MATH 334 - DIFFERENTIAL
EQUATIONS - Initial value problems
Question Bank - Set 1
Liberty University
Question 1
Question
Solve the initial value problem:
dy
dx =x2+y2, y(0) = 1
Solution
Step 1: First, rewrite the initial value problem as a separable first-order differ-
ential equation by moving all terms involving yto the left side:
dy
dx −y2=x2
Step 2: This separable differential equation can be solved by dividing both
sides by (x2+y2):
1
x2+y2
dy
dx −y2
x2+y2= 1
Step 3: Recognize that the left side can be rewritten using the chain rule for
derivatives as: d
dx arctan y
x= 1
Step 4: Integrate both sides with respect to xto solve for y:
arctan y
x=x+C
Step 5: Apply the initial condition y(0) = 1 to find the value of the constant
C:
arctan 1
0= 0 + C⇒π
2=C
Step 6: Substitute C=π
2back into the equation to find the particular
solution:
arctan y
x=x+π
2
Step 7: To express yexplicitly, take the tangent of both sides:
y
x= tanx+π
2
Step 8: Multiply by xto solve for y:
y=xtanx+π
2
Therefore, the solution to the initial value problem is y=xtanx+π
2.
Question 2
Question
Solve the initial value problem:
dy
dx =2x
y, y(0) = 1
Solution
Step 1: Rewrite the differential equation Step 2: Separate variables Step 3:
Integrate both sides Step 4: Solve for y by rearranging terms Step 5: Find the
particular solution using the initial condition
Step 1: Rewrite the differential equation The given differential equation is:
dy
dx =2x
y
Step 2: Separate variables Separating variables, we get:
y dy = 2x dx
Step 3: Integrate both sides Integrating both sides gives:
Zy dy =Z2x dx
Step 4: Solve for y by rearranging terms This simplifies to:
1
2y2=x2+C
where Cis the constant of integration.
2
Step 5: Find the particular solution using the initial condition Applying
the initial condition y(0) = 1, we have:
1
2(1)2= 0 + C
1
2=C
Therefore, the particular solution is:
1
2y2=x2+1
2
y2= 2x2+ 1
y=±p2x2+ 1
Hence, the solution to the initial value problem is y=±√2x2+ 1 with the
initial condition y(0) = 1.
Question 3
Question
Solve the initial value problem:
dy
dx = 2x+ 3y, y(0) = 1
Solution
Step 1: First, we write the differential equation in standard form: dy
dx −3y= 2x
Step 2: The integrating factor is given by eR−3dx =e−3x.
Step 3: Multiplying the integrating factor throughout gives:
e−3xdy
dx −3e−3xy= 2xe−3x
Step 4: Recognizing the left-hand side as a derivative, we can rewrite the
equation as: d
dx e−3xy= 2xe−3x
Step 5: Integrating both sides with respect to x, we get:
Zd
dx e−3xydx =Z2xe−3xdx
Step 6: Integrating the right-hand side gives:
e−3xy=−2
3xe−3x−2
9e−3x+C
3
Step 7: Multiplying through by e3xgives the general solution:
y=−2
3x−2
9+Ce3x
Step 8: Applying the initial condition y(0) = 1, we find C:
1 = −2
9+C⇒C=11
9
Step 9: Therefore, the particular solution to the initial value problem is:
y=−2
3x−2
9+11
9e3x
Question 4
Question
Consider the initial value problem:
dy
dx =e2x−y, y(0) = 1
Find the particular solution to the initial value problem.
Solution
Step 1: First, let’s rewrite the initial value problem in the form of a separable
differential equation: dy
dx +y=e2x
Step 2: Now, we multiply through by the integrating factor eR1dx =ex:
exdy
dx +exy=e3x
Step 3: Notice that the left side can be rewritten as the derivative of the
product exy:d
dx(exy) = e3x
Step 4: Integrating both sides with respect to x, we get:
exy=e3x
3+C
where Cis the constant of integration.
