t+C2
|3y−6|=Ce3t(where C=eC2)
3y−6 = Ce3tor 3y=Ce3t+ 6
y=1
3Ce3t+ 6
Step 3: Next, we apply the initial condition y(0) = 2 to find the value of C:
2 = 1
3C·e0+ 6
2 = 1
3(C+ 6)
6 = C+ 6
C= 0
Step 4: Now that we have C= 0, the solution to the initial value problem
is:
y=1
3·6=2e3t
23
Question 24
Question
Solve the initial value problem:
dy
dx =3x2
y, y(1) = 2
Solution
Step 1: Rewrite the differential equation in a separable form.
y dy = 3x2dx
Step 2: Integrate both sides with respect to their respective variables.
Zy dy =Z3x2dx
Step 3: Solve the integrals.
1
2y2=x3+C
where Cis the constant of integration.
Step 4: Apply the initial condition y(1) = 2 to find the value of C.
1
2(2)2= 13+C=⇒2 = 1 + C=⇒C= 1
Step 5: Substitute the value of Cback into the equation.
1
2y2=x3+ 1
Step 6: Solve for yin terms of x.
y2= 2x3+ 2 =⇒y=±p2x3+ 2
Therefore, the solution to the initial value problem is y=√2x3+ 2 since
y(1) = 2.
Question 25
Question
Solve the initial value problem:
dy
dx = 2x+y, y(0) = 1
24
Solution
Step 1: Write the given initial value problem as:
dy
dx −y= 2x, y(0) = 1
Step 2: This is a first-order linear differential equation. To solve it, we first
find the integrating factor µ(x) which is given by µ(x) = eR−1dx.
µ(x) = eR−1dx =e−x
Step 3: Multiply both sides of the differential equation by the integrating
factor µ(x):
e−xdy
dx −e−xy= 2xe−x
Step 4: Recognize the left side as the derivative of the product of e−xand y
(this step involves integrating factors):
d
dx(ye−x) = 2xe−x
Step 5: Integrate both sides with respect to x:
Zd
dx(ye−x)dx =Z2xe−xdx
Step 6: Integrate to get:
ye−x=−2xe−x−2e−x+C
Step 7: Solve for yby multiplying both sides by ex:
y=−2x−2 + Cex
Step 8: Use the initial condition y(0) = 1 to find C:
1 = −2(0) −2 + C
C= 3
Step 9: Substitute C= 3 back into the general solution to get the particular
solution:
y=−2x−2+3ex
Therefore, the solution to the initial value problem is y=−2x−2 + 3ex
with the initial condition y(0) = 1.
Question 26
Question
Solve the initial value problem:
y′= 2xy2, y(0) = 1.
25
Solution
Step 1: Let’s rewrite the differential equation in a form suitable for separation
of variables: dy
dx = 2xy2.
Step 2: Separate variables by dividing both sides by y2and multiplying by
dx:dy
y2= 2xdx.
Step 3: Integrate both sides:
Z1
y2dy = 2 Zxdx.
Step 4: Solving the integrals gives:
−1
y=x2+C,
where Cis the constant of integration.
Step 5: To find the value of the constant C, we use the initial condition
y(0) = 1:
−1
1= 02+C⇒C=−1.
Step 6: Substituting Cback into the equation gives:
−1
y=x2−1.
Step 7: Solve for y:
y=1
1−x2.
Therefore, the solution to the initial value problem is y=1
1−x2with the
initial condition y(0) = 1.
Question 27
Question
Solve the initial value problem:
dy
dx =2x
y2, y(0) = 1
26
Solution
Step 1: Separate variables by writing the given differential equation as:
y2dy = 2xdx
Step 2: Integrate both sides with respect to their respective variables:
Zy2dy =Z2xdx
Step 3: Integrate to solve for y:
y3
3=x2+C
where Cis the constant of integration.
Step 4: Applying the initial condition y(0) = 1 to solve for C:
13
3= 0 + C=⇒C=1
3
Step 5: Substitute the value of Cback into the equation to find the particular
solution: y3
3=x2+1
3=⇒y3= 3x2+ 1
Step 6: Solve for yby taking the cube root of both sides:
y=3
p3x2+ 1
Therefore, the solution to the initial value problem is y=3
√3x2+ 1 with
the initial condition y(0) = 1.
Question 28
Question
Consider the initial value problem:
dy
dx =2x
y, y(0) = 2.
Solution
Step 1: Rewrite the differential equation in a separable form by isolating vari-
ables xand y:
y dy = 2x dx.
Step 2: Integrate both sides of the equation:
Zy dy =Z2x dx.
27
Step 3: Solve the integrals:
y2
2=x2+C,
where Cis the constant of integration.
Step 4: Use the initial condition y(0) = 2 to find the value of the constant
C:22
2= 0 + C=⇒C= 2.
Step 5: Substitute the value of Cback into the equation:
y2
2=x2+ 2.
Step 6: Solve for yin terms of x:
y2= 2(x2+ 2) =⇒y=±p2x2+ 4.
Therefore, the solution to the initial value problem is y=√2x2+ 4 since
y(0) = 2.
Question 29
Question
Solve the initial value problem:
dy
dx =4x+ 3y−6
3x+ 2y−5, y(0) = 2
Solution
Step 1: First, we rewrite the given differential equation in terms of dy and dx
by multiplying both sides by dx:
dx ·dy
dx =4x+ 3y−6
3x+ 2y−5·dx
Step 2: To make it easier to solve, we rewrite the equation as follows:
3x+ 2y−5
4x+ 3y−6dy =dx
Step 3: We now integrate both sides, and find a constant of integration, C:
Z3x+ 2y−5
4x+ 3y−6dy =Zdx
28
Step 4: After integrating, we get:
1
2ln |4x+ 3y−6|=x+C
Step 5: We can rewrite the expression using exponential functions:
|4x+ 3y−6|=e2x+2C
Step 6: Since y(0) = 2, we substitute x= 0 and y= 2 into the equation:
|4(0) + 3(2) −6|=e2(0)+2C
Step 7: Simplifying, we get:
|6|=e2C
Step 8: Thus, 6 = e2Cor e2C= 6. Taking the natural logarithm of both
sides, we get:
2C= ln 6
Step 9: Solving for C, we have:
C=1
2ln 6
Step 10: Substitute the value of Cback into the equation we obtained:
|4x+ 3y−6|=e2x+ln 6
Step 11: Finally, we simplify the equation to get the solution to the initial
value problem:
4x+ 3y−6 = 6e2x
Question 30
Question
Consider the initial value problem:
dy
dx = 2x3+ 3x2+x−1, y(0) = 5
Use Euler’s method with step size h= 0.1 to approximate the value of y(0.2).
29
Solution
Step 1: Define the step size h, the initial condition, and the function f(x, y).
Step size: h= 0.1
Initial condition: y(0) = 5
Function f(x, y)=2x3+ 3x2+x−1
Step 2: Perform Euler’s method to approximate y(0.1), y(0.2), . . . until x=
0.2.
x1=x0+h= 0 + 0.1=0.1
y1=y0+hf(x0, y0)
= 5 + 0.1(2(0)3+ 3(0)2+ 0 −1)
= 5 −0.1 = 4.9
x2=x1+h= 0.1+0.1=0.2
y2=y1+hf(x1, y1)
= 4.9+0.1(2(0.1)3+ 3(0.1)2+ 0.1−1)
= 4.9+0.1(0.002 + 0.03 + 0.1−1)
= 4.9+0.1(−0.968)
= 4.9−0.0968
= 4.8032
Therefore, the approximate value of y(0.2) using Euler’s method with step
size h= 0.1 is y(0.2) ≈4.8032.
Question 31
Question
Solve the initial value problem:
dy
dx =2x
y, y(0) = 1
Solution
Step 1: Rewrite the differential equation in terms of yand dy:
y dy = 2x dx
Step 2: Integrate both sides:
Zy dy =Z2x dx
y2
2=x2+C1
30
Step 3: Apply the initial condition y(0) = 1:
12
2= 0 + C1
C1=1
2
Step 4: Substitute C1=1
2back into the equation:
y2
2=x2+1
2
y2= 2x2+ 1
Step 5: Solve for yby taking the square root:
y=±p2x2+ 1
Step 6: Apply the initial condition y(0) = 1 to determine the specific solu-
tion:
1 = p2(0)2+ 1
1 = √1
y= 1
Therefore, the solution to the initial value problem is y= 1.
Question 32
Question
Solve the initial value problem: dy
dx =y2+x2,y(0) = 1.
Solution
Step 1: Rewrite the equation as a separable first-order differential equation by
bringing all terms involving yto the left side and all terms involving xto the
right side. dy
dx =y2+x2
dy
y2=dx
Step 2: Integrate both sides with respect to yand xseparately.
Z1
y2dy =Zdx
−1
y=x+C1
31
where C1is the constant of integration.
Step 3: Solve for yin terms of xby isolating y.
−1
y=x+C1
−1 = y(x+C1)
y=1
1−x−C1
Step 4: Use the initial condition y(0) = 1 to find the value of the constant
C1.
1 = 1
1−0−C1
1 = 1
1−C1
1−C1= 1
C1= 0
Step 5: Substitute C1= 0 back into the expression for yto obtain the
particular solution.
y=1
1−x
Therefore, the solution to the initial value problem is y=1
1−x.
Question 33
Question
Solve the initial value problem:
dy
dx = 2x(1 −y), y(0) = 0
Solution
Step 1: First, rewrite the differential equation in a separable form by moving
all terms involving yto one side and all terms involving xto the other side:
dy
1−y= 2x dx
Step 2: Integrate both sides with respect to their respective variables:
Z1
1−ydy =Z2x dx
32
Step 3: Solve the integrals:
−ln |1−y|=x2+C1
where C1is the constant of integration.
Step 4: Exponentiate both sides to eliminate the natural logarithm:
|1−y|=e−x2−C1=Ce−x2
where C=±e−C1.
Step 5: Rewrite the equation without absolute values:
1−y=Ce−x2
or
y= 1 −Ce−x2
Step 6: Apply the initial condition y(0) = 0 to solve for the constant C:
0=1−Ce−0
C= 1
Step 7: Substitute C= 1 back into the solution:
y= 1 −e−x2
Therefore, the solution to the initial value problem is y= 1 −e−x2.
Question 34
Question
Solve the initial value problem:
dy
dx =1
x+y, y(1) = 1
Solution
Step 1: Rewrite the differential equation in standard form by separating vari-
ables:
(1 + y)dy =dx
Step 2: Integrate both sides with respect to their respective variables:
Z(1 + y)dy =Zdx
Z1dy +Zy dy =Zdx
33
y+y2
2=x+C
Step 3: Use the initial condition y(1) = 1 to find the value of the constant
C:
1 + 1
2= 1 + C
C=1
2
Step 4: Substitute C=1
2back into the equation to find the particular
solution:
y+y2
2=x+1
2
Step 5: Use the quadratic formula to solve for y:
y2+ 2y−2x−1 = 0
y=−2±p22−4·1·(−2x−1)
2·1
y=−2±√4+8x+ 4
2
y=−2±√8x+ 8
2
y=−1±√2x+ 2
Step 6: Use the initial condition y(1) = 1 to determine the specific solution:
1 = −1±√2·1+2
2 = √4
Since √4=2>1,we choose the positive sign:
y=−1 + √2x+ 2
Therefore, the solution to the initial value problem is y=−1 + √2x+ 2.
Question 35
Question
Consider the initial value problem:
dy
dx = 2y, y(0) = 3
Find the solution to this initial value problem.
34
Step 4: Solve the integral on the right-hand side:
ye−3x=e−3x(−2
3x−2
9) + C
y=−2
3x−2
9+Ce3x
Step 5: Use the initial condition y(0) = 1 to find the value of C:
1 = −2
9+C
C= 1 + 2
9=11
9
Step 6: Substitute C=11
9back into the particular solution:
y=−2
3x−2
9+11
9e3x
Therefore, the particular solution to the initial value problem is y=−2
3x−
2
9+11
9e3x.