Step 5: Using the initial condition y(0) = 1, we find C:
e0·1 = e0
3+C=⇒1 = 1
3+C=⇒C=2
3
4
Step 6: Substituting back into our equation, we have:
exy=e3x
3+2
3
Step 7: Finally, we solve for y:
y=e3x
3ex+2
3=e2x
3+2
3
Therefore, the particular solution to the initial value problem is y=e2x
3+2
3.
Question 5
Question
Solve the initial value problem:
y′= 2y+ 3e2x, y(0) = 1
Solution
Step 1: First, we find the general solution to the homogeneous equation y′= 2y.
The homogeneous equation can be solved by separating variables:
dy
dx = 2y
1
ydy = 2dx
Integrating both sides:
ln |y|= 2x+C1
where C1is the constant of integration.
Step 2: Exponentiating both sides to solve for y:
y=e2x+C1=Ce2x
where C=eC1.
Step 3: Now, we find the particular solution to the non-homogeneous equa-
tion y′= 2y+ 3e2xusing the method of undetermined coefficients. Assume
the particular solution has the form yp=Ae2x. Substitute ypinto the non-
homogeneous equation and solve for A:
2Ae2x= 2(Ae2x)+3e2x
A=3
2
5
Step 4: The general solution to the non-homogeneous equation is the sum
of the general solution yh=Ce2xand the particular solution yp=3
2e2x:
y=Ce2x+3
2e2x= (C+3
2)e2x
Step 5: Applying the initial condition y(0) = 1:
1=(C+3
2)e0
1 = C+3
2
C=−1
2
Therefore, the solution to the initial value problem is:
y= (−1
2+3
2)e2x=e2x
Question 6
Question
Consider the initial value problem: y′= 2y+e−x,y(0) = 1. Find the solution
to this initial value problem.
Solution
Step 1: First, we find the integrating factor. The integrating factor µ(x) is given
by µ(x) = eR2dx =e2x.
Step 2: Next, we multiply both sides of the differential equation by the
integrating factor:
e2xy′= 2e2xy+e2xe−x
Step 3: The left side of the equation can be rewritten using the product rule
as: d
dx(e2xy)=2e2xy+ex
Step 4: Integrating both sides with respect to x, we get:
e2xy=Z(2e2xy+ex)dx
Step 5: Solving the integral on the right side, we get:
e2xy=e2xy+ex+C
6
Step 6: Next, we apply the initial condition y(0) = 1:
e0·1 = e0·1 + e0+C
1 = 1 + 1 + C
C=−1
Step 7: Substitute C=−1 back into the equation:
e2xy=e2xy+ex−1
Step 8: Simplifying, we get the solution to the initial value problem:
y=e−x−1
Question 7
Question
Consider the initial value problem:
dy
dx =2x
1 + y2, y(0) = 1
Solve the initial value problem and determine the interval on which the
solution is defined.
Solution
Step 1: Rewrite the initial value problem as:
dy
dx =2x
1 + y2
Step 2: Separate variables:
(1 + y2)dy = 2x dx
Step 3: Integrate both sides:
Z(1 + y2)dy =Z2x dx
Step 4: Simplify the integrals:
y+y3
3=x2+C
Step 5: Apply the initial condition y(0) = 1:
1 + 1
3= 0 + C=⇒C=4
3
7
Step 6: Substitute Cback into the general solution:
y+y3
3=x2+4
3
Step 7: Determine the interval for the solution: Since ymust be real-valued,
1 + y2>0⇒y2>−1⇒y∈R. Therefore, the solution is defined for all real
numbers y.
Step 8: Thus, the solution to the initial value problem is:
y+y3
3=x2+4
3
Question 8
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: First, we rewrite the initial value problem as a separable differential
equation: dy
dx −3y= 2x
Step 2: To solve the separable differential equation, we start by multiplying
both sides by the integrating factor e−3x:
e−3xdy
dx −3e−3xy= 2xe−3x
Step 3: Recognizing that the left side is the derivative of the product e−3xy,
we can rewrite the equation as:
d
dx(e−3xy) = 2xe−3x
Step 4: Integrating both sides with respect to x, we get:
e−3xy=−2
3xe−3x+C
where Cis the constant of integration.