Question 2
Question
Solve the initial value problem:
y′+ 2y= 4e−2x, y(0) = 3
Solution
Step 1: First, we find the integrating factor: Let P(x) = 2, then the integrating
factor is given by:
µ(x) = eRP(x)dx =eR2dx =e2x
Step 2: Next, we multiply the given equation by the integrating factor µ(x):
e2xy′+ 2e2xy= 4e2xe−2x
e2xy′+ 2e2xy= 4
Step 3: Rewrite the left side of the equation as the derivative of the product
y(x)e2x:
(e2xy)′= 4
Step 4: Integrate both sides with respect to x:
Z(e2xy)′dx =Z4dx
2
e2xy= 4x+C
Step 5: Apply the initial condition y(0) = 3 to find the value of the constant
C:
e2·0·3=4·0 + C
3 = C
Step 6: Write the final solution by substituting the value of Cback into the
equation:
e2xy= 4x+ 3
y=4x+ 3
e2x=4x+ 3
e2x
Question 3
Question
Solve the initial value problem: y′= 2y−4 with the initial condition y(0) = 3.
Solution
Step 1: First, we solve the differential equation y′= 2y−4. This is a first-
order linear differential equation. To solve it, we rewrite it in the standard form
y′−2y=−4.
Step 2: To solve the homogeneous part of the equation, we find the integrat-
ing factor. The integrating factor is given by eR−2dx =e−2x.
Step 3: Multiply the integrating factor by the given equation to get e−2xy′−
2e−2xy=−4e−2x.
Step 4: Rewriting the left side as the derivative of e−2xy, we have d
dx (e−2xy) =
−4e−2x.
Step 5: Integrating both sides with respect to x gives e−2xy= 2e−2x+C,
where Cis the constant of integration.
Step 6: Solving for y, we have y= 2 + Ce2x.
Step 7: To find the particular solution, we use the initial condition y(0) = 3.
Substitute x= 0 and y= 3 into the equation y= 2 + Ce2xto find the value of
C.
Step 8: We have 3 = 2 + C, which implies C= 1.
Step 9: Substituting C= 1 back into the equation y= 2 + Ce2xgives the
particular solution y= 2 + e2x.
Therefore, the solution to the initial value problem y′= 2y−4 with y(0) = 3
is y= 2 + e2x.
3
Question 4
Question
Solve the initial value problem:
y′+ 2xy = 2x, y(0) = 1
Solution
Step 1: First, we find the integrating factor by calculating eR2xdx.
Z2xdx =x2
Therefore, the integrating factor is ex2.
Step 2: Multiply both sides of the differential equation by the integrating
factor and rewrite the left-hand side as a product rule:
ex2y′+ 2xex2y= 2xex2
Step 3: Recognize that the left-hand side is now the derivative of a product:
(ex2y)′= 2xex2
Step 4: Integrate both sides with respect to x:
Z(ex2y)′dx =Z2xex2dx
ex2y=Z2xex2dx +C
Step 5: Solve the integral on the right-hand side:
ex2y=ex2+C
Step 6: Solve for y by dividing by ex2:
y= 1 + Ce−x2
Step 7: Use the initial condition y(0) = 1 to find the value of the constant
C:
1 = 1 + C·e0
C= 0
Step 8: Substitute the value of C back into the equation to obtain the
particular solution:
y= 1
Therefore, the solution to the initial value problem is y= 1.
4
Question 5
Question
Solve the initial value problem:
dy
dx = 3y−6x, y(0) = 4
Solution
Step 1: First, we identify the differential equation as a first-order linear equation
and rewrite it in standard form dy/dx +P(x)y=Q(x), where P(x) and Q(x)
are functions of x.
dy
dx = 3y−6x
dy
dx −3y=−6x
Step 2: Next, we identify P(x) = −3 and Q(x) = −6x. We now find the
integrating factor I(x) = eRP(x)dx.
I(x) = eR−3dx
=e−3x
Step 3: We multiply the differential equation by the integrating factor I(x).
e−3xdy
dx −3e−3xy=−6xe−3x
d
dx(e−3xy) = −6xe−3x
Step 4: Integrate both sides of the equation with respect to x.
Zd
dx(e−3xy)dx =Z−6xe−3xdx
e−3xy=Z−6xe−3xdx
Step 5: Solve the integral on the right side.
e−3xy= 3xe−3x−Z3e−3xdx
e−3xy= 3xe−3x+C
Step 6: Multiply through by e3x.
y= 3x+Ce3x
5
Step 7: Apply the initial condition y(0) = 4 to solve for C.
4 = 3(0) + C(1)
C= 4
Step 8: Substitute C= 4 back into the general solution.
y= 3x+ 4e3x
Therefore, the solution to the initial value problem is y= 3x+ 4e3x.
Question 6
Question
Consider the initial value problem:
dy
dx =y2, y(0) = 1.
Find the particular solution to the differential equation that satisfies the
initial condition.
Solution
Step 1: Separate the variables:
1
y2
dy
dx = 1
Step 2: Integrate both sides with respect to x:
Z1
y2
dy
dx dx =Z1dx
Step 3: Simplify the left-hand side and integrate:
Z1
y2dy =Z1dx
Step 4: Integrate R1
y2dy on the left-hand side:
−1
y=x+C
where Cis the constant of integration.
Step 5: Solve for yby multiplying both sides by -1:
1
y=−x−C
6
Step 6: Apply the initial condition y(0) = 1 to find the value of the constant
C:1
1=−0−C=⇒C=−1
Step 7: Substitute the value of Cback into the equation:
1
y=−x+ 1
Step 8: Solve for y:
y=1
−x+ 1
Therefore, the particular solution to the initial value problem is y=1
1−x.
Question 7
Question
Consider the initial value problem:
dy
dx =x+y
x−y, y(0) = 1.
Find the particular solution to the initial value problem on the interval
(−∞,∞).
Solution
To solve this initial value problem, we will first write the given differential
equation in a more standard form and then proceed to separate variables to
solve it.
Step 1: Rewrite the differential equation. We rewrite the given differential
equation as:
dy
dx =x+y
x−y=⇒dy
dx =1 + y
x
1−y
x
.
Step 2: Separate variables. Separating variables in the differential equation,
we get:
1−y
x
1 + y
x
dy =dx =⇒x−y
x+ydy =dx.
Step 3: Integrate both sides. Integrating both sides of the equation, we
have: Zx−y
x+ydy =Zdx.
Step 4: Perform the integration. Integrating the left side, we get:
Zx−y
x+ydy =Zx
x+ydy −Zy
x+ydy.
7
Solving these integrals and simplifying, we obtain:
xln |x+y| − yln |x+y|=x+C,
where Cis the constant of integration.
Step 5: Apply the initial condition. Using the initial condition y(0) = 1,
we substitute x= 0 and y= 1 into the equation above:
0 ln |0+1| − 1 ln |0+1|= 0 + C.
This simplifies to:
−ln(1) = C=⇒C= 0.
Step 6: Final solution. Substitute C= 0 back into the solution to get the
particular solution:
xln |x+y| − yln |x+y|=x.
Question 8
Question
Solve the initial value problem: y′= 2y+ 3xwith y(0) = 1.
Solution
Step 1: First, we need to find the general solution to the differential equation
y′= 2y+ 3x. This is a first-order linear differential equation, which can be
written in the standard form as dy
dx −2y= 3x. The integrating factor is eR−2dx =
e−2x.
Step 2: Multiply both sides of the standard form by the integrating factor
e−2xto get: e−2xdy
dx −2e−2xy= 3xe−2x.
Step 3: Rewrite the left-hand side using the product rule: d
dx (e−2xy) =
3xe−2x.
Step 4: Integrate both sides with respect to x:Rd
dx (e−2xy)dx =R3xe−2xdx.
This gives e−2xy=−3
2xe−2x+C, where Cis the constant of integration.
Step 5: Solve for yby multiplying by e2x:y=−3
2x+Ce2x.
Step 6: Apply the initial condition y(0) = 1 to find the value of C: 1 =
−3
2(0) + Ce2(0). So, C= 1.
Step 7: Substitute C= 1 back into the general solution to obtain the par-
ticular solution: y=−3
2x+e2x.
Therefore, the solution to the initial value problem y′= 2y+3xwith y(0) = 1
is y=−3
2x+e2x.
8
Question 9
Question
Solve the initial value problem:
y′+y=e−2x, y(0) = 3
Solution
Step 1: First, let’s find the integrating factor µ(x) by multiplying both sides of
the differential equation by it:
µ(x) = eR1dx =ex
Step 2: Now, multiply the differential equation by the integrating factor:
exy′+exy=e−x·e−2x
Step 3: Notice that the left side can be rewritten as the derivative of the
product exy:
(exy)′=e−3x
Step 4: Integrate both sides with respect to x:
Z(exy)′dx =Ze−3xdx
Step 5: Solve the integrals:
exy=−1
3e−3x+C
Step 6: Next, solve for yby dividing by ex:
y=−1
3e−4x+Ce−x
Step 7: Use the initial condition y(0) = 3 to find the value of C:
3 = −1
3+C⇒C=10
3
Step 8: Finally, the solution to the initial value problem is:
y=−1
3e−4x+10
3e−x
9
Question 10
Question
Consider the initial value problem
dy
dx =x2+y, y(0) = 1.
Find the particular solution to the initial value problem using the method of
integrating factors.
Solution
Step 1: First, we rewrite the given differential equation in the form y′+P(x)y=
Q(x) where P(x) and Q(x) are functions of x.
dy
dx =x2+y
dy
dx −y=x2.
Step 2: Next, we identify P(x) and Q(x):
P(x) = −1, Q(x) = x2.
Step 3: To find the integrating factor I(x), we use the formula I(x) =
eRP(x)dx.
I(x) = eR−1dx
=e−x.
Step 4: Multiply the integrating factor by the differential equation.
e−xdy
dx −e−xy=x2e−x
d
dx(e−xy) = x2e−x.
Step 5: Integrate both sides with respect to x.
Zd
dx(e−xy)dx =Zx2e−xdx
e−xy= (−x2−2x−2)e−x+C,
where Cis the constant of integration.
Step 6: Solve for y.
y=−x2−2x−2 + Cex.
10
Step 7: To find the particular solution, we substitute x= 0 and y= 1 into
the equation.
1 = −02−2(0) −2 + Ce0
1 = −2 + C
C= 3.
Therefore, the particular solution to the initial value problem is y=−x2−
2x−2+3ex.
Question 11
Question
Consider the initial value problem:
dy
dx =x2+y2
xy , y(1) = 2
Use the method of separation of variables to find the particular solution to
this initial value problem.
Solution
Step 1: Rewrite the differential equation by separating variables:
dy
dx =x2+y2
xy =⇒y
y2dy =x
x2dx
Step 2: Integrate both sides with respect to their respective variables:
Z1
ydy =Z1
xdx
Step 3: Solve the integrals:
ln |y|= ln |x|+C
where Cis the constant of integration.
Step 4: Exponentiate both sides to remove the natural logarithm:
y=eln |x|+C=eln |x|·eC=Celn |x|=C|x|
Step 5: Apply the initial condition y(1) = 2 to find the value of the constant
C:
2 = C·1 =⇒C= 2
Step 6: Substitute C= 2 back into the general solution to find the particular
solution:
y= 2|x|
Therefore, the particular solution to the initial value problem is y= 2|x|.
11
Question 12
Question
Solve the initial value problem:
dy
dx + 2y= 4x, y(0) = 3
Solution
Step 1: First, we identify the integrating factor, which is given by eR2dx =e2x.
Step 2: Multiply the integrating factor by the differential equation:
e2xdy
dx + 2e2xy= 4xe2x
Step 3: Recognize that the left-hand side is the derivative of the product of
yand the integrating factor:
d
dx(ye2x) = 4xe2x
Step 4: Integrate with respect to x:
Zd
dx(ye2x)dx =Z4xe2xdx
Step 5: Solve the integrals:
ye2x=Z4xe2xdx
ye2x= 2xe2x−Z2e2xdx
ye2x= 2xe2x−e2x+C1
Step 6: Solve for y:
y= 2x−1 + C1e−2x
Step 7: Apply the initial condition y(0) = 3 to determine the value of C1:
3 = 2(0) −1 + C1
C1= 4
Step 8: The solution to the initial value problem is:
y= 2x−1+4e−2x
12
Question 13
Question
Consider the initial value problem
dy
dx = 2x(3 −y2), y(0) = 1.
Find the particular solution to this initial value problem.
Solution
Step 1: First, let’s separate variables in the given differential equation.
dy
3−y2= 2x dx.
Step 2: Next, we integrate both sides of the equation.