Step 5: To find the particular solution, we use the initial condition y(0) = 1:
e0·1 = −2
3·0·e0+C⇒C= 1
8
Step 6: Substitute C= 1 back into the equation to find the particular
solution:
e−3xy=−2
3xe−3x+ 1
Step 7: Finally, solving for y, we have:
y=−2
3x+e3x
Therefore, the particular solution to the initial value problem is y=−2
3x+
e3x.
Question 9
Question
Solve the initial value problem:
dy
dx + 2y=e−2x, y(0) = 1
Solution
Step 1: First, we find the integrating factor. The integrating factor I(x) is given
by I(x) = eR2dx =e2x.
Step 2: Next, we multiply both sides of the differential equation by the
integrating factor:
e2xdy
dx + 2e2xy=e2xe−2x
Step 3: Rewrite the equation in terms of the product rule:
d
dx ye2x= 1
Step 4: Integrate both sides with respect to x:
Zd
dx ye2xdx =Z1dx
Step 5: Simplify the integrals and constant of integration:
ye2x=x+C
Step 6: To find the value of the constant C, we use the initial condition
y(0) = 1:
1 = 0 + C⇒C= 1
Step 7: Thus, the solution to the initial value problem is:
y=x+ 1
e2x
9
Question 10
Question
Consider the initial value problem:
dy
dx = 2y+x2, y(0) = 1
Solution
To solve this initial value problem, we will use the method of integrating factors.
Step 1: Find the integrating factor The integrating factor is given by
µ(x) = eR2dx =e2x.
Step 2: Multiply the differential equation by the integrating factor
Multiplying both sides of the differential equation by µ(x), we get:
e2xdy
dx = 2e2xy+x2e2x
Step 3: Integrate both sides Integrating both sides with respect to x,
we have: Ze2xdy
dx dx =Z(2e2xy+x2e2x)dx
Step 4: Solve the integrals Integrating the left side with respect to x, we
get:
e2xy= 2 Zxe2xdx +Zx2e2xdx
Step 5: Solve the integrals on the right side Integrating Rxe2xdx by
parts, we find:
Zxe2xdx =1
2xe2x−1
4e2x
Integrating Rx2e2xdx by parts twice, we find:
Zx2e2xdx =1
2x2e2x−xe2x+1
2e2x
Step 6: Substitute back and solve for ySubstitute the results back
into the equation e2xy= 2 Rxe2xdx +Rx2e2xdx and simplify to solve for y(x).
Remember to apply the initial condition y(0) = 1 to find the specific solution.
Question 11
Question
Solve the initial value problem:
dy
dx = 2y+x, y(0) = 1
10
Solution
Step 1: First, let’s solve the differential equation dy
dx = 2y+x. This is a first-order
linear differential equation in standard form dy
dx + (−2y) = x. The integrating
factor µ(x) is given by µ(x) = eR−2dx =e−2x.
Step 2: Multiply both sides of the differential equation by the integrating
factor µ(x):
e−2xdy
dx + (−2e−2xy) = xe−2x
Step 3: Rewrite the left side of the equation as the derivative of (ye−2x):
d
dx(ye−2x) = xe−2x
Step 4: Integrate both sides with respect to x:
Zd
dx(ye−2x)dx =Zxe−2xdx
Step 5: Integrate both sides to get the general solution:
ye−2x=Zxe−2xdx
Step 6: Solve the integral on the right side using integration by parts: Let
u=xand dv =e−2xdx. Then, du =dx and v=−1
2e−2x. The integral
becomes: Zxe−2xdx =−1
2xe−2x+1
2Ze−2xdx
Step 7: Integrate the remaining integral and substitute back into the general
solution: Zxe−2xdx =−1
2xe−2x+1
4e−2x+C
So, the general solution is:
ye−2x=−1
2xe−2x+1
4e−2x+C
Step 8: Apply the initial condition y(0) = 1 to find the value of C:
1 = −1
2(0)e0+1
4e0+C
1 = 1
4+C
C=3
4
Step 9: Substitute C=3
4back into the general solution:
ye−2x=−1
2xe−2x+1
4e−2x+3
4
11
Step 10: Finally, solve for yto get the particular solution:
y=−1
2x+1
4+ 3e2x
Therefore, the solution to the initial value problem is y=−1
2x+1
4+ 3e2x.
Question 12
Question
Solve the initial value problem: dy
dx = 2x−5ywith y(0) = 1.
Solution
Step 1: First, let’s rewrite the differential equation in the standard form of a
first-order linear differential equation:
dy
dx + 5y= 2x.
Step 2: To solve this differential equation, we will use an integrating factor.
The integrating factor is given by µ(x) = eR5dx =e5x.
Step 3: Multiplying both sides of the differential equation by the integrating
factor, we get:
e5xdy
dx + 5e5xy= 2xe5x.
Step 4: Recognizing the left side as the derivative of the product of yand
the integrating factor, we can rewrite the equation as:
d
dx(ye5x)=2xe5x.
Step 5: Integrating both sides with respect to x, we have:
Zd
dx(ye5x)dx =Z2xe5xdx.
Step 6: Simplifying the integrals gives:
ye5x=Z2xe5xdx +C,
where Cis the constant of integration.
Step 7: Solving the integral on the right side, we get:
ye5x= 2 Zxe5xdx +C.
12
Step 8: By using integration by parts, we find:
ye5x= 2 x
5e5x−1
5Ze5xdx+C.
Step 9: Further simplifying gives:
ye5x=2x
5e5x−2
25e5x+C.
Step 10: Finally, we can solve for yby dividing through by e5x:
y=2x
5−2
25 +Ce−5x.
Step 11: To determine the value of the constant C, we use the initial condi-
tion y(0) = 1:
1 = 2(0)
5−2
25 +Ce0.
Step 12: Solving for Cgives:
1 = −2
25 +C⇒C=27
25.
Therefore, the solution to the initial value problem is:
y=2x
5−2
25 +27
25e−5x.
Question 13
Question
Solve the initial value problem:
dy
dx =x2−y, y(0) = 1
Solution
We can solve this first-order ordinary differential equation using the method of
integrating factors.
Step 1: Find the integrating factor The integrating factor for the given
differential equation dy
dx =P(x)y+Q(x) is given by
µ(x) = eRP(x)dx
In this case, P(x) = −1, Q(x) = x2. Thus,
µ(x) = eR−1dx =e−x
13
Step 2: Multiply the given differential equation by the integrating
factor Multiply both sides of the differential equation dy
dx =x2−yby the
integrating factor e−x:
e−xdy
dx =e−x(x2−y)
This simplifies to d
dx(e−xy) = x2e−x
Step 3: Integrate both sides of the equation Integrating both sides
with respect to x gives
Zd
dx(e−xy)dx =Zx2e−xdx
Solving the left-hand side gives
e−xy=Zx2e−xdx
Step 4: Solve the integral The integral on the right-hand side can be
solved using integration by parts or by recognizing it as a product rule for
differentiation. Let’s use integration by parts:
u=x2, dv =e−xdx
du = 2x dx, v =−e−x
Then,
Zx2e−xdx =−x2e−x−Z(−2x)e−xdx =−x2e−x+ 2 Zxe−xdx
This integral can by solved by integrating xe−xby parts again.