Z1
3−y2dy = 2 Zx dx.
Step 3: To integrate the left side, we use partial fraction decomposition. Let
Aand Bbe constants such that
1
3−y2=A
√3 + y+B
√3−y.
Step 4: Multiplying both sides by 3 −y2gives
1 = A(√3−y) + B(√3 + y).
Step 5: Set y=√3 to find Aand set y=−√3 to find B. Solving the system
of equations gives A=−1
2√3and B=1
2√3.
Step 6: Now, we can rewrite the integral as
−1
2√3Z1
√3 + ydy +1
2√3Z1
√3−ydy = 2 Zx dx.
Step 7: After integrating both sides, we get
−1
2√3ln |√3 + y|+1
2√3ln |√3−y|=x2+C,
where Cis the constant of integration.
Step 8: Simplifying the left side and using the initial condition y(0) = 1, we
find C= 0.
Step 9: Thus, the particular solution to the initial value problem is
−1
2√3ln |√3 + y|+1
2√3ln |√3−y|=x2.
13
Question 14
Question
Solve the initial value problem:
dy
dx =2y−x
x+y, y(0) = 1.
Solution
Step 1: Rewrite the differential equation in a more standard form by dividing
both sides by x+y:
dy
dx =2y−x
x+y=⇒dy
dx =2y/x −1
1 + y/x .
Step 2: Let u=y/x, then y=ux and dy
dx =u+xdu
dx . Substituting this into
the differential equation gives:
u+xdu
dx =2u−1
1 + u=⇒xdu
dx =u−1
1 + u.
Step 3: Separate variables and integrate both sides:
Z1 + u
u−1du =Z1
xdx.
Step 4: Solve the integral on the left side:
Z1 + u
u−1du =Z(u−1) + 2
u−1du =Z1 + 2
u−1du =u+ 2 ln |u−1|+C.
Step 5: Substitute back u=y/x and solve for y:
y
x+ 2 ln |y/x −1|= ln |x|+C.
Step 6: Use the initial condition y(0) = 1 to find the value of C:
1
0+ 2 ln |1−1|= ln |0|+C=⇒0 = −∞ +C=⇒C=∞.
Step 7: The particular solution is then:
y
x+ 2 ln |y/x −1|= ln |x|+∞.
Therefore, the solution to the initial value problem is y
x+ 2 ln |y/x −1|=
ln |x|+∞.
14
Question 15
Question
Consider the initial value problem:
dy
dx =2x
y, y(0) = 1
Solution
Step 1: Rewrite the initial value problem as a separable first-order differential
equation.
y dy = 2x dx
Step 2: Integrate both sides of the equation.
Zy dy =Z2x dx
Step 3: Solve the integrals.
1
2y2=x2+C
where Cis the constant of integration.
Step 4: Use the initial condition y(0) = 1 to find the value of the constant
C.1
2·12= 02+C=⇒C=1
2
Step 5: Substitute the value of Cback into the equation.
1
2y2=x2+1
2
Step 6: Solve for yin terms of x.
y2= 2x2+ 1 ⇒y=±p2x2+ 1
Therefore, the solution to the initial value problem is y=±√2x2+ 1, but
since y(0) = 1, the solution is y=√2x2+ 1.
Question 16
Question
Solve the initial value problem:
dy
dx = sin(2x) + 1
2y, y(0) = 1
15
Solution
Step 1: Rewrite the initial value problem:
dy
dx = sin(2x) + 1
2ywith y(0) = 1
Step 2: This is a first-order linear differential equation in the form dy
dx +
p(x)y=q(x), where p(x) = 1
2and q(x) = sin(2x).
Step 3: We start by finding the integrating factor I(x) = eRp(x)dx:
I(x) = eR1
2dx =e1
2x
Step 4: Multiply both sides of the differential equation by the integrating
factor I(x):
e1
2xdy
dx +e1
2x·1
2y=e1
2xsin(2x)
Step 5: Rewrite the left-hand side as the derivative of the product I(x)y
using the product rule:
d
dx e1
2xy=e1
2xsin(2x)
Step 6: Integrate both sides with respect to x:
Zd
dx e1
2xydx =Ze1
2xsin(2x)dx
Step 7: Integrating the right-hand side with respect to x, we get:
e1
2xy=−1
2e1
2xcos(2x) + C
Step 8: Solve for yby multiplying through by e−1
2x:
y=−1
2cos(2x) + Ce−1
2x
Step 9: Use the initial condition y(0) = 1 to solve for the constant C:
1 = −1
2cos(0) + Ce−1
2·0=⇒1 = −1
2+C=⇒C=3
2
Step 10: Substitute the value of Cback into the general solution to find the
particular solution:
y=−1
2cos(2x) + 3
2e−1
2x
Therefore, the solution to the initial value problem dy
dx = sin(2x) + 1
2ywith
y(0) = 1 is y=−1
2cos(2x) + 3
2e−1
2x.
16
Question 17
Question
Solve the initial value problem:
y′= 4y−e5x, y(0) = 3
Solution
Step 1: First, we find the general solution to the given differential equation.
The standard form of the linear first-order differential equation is y′+
P(x)y=Q(x). Comparing the given equation y′= 4y−e5xwith the stan-
dard form, we have P(x) = 4 and Q(x) = e5x.
The integrating factor I(x) is given by:
I(x) = eRP(x)dx =eR4dx =e4x
Multiplying both sides of the differential equation by the integrating factor,
we get:
e4xy′+ 4e4xy=e9x
Now, we can rewrite the left side of the above equation as the derivative of
the product of e4xand yusing the product rule:
d
dx(e4xy) = e9x
Integrating both sides with respect to xgives us:
e4xy=1
9e9x+C
(Where Cis the constant of integration)
Step 2: Next, we apply the initial condition y(0) = 3 to find the value of the
constant C:
e4(0) ·3 = 1
9e9(0) +C
3 = 1
9+C
C= 3 −1
9=26
9
Therefore, the particular solution to the initial value problem is:
e4xy=1
9e9x+26
9
y=1
9e5x+26
9e−4x
17
Question 18
Question
Consider the initial value problem given by
dy
dx =x+y
x−y, y(0) = 1.
Find the particular solution to the initial value problem.