Step 5: Finish solving the integral For the integral Rxe−xdx, let u=
x, dv =e−xdx:
du =dx, v =−e−x
Then,
Zxe−xdx =−xe−x−Z(−e−x)dx =−xe−x+Ze−xdx
Thus,
Zx2e−xdx =−x2e−x+ 2(−xe−x−e−x) + C
Step 6: Reintroduce y and solving for C Substitute this back into the
equation µ(x)y=Rx2e−xdx to get
e−xy=−x2e−x+ 2(−xe−x−e−x) + C
14
Applying the initial condition y(0) = 1, we find
e−0·1 = 0 + 2(0 −1) + C=⇒1 = −2 + C=⇒C= 3
Therefore, the solution to the initial value problem is
e−xy=−x2e−x+ 2(−xe−x−e−x)+3
Or equivalently,
y=−x2+ 2(−x−1) + 3ex
Question 14
Question
Solve the initial value problem:
y′+ 2y= 4xwith y(0) = 3
Solution
Step 1: First, we find the integrating factor by computing eR2dx.
Z2dx = 2x
e2x
Step 2: Multiply the given differential equation by the integrating factor
found in Step 1.
e2x(y′+ 2y) = e2x(4x)
(e2xy)′= 4xe2x
Step 3: Integrate both sides of the equation with respect to x.
Z(e2xy)′dx =Z4xe2xdx
e2xy= 2xe2x−Z2e2xdx
Step 4: Solve the integral on the right side.
e2xy= 2xe2x−2e2x+C
y= 2x−2 + Ce−2x
Step 5: Use the initial condition y(0) = 3 to solve for the constant C.
3=0−2 + Ce0
C= 5
Step 6: Substitute the value of Cback into the general solution to get the
particular solution.
y= 2x−2+5e−2x
Therefore, the solution to the initial value problem is y= 2x−2+5e−2x.
15
Question 15
Question
Solve the initial value problem:
dy
dx = 3x2−2x, y(0) = 4.
Solution
Step 1: First, we find the general solution to the differential equation by inte-
grating both sides with respect to x.
Zdy
dx dx =Z(3x2−2x)dx.
Step 2: Integrating, we get
y=x3−x2+C,
where Cis the constant of integration.
Step 3: To find the specific solution, we use the initial condition y(0) = 4.
4=0−0 + C.
Step 4: Solving for C, we get C= 4.
Step 5: Therefore, the solution to the initial value problem is
y=x3−x2+ 4.
Question 16
Question
Solve the initial value problem:
dy
dx = 3x2−y, y(0) = 1
Solution
Step 1: We notice that the given differential equation is a first-order linear
homogeneous ordinary differential equation. We can rewrite it in standard form
as: dy
dx +y= 3x2
Step 2: To solve this equation, we will use an integrating factor which is
given by eR1dx =ex.
16
Step 3: Multiply both sides of the differential equation by the integrating
factor:
exdy
dx +exy= 3x2ex
Step 4: Notice that the left-hand side can be rewritten using the product
rule as: d
dx(exy)=3x2ex
Step 5: Integrate both sides with respect to x:
Zd
dx(exy)dx =Z3x2exdx
Step 6: Simplify the integrals:
exy=Z3x2exdx
Step 7: Solve the integral on the right-hand side:
exy= 3 Zx2exdx
Step 8: Use integration by parts to find the integral: Let u=x2and dv =
exdx. Then, du = 2xdx and v=ex.
exy= 3(uv −Zvdu)
Step 9: Substitute back the expressions of uand v:
exy= 3(x2ex−Z2xexdx)
Step 10: Solve the remaining integral:
exy= 3(x2ex−2Zxexdx)
Step 11: Use integration by parts again to find the remaining integral: Let
u=xand dv =exdx. Then, du =dx and v=ex.
exy= 3(x2ex−2(uv −Zvdu))
Step 12: Simplify the equation:
exy= 3(x2ex−2(xex−Zexdx))
Step 13: Solve the remaining integral:
exy= 3(x2ex−2(xex−ex) + C)
17
Step 14: Simplify the right-hand side:
exy= 3(x2ex−2xex+ 2ex+C)
Step 15: Divide by exto solve for y:
y= 3x2−2x+2+Ce−x
Step 16: Use the initial condition y(0) = 1 to find the value of C:
1 = 3(02)−2(0) + 2 + C
C=−1
Step 17: Finally, substitute C=−1 back into the general solution to obtain
the particular solution:
y= 3x2−2x+ 1 −e−x
Therefore, the solution to the initial value problem is y= 3x2−2x+ 1 −e−x.