Solution
Step 1: Rewrite the initial value problem in a separable form.
dy
dx =x+y
x−y
dy
dx =x+y
x−y
(x−y)dy = (x+y)dx
Step 2: Integrate both sides to solve for y.
Z(x−y)dy =Z(x+y)dx
Zx dy −Zy dy =Zx dx +Zy dx
xy −y2
2=x2
2+y2
2+C1
where C1is the constant of integration.
Step 3: Use the initial condition y(0) = 1 to find the value of C1.
(0)(1) −(1)2
2=(0)2
2+(1)2
2+C1
−1
2=1
2+C1
−1 = 1 + 2C1
C1=−1
Step 4: Substitute C1=−1 back into the equation to find the particular
solution.
xy −y2
2=x2
2+y2
2−1
2xy −2y2=x2+y2−2
2xy −2y2=x2+y2−2
2xy +y2=x2+ 2
Therefore, the particular solution to the initial value problem is 2xy +y2=
x2+ 2.
18
Question 19
Question
Consider the initial value problem:
dy
dx = 2e3x−5y, y(0) = 3
Solution
Step 1: First we rewrite the differential equation as a separable equation:
dy
dx + 5y= 2e3x
Step 2: To solve the separable equation, we separate variables:
dy
dx + 5y= 2e3x=⇒dy
dx = 2e3x−5y
Step 3: We now solve the separable equation: Separating variables, we have:
dy
2e3x−5y=dx
Step 4: Next, we integrate both sides of the equation:
Z1
2e3x−5ydy =Zdx
Step 5: By substituting u= 2e3x−5y,du =−5dy, we get:
Z−1
5udu =Zdx
Step 6: Integrating both sides, we have:
−1
5ln |u|=x+C
Step 7: Substituting back u= 2e3x−5yand rearranging, we get:
ln |2e3x−5y|=−5x−5C
Step 8: Next, we solve for y:
2e3x−5y=e−5x−5C
Step 9: Applying the initial condition y(0) = 3, we can solve for C:
2e0−5(3) = e0=⇒2−15 = 1 =⇒C=−13
Step 10: Substituting this value of Cback in, we have:
2e3x−5y=e−5x−5(−13)
Step 11: Simplifying, we get the solution to the initial value problem:
2e3x−5y=e−5x+65
19
Question 20
Question
Solve the initial value problem:
dy
dx =x
y, y(0) = 1
Solution
Step 1: Rewrite the differential equation as dy
dx =x
y. Step 2: Rearrange the
terms to separate variables: y dy =x dx. Step 3: Integrate both sides of the
equation:
Zy dy =Zx dx
Step 4: Solve the integrals to get:
y2
2=x2
2+C
where Cis the constant of integration. Step 5: Find the value of the constant
Cusing the initial condition y(0) = 1:
12
2=02
2+C=⇒1
2=C
So, the equation becomes y2
2=x2
2+1
2. Step 6: Solve for yin terms of x:
y2=x2+ 1 =⇒y=±px2+ 1
Step 7: Since y(0) = 1, we choose y=√x2+ 1. Therefore, the solution to the
initial value problem is y=√x2+ 1.
Question 21
Question
Consider the initial value problem:
dy
dx = 2y+x, y(0) = 1
Find the particular solution to the differential equation that satisfies the initial
condition.
20
Solution
Step 1: First, let’s solve the differential equation dy
dx = 2y+xusing the method
of integrating factors. The integrating factor is given by eR2dx =e2x.
d
dx(e2xy) = e2x(2y+x)
e2xy=Ze2x(2y+x)dx
e2xy=Z2e2xydx +Zxe2xdx
Step 2: We can solve the integrals on the right side of the equation to get:
e2xy= 2 Ze2xydx +Zxe2xdx
e2xy= 2e2xy−Ze2xdy +1
2xe2x−1
2Ze2xdx
e2xy= 2e2xy−e2x+1
2xe2x−1
4e2x+C
Step 3: Now, applying the initial condition y(0) = 1, we get:
e0·1 = 2e0·1−e0+1
2·0·e0−1
4·e0+C
1 = 2 −1−1
4
1 = 1
4+C
C=3
4
Step 4: Substituting C=3
4back into our equation, we find:
e2xy= 2e2xy−e2x+1
2xe2x−1
4e2x+3
4
e2xy−2e2xy=−e2x+1
2xe2x−1
4e2x+3
4
−e2xy=−e2x+1
2xe2x−1
4e2x+3
4
y= 1 −1
2x+1
4+3
4e−2x
Therefore, the particular solution to the differential equation that satisfies
the initial condition y(0) = 1 is y= 1 −1
2x+1
4+3
4e−2x.
21
Question 22
Question
Consider the initial value problem:
dy
dx = 2x+ 3y, y(0) = 1.
Find the particular solution to the initial value problem.
Solution
Step 1: Rewrite the given initial value problem as:
dy
dx −3y= 2x, y(0) = 1.
Step 2: This is a first-order linear differential equation of the form dy
dx +
P(x)y=Q(x). The integrating factor is given by eRP(x)dx.
Step 3: In this case, P(x) = −3, so the integrating factor is eR−3dx =e−3x.
Step 4: Multiply both sides of the differential equation by the integrating
factor:
e−3xdy
dx −3e−3xy= 2xe−3x.
Step 5: Rewrite the equation as the derivative of a product:
d
dx(ye−3x)=2xe−3x.
Step 6: Integrate both sides with respect to x:
Zd
dx(ye−3x)dx =Z2xe−3xdx.
Step 7: This simplifies to:
ye−3x=−2
3xe−3x−2
9e−3x+C,
where Cis the constant of integration.
Step 8: Applying the initial condition y(0) = 1, we have:
1=0−2
9+C⇒C=11
9.
Step 9: Substitute C=11
9back into the equation to find the particular
solution:
ye−3x=−2
3xe−3x−2
9e−3x+11
9.
Thus, the particular solution to the initial value problem is y=−2
3x−2
9+
11
9e3x.