Question 17
Question
Consider the initial value problem:
dy
dx =2xy
x2+y2, y(0) = 1
Solution
Step 1: We rewrite the given differential equation in terms of differentials:
dy
dx =2xy
x2+y2.
Step 2: We separate variables by multiplying both sides by (x2+y2):
(x2+y2)dy = 2xydx.
Step 3: We integrate both sides:
Z(x2+y2)dy =Z2xydx.
Step 4: Integrating the left side with respect to yand the right side with
respect to xgives:
x2y+1
3y3=x2y+C,
where Cis the constant of integration.
18
Step 5: Simplifying, we find:
1
3y3=C.
Step 6: Applying the initial condition y(0) = 1, we find:
1
3=C.
Step 7: Therefore, the particular solution to the initial value problem is
given by: 1
3y3=1
3.
Step 8: Solving for y, we get:
y3= 1.
Thus, the solution to the initial value problem is y= 1.
Question 18
Question
Solve the initial value problem:
y′= 2√y, y(0) = 1.
Solution
Step 1: Rewrite the differential equation in a more manageable form by sepa-
rating variables: dy
dx = 2√y.
Step 2: Divide both sides by √yand multiply both sides by dx to separate
variables: dy
√y= 2dx.
Step 3: Integrate both sides with respect to their respective variables:
Z1
√ydy =Z2dx.
Step 4: Simplify the integrals:
2√y= 2x+C,
where Cis the constant of integration.
19
Step 5: Apply the initial condition y(0) = 1:
2√1 = 2(0) + C.
2 = C.
Step 6: Substitute C= 2 back into the equation from Step 4:
2√y= 2x+ 2.
Step 7: Solve for y:√y=x+ 1.
y= (x+ 1)2.
Therefore, the solution to the initial value problem is y= (x+ 1)2.
Question 19
Question
Consider the initial value problem:
dy
dx =2x+y
2y, y(0) = 1
Solve the initial value problem using the method of separation of variables.
Solution
Step 1: Rewrite the differential equation by separating variables:
2y dy = (2x+y)dx
Step 2: Integrate both sides with respect to their respective variables:
Z2y dy =Z(2x+y)dx
Step 3: Perform the integrations:
y2=x2+xy +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 = 0 + 0 + C=⇒C= 1
Step 5: Substitute the value of Cback into the equation:
y2=x2+xy + 1
20
Step 6: Solve for yin terms of x:
y=−x±√x2+ 4
2
Therefore, the solution to the initial value problem is y=−x±√x2+4
2, where
y(0) = 1.
Question 20
Question
Solve the initial value problem:
dy
dx =2x+y
x+ 2y, y(0) = 1
Solution
Step 1: First, rewrite the given differential equation in a standard form. Let
v=y/x. Then, y=vx, and dy
dx =xdv
dx +v. Substituting these into the
differential equation gives:
xdv
dx +v=2x+vx
x+ 2vx
Step 2: Simplify the equation by multiplying both sides by (x+ 2vx) to get
rid of the denominator:
x(x+ 2vx)dv
dx + (x+ 2vx)v= 2x+vx
Step 3: Expand the left side of the equation and simplify:
x2dv
dx + 2xv2+xv = 2x+vx
Step 4: Rearrange the equation and divide by x:
x2dv
dx + (v−2)v= 2 −v
Step 5: This is now a separable differential equation. Separate the variables
and integrate to solve for v:
dv
(v−2)v=2−v
x2dx
Z1
v−2−1
vdv =Z2−v
x2dx
21
ln |v−2| − ln |v|=−1
x+C
Step 6: Combine the logarithms and simplify:
ln
v−2
v
=−1
x+C
ln |1−2/v|=−1
x+C
1−2
v=Ce−1/x
Step 7: Reintroduce yby substituting v=y/x and solve for y:
1−2x
y=Ce−1/x
y=2x
1−Ce−1/x
Step 8: Apply the initial condition y(0) = 1 to find the value of C:
1 = 2·0
1−C
1=0
Step 9: Since the initial condition leads to a contradiction, there must have
been a mistake made in the solution process. Double-check the steps to find
and correct the error.
Question 21
Question
Consider the initial value problem
dy
dx = 3y+ex, y(0) = 1.
Find the solution to the initial value problem.
Solution
Step 1: First, we find the general solution to the corresponding homogeneous
equation dy
dx = 3y. Since this is a separable differential equation, we can rewrite
it as dy
y= 3dx and integrate both sides:
Z1
ydy =Z3dx.
22
This gives us ln |y|= 3x+C1, where C1is the constant of integration. Expo-
nentiating both sides, we obtain |y|=e3x+C1=Ce3x, where C=±eC1.
Step 2: Now, we consider the non-homogeneous part of the equation. Be-
cause y=Ce3xis also a solution to dy
dx = 3y, we seek a particular solution to
the non-homogeneous equation in the form yp=Axex, where Ais a constant
we need to find. Computing y′
pand plugging it into the differential equation,
we get
Axex= 3(Axex) + ex.
This simplifies to A=1
3.
Step 3: The general solution to the non-homogeneous equation is the sum of
the general solution to the homogeneous equation and the particular solution:
y=Ce3x+1
3xex.
Now we use the initial condition y(0) = 1 to find C:
1 = C+ 0 ⇒C= 1.
Therefore, the solution to the initial value problem is
y=e3x+1
3xex.
Question 22
Question
Solve the initial value problem: dy
dx = 2y−x, with y(0) = 3.
Solution
Step 1: First, let’s rewrite the differential equation as dy
dx −2y=−x.
Step 2: Next, we find the integrating factor by computing eR−2dx =e−2x.
Step 3: Multiply the integrating factor by both sides of the differential equa-
tion to get:
e−2xdy
dx −2e−2xy=−xe−2x
.
Step 4: Recognize that the left side is the derivative of the product e−2xy,
which simplifies to: d
dx(e−2xy) = −xe−2x
.
Step 5: Integrate the equation with respect to xto get:
e−2xy=Z−xe−2xdx
23
.
Step 6: Solve the integral to obtain:
e−2xy=1
4xe−2x+C
.
Step 7: Now, apply the initial condition y(0) = 3 to find the value of the
constant C:
e0·3 = 1
4·0 + C
.
3 = C
.
Step 8: Substitute C= 3 back into the equation to find the particular
solution:
e−2xy=1
4xe−2x+ 3
.
y=1
4x+ 3e2x
.
Hence, the solution to the initial value problem is y=1
4x+ 3e2x.
Question 23
Question
Solve the initial value problem:
dy
dx + 2y=e−x, y(0) = 1
Solution
Step 1: First, we find the integrating factor by solving the auxiliary equation
v′+ 2v= 0.
r+ 2 = 0 =⇒r=−2
So, the integrating factor is v(x) = e−2x.
Step 2: We multiply both sides of the given differential equation by the
integrating factor v(x).
e−2xdy
dx + 2e−2xy=e−3x
Step 3: Rewrite the equation in terms of the product rule for differentiation.
d
dx(e−2xy) = e−3x
24
Step 4: Integrate both sides with respect to x.
Zd
dx(e−2xy)dx =Ze−3xdx
e−2xy=−1
3e−3x+C
where Cis the constant of integration.
Step 5: Solve for yby multiplying through by e2x.
y=−1
3e−x+Ce2x
Step 6: Use the initial condition y(0) = 1 to solve for C.
1 = −1
3−C=⇒C=4
3
Step 7: Substitute C=4
3back into the general solution to get the particular
solution.
y=−1
3e−x+4
3e2x
Therefore, the solution to the initial value problem is y=−1
3e−x+4
3e2x.
Question 24
Question
Solve the initial value problem
y′= sin(2x) + y2, y(0) = 1.
Solution
To solve the initial value problem, we will use the method of integrating factors.