22
Question 23
Question
Solve the initial value problem:
y′= 3y−6, y(0) = 2
Solution
Step 1: First, let’s solve the differential equation y′= 3y−6 by separating
variables: dy
dt = 3y−6
dy
3y−6=dt
Step 2: Integrate both sides with respect to yand t:
Z1
3y−6dy =Zdt
1
3ln |3y−6|=t+C1
ln |3y−6|= 3t+C2
|3y−6|=e3t+C2
|3y−6|=Ce3t(where C=eC2)
3y−6 = Ce3tor 3y=Ce3t+ 6
y=1
3Ce3t+ 6
Step 3: Next, we apply the initial condition y(0) = 2 to find the value of C:
2 = 1
3C·e0+ 6
2 = 1
3(C+ 6)
6 = C+ 6
C= 0
Step 4: Now that we have C= 0, the solution to the initial value problem
is:
y=1
3·6=2e3t
23
Question 24
Question
Solve the initial value problem:
dy
dx =3x2
y, y(1) = 2
Solution
Step 1: Rewrite the differential equation in a separable form.
y dy = 3x2dx
Step 2: Integrate both sides with respect to their respective variables.
Zy dy =Z3x2dx
Step 3: Solve the integrals.
1
2y2=x3+C
where Cis the constant of integration.
Step 4: Apply the initial condition y(1) = 2 to find the value of C.
1
2(2)2= 13+C=⇒2 = 1 + C=⇒C= 1
Step 5: Substitute the value of Cback into the equation.
1
2y2=x3+ 1
Step 6: Solve for yin terms of x.
y2= 2x3+ 2 =⇒y=±p2x3+ 2
Therefore, the solution to the initial value problem is y=√2x3+ 2 since
y(1) = 2.
Question 25
Question
Solve the initial value problem:
dy
dx = 2x+y, y(0) = 1
24
Solution
Step 1: Write the given initial value problem as:
dy
dx −y= 2x, y(0) = 1
Step 2: This is a first-order linear differential equation. To solve it, we first
find the integrating factor µ(x) which is given by µ(x) = eR−1dx.
µ(x) = eR−1dx =e−x
Step 3: Multiply both sides of the differential equation by the integrating
factor µ(x):
e−xdy
dx −e−xy= 2xe−x
Step 4: Recognize the left side as the derivative of the product of e−xand y
(this step involves integrating factors):
d
dx(ye−x) = 2xe−x
Step 5: Integrate both sides with respect to x:
Zd
dx(ye−x)dx =Z2xe−xdx
Step 6: Integrate to get:
ye−x=−2xe−x−2e−x+C
Step 7: Solve for yby multiplying both sides by ex:
y=−2x−2 + Cex
Step 8: Use the initial condition y(0) = 1 to find C:
1 = −2(0) −2 + C
C= 3
Step 9: Substitute C= 3 back into the general solution to get the particular
solution:
y=−2x−2+3ex
Therefore, the solution to the initial value problem is y=−2x−2 + 3ex
with the initial condition y(0) = 1.
Question 26
Question
Solve the initial value problem:
y′= 2xy2, y(0) = 1.
25
Solution
Step 1: Let’s rewrite the differential equation in a form suitable for separation
of variables: dy
dx = 2xy2.
Step 2: Separate variables by dividing both sides by y2and multiplying by
dx:dy
y2= 2xdx.
Step 3: Integrate both sides:
Z1
y2dy = 2 Zxdx.
Step 4: Solving the integrals gives:
−1
y=x2+C,
where Cis the constant of integration.
Step 5: To find the value of the constant C, we use the initial condition
y(0) = 1:
−1
1= 02+C⇒C=−1.
Step 6: Substituting Cback into the equation gives:
−1
y=x2−1.
Step 7: Solve for y:
y=1
1−x2.
Therefore, the solution to the initial value problem is y=1
1−x2with the
initial condition y(0) = 1.
Question 27
Question
Solve the initial value problem:
dy
dx =2x
y2, y(0) = 1
26
Solution
Step 1: Separate variables by writing the given differential equation as:
y2dy = 2xdx
Step 2: Integrate both sides with respect to their respective variables:
Zy2dy =Z2xdx
Step 3: Integrate to solve for y:
y3
3=x2+C
where Cis the constant of integration.
Step 4: Applying the initial condition y(0) = 1 to solve for C:
13
3= 0 + C=⇒C=1
3
Step 5: Substitute the value of Cback into the equation to find the particular
solution: y3
3=x2+1
3=⇒y3= 3x2+ 1
Step 6: Solve for yby taking the cube root of both sides:
y=3
p3x2+ 1
Therefore, the solution to the initial value problem is y=3
√3x2+ 1 with
the initial condition y(0) = 1.
Question 28
Question
Consider the initial value problem:
dy
dx =2x
y, y(0) = 2.
Solution
Step 1: Rewrite the differential equation in a separable form by isolating vari-
ables xand y:
y dy = 2x dx.
Step 2: Integrate both sides of the equation:
Zy dy =Z2x dx.
27
Step 3: Solve the integrals:
y2
2=x2+C,
where Cis the constant of integration.
Step 4: Use the initial condition y(0) = 2 to find the value of the constant
C:22
2= 0 + C=⇒C= 2.
Step 5: Substitute the value of Cback into the equation:
y2
2=x2+ 2.
Step 6: Solve for yin terms of x:
y2= 2(x2+ 2) =⇒y=±p2x2+ 4.
Therefore, the solution to the initial value problem is y=√2x2+ 4 since
y(0) = 2.
Question 29
Question
Solve the initial value problem:
dy
dx =4x+ 3y−6
3x+ 2y−5, y(0) = 2
Solution
Step 1: First, we rewrite the given differential equation in terms of dy and dx
by multiplying both sides by dx:
dx ·dy
dx =4x+ 3y−6
3x+ 2y−5·dx
Step 2: To make it easier to solve, we rewrite the equation as follows:
3x+ 2y−5
4x+ 3y−6dy =dx
Step 3: We now integrate both sides, and find a constant of integration, C:
Z3x+ 2y−5
4x+ 3y−6dy =Zdx
28
Step 4: After integrating, we get:
1
2ln |4x+ 3y−6|=x+C
Step 5: We can rewrite the expression using exponential functions:
|4x+ 3y−6|=e2x+2C
Step 6: Since y(0) = 2, we substitute x= 0 and y= 2 into the equation:
|4(0) + 3(2) −6|=e2(0)+2C
Step 7: Simplifying, we get:
|6|=e2C
Step 8: Thus, 6 = e2Cor e2C= 6. Taking the natural logarithm of both
sides, we get:
2C= ln 6
Step 9: Solving for C, we have:
C=1
2ln 6
Step 10: Substitute the value of Cback into the equation we obtained:
|4x+ 3y−6|=e2x+ln 6
Step 11: Finally, we simplify the equation to get the solution to the initial
value problem:
4x+ 3y−6 = 6e2x
Question 30
Question
Consider the initial value problem:
dy
dx = 2x3+ 3x2+x−1, y(0) = 5
Use Euler’s method with step size h= 0.1 to approximate the value of y(0.2).
29
Solution
Step 1: Define the step size h, the initial condition, and the function f(x, y).
Step size: h= 0.1
Initial condition: y(0) = 5
Function f(x, y)=2x3+ 3x2+x−1
Step 2: Perform Euler’s method to approximate y(0.1), y(0.2), . . . until x=
0.2.
x1=x0+h= 0 + 0.1=0.1
y1=y0+hf(x0, y0)
= 5 + 0.1(2(0)3+ 3(0)2+ 0 −1)
= 5 −0.1 = 4.9
x2=x1+h= 0.1+0.1=0.2
y2=y1+hf(x1, y1)
= 4.9+0.1(2(0.1)3+ 3(0.1)2+ 0.1−1)
= 4.9+0.1(0.002 + 0.03 + 0.1−1)
= 4.9+0.1(−0.968)
= 4.9−0.0968
= 4.8032
Therefore, the approximate value of y(0.2) using Euler’s method with step
size h= 0.1 is y(0.2) ≈4.8032.
Question 31
Question
Solve the initial value problem:
dy
dx =2x
y, y(0) = 1
Solution
Step 1: Rewrite the differential equation in terms of yand dy:
y dy = 2x dx
Step 2: Integrate both sides:
Zy dy =Z2x dx
y2
2=x2+C1
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Step 3: Apply the initial condition y(0) = 1:
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2= 0 + C1
C1=1
2
Step 4: Substitute C1=1
2back into the equation:
y2
2=x2+1
2
y2= 2x2+ 1
Step 5: Solve for yby taking the square root:
y=±p2x2+ 1
Step 6: Apply the initial condition y(0) = 1 to determine the specific solu-
tion:
1 = p2(0)2+ 1
1 = √1
y= 1
Therefore, the solution to the initial value problem is y= 1.
Question 32
Question
Solve the initial value problem: dy
dx =y2+x2,y(0) = 1.
Solution
Step 1: Rewrite the equation as a separable first-order differential equation by
bringing all terms involving yto the left side and all terms involving xto the
right side. dy
dx =y2+x2
dy
y2=dx
Step 2: Integrate both sides with respect to yand xseparately.
Z1
y2dy =Zdx
−1
y=x+C1
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where C1is the constant of integration.
Step 3: Solve for yin terms of xby isolating y.
−1
y=x+C1
−1 = y(x+C1)
y=1
1−x−C1
Step 4: Use the initial condition y(0) = 1 to find the value of the constant
C1.
1 = 1
1−0−C1
1 = 1
1−C1
1−C1= 1
C1= 0
Step 5: Substitute C1= 0 back into the expression for yto obtain the
particular solution.
y=1
1−x
Therefore, the solution to the initial value problem is y=1
1−x.
Question 33
Question
Solve the initial value problem:
dy
dx = 2x(1 −y), y(0) = 0
Solution
Step 1: First, rewrite the differential equation in a separable form by moving
all terms involving yto one side and all terms involving xto the other side:
dy
1−y= 2x dx
Step 2: Integrate both sides with respect to their respective variables:
Z1
1−ydy =Z2x dx
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Step 3: Solve the integrals:
−ln |1−y|=x2+C1
where C1is the constant of integration.
Step 4: Exponentiate both sides to eliminate the natural logarithm:
|1−y|=e−x2−C1=Ce−x2
where C=±e−C1.
Step 5: Rewrite the equation without absolute values:
1−y=Ce−x2
or
y= 1 −Ce−x2
Step 6: Apply the initial condition y(0) = 0 to solve for the constant C:
0=1−Ce−0
C= 1
Step 7: Substitute C= 1 back into the solution:
y= 1 −e−x2
Therefore, the solution to the initial value problem is y= 1 −e−x2.
Question 34
Question
Solve the initial value problem:
dy
dx =1
x+y, y(1) = 1
Solution
Step 1: Rewrite the differential equation in standard form by separating vari-
ables:
(1 + y)dy =dx
Step 2: Integrate both sides with respect to their respective variables:
Z(1 + y)dy =Zdx
Z1dy +Zy dy =Zdx
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y+y2
2=x+C
Step 3: Use the initial condition y(1) = 1 to find the value of the constant
C:
1 + 1
2= 1 + C
C=1
2
Step 4: Substitute C=1
2back into the equation to find the particular
solution:
y+y2
2=x+1
2
Step 5: Use the quadratic formula to solve for y:
y2+ 2y−2x−1 = 0
y=−2±p22−4·1·(−2x−1)
2·1
y=−2±√4+8x+ 4
2
y=−2±√8x+ 8
2
y=−1±√2x+ 2
Step 6: Use the initial condition y(1) = 1 to determine the specific solution:
1 = −1±√2·1+2
2 = √4
Since √4=2>1,we choose the positive sign:
y=−1 + √2x+ 2
Therefore, the solution to the initial value problem is y=−1 + √2x+ 2.
Question 35
Question
Consider the initial value problem:
dy
dx = 2y, y(0) = 3
Find the solution to this initial value problem.
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Solution
Step 1: We start by separating variables in the differential equation.
dy
dx = 2y⇒dy
y= 2dx
Step 2: Next, we integrate both sides.
Z1
ydy =Z2dx ⇒ln |y|= 2x+C
Step 3: Apply the initial condition y(0) = 3 to solve for the constant C.
ln |3|= 2 ·0 + C⇒ln |3|=C
Step 4: Substitute back the value of Cinto the equation obtained in Step 2.
ln |y|= 2x+ ln |3|
Step 5: Solve for yby removing the natural logarithm.
|y|=e2x+ln |3|=e2x·3
Step 6: Consider both the positive and negative cases for y.
y=±3e2x
Therefore, the solution to the initial value problem is y= 3e2xor y=−3e2x.
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