Step 1: Rewrite the initial value problem in the form y′+P(x)y=Q(x).
The given initial value problem can be rewritten as
y′−y2= sin(2x).
Step 2: Find the integrating factor µ(x) = eRP(x)dx. Here, P(x) = −1.
Thus, the integrating factor is
µ(x) = eR−1dx =e−x.
Step 3: Multiply the integrating factor by the differential equation. Multi-
plying both sides of the differential equation by µ(x), we get
e−xy′−e−xy2=e−xsin(2x).
25
Step 4: Rewrite the left side as the derivative of a product. The left side
can be rewritten using the product rule as
(e−xy)′=e−xsin(2x).
Step 5: Integrate both sides with respect to x. Integrating both sides, we
have Z(e−xy)′dx =Ze−xsin(2x)dx.
Integrating, we get
e−xy=Ze−xsin(2x)dx.
Step 6: Solve for y. To find y, we solve the integral on the right side and
then solve for y.
e−xy=−1
5cos(2x) + C,
where Cis the constant of integration.
Thus, the general solution is
y(x) = −1
5excos(2x) + Cex.
Step 7: Use the initial condition to find the value of C. Applying the initial
condition y(0) = 1, we get
1 = −1
5cos(0) + C.
This simplifies to 1 = −1
5+C, so C=6
5.
Step 8: Write the final solution. Therefore, the solution to the initial value
problem is
y(x) = −1
5excos(2x) + 6
5ex.
Question 25
Question
Solve the initial value problem:
dy
dx =y
x+ex, y(1) = 2
26
Solution
Step 1: Rewrite the differential equation in standard form.
dy
dx −y
x=ex
Step 2: Identify the integrating factor µ(x), which is given by µ(x) = eR−1
xdx.
µ(x) = eR−1
xdx =e−ln |x|=1
x
Step 3: Multiply the integrating factor µ(x) throughout the standard form
equation. 1
x
dy
dx −1
x2y=ex·1
x
Step 4: Rewrite the left-hand side of the equation using the product rule,
and simplify the right-hand side.
d
dx y
x=ex·1
x
Step 5: Integrate both sides with respect to x.
Zd
dx y
xdx =Zex·1
xdx
Step 6: Solve the integrals.
y
x=Zex·1
xdx =Z1
xexdx =ex+C
Step 7: Multiply by xto solve for y.
y=x(ex+C)
Step 8: Use the initial condition y(1) = 2 to solve for C.
2 = 1(e1+C) =⇒C= 2 −e
Step 9: Substitute Cback into the equation to obtain the final solution.
y=x(ex+ 2 −e)
Question 26
Question
Solve the initial value problem:
y′=y
x+xy, y(1) = 1.
27
Solution
Step 1: This is a first-order linear differential equation in the form y′−P(x)y=
Q(x). We can rewrite the given equation as:
y′−1
xy=x2.
Step 2: To solve the homogeneous part of the equation, we first find the
integrating factor I(x) = eR−1
xdx =e−ln x=1
x.
Step 3: Multiply the homogeneous equation by the integrating factor:
1
xy′−1
x2y=x
d
dx 1
xy=x
Zd
dx 1
xydx =Zxdx
1
xy=1
2x2+C,
where Cis an arbitrary constant.
Step 4: Now, we add the particular solution to the general solution:
y=x2+Cx
y(1) = 1 + C= 1
C= 0.
Therefore, the solution to the initial value problem is y=x2.
Question 27
Question
Solve the initial value problem: dy
dx =√y, with y(0) = 4.
Solution
Step 1: Rewrite the differential equation in a separable form:
dy
dx =√y
dy
√y=dx
Step 2: Integrate both sides:
Z1
√ydy =Zdx
28
2√y=x+C1
Step 3: Solve for yusing the initial condition y(0) = 4:
2√4 = 0 + C1
C1= 4
Step 4: Substitute C1= 4 back into the general solution:
2√y=x+ 4
√y=1
2x+ 2
Step 5: Solve for y:
y=1
2x+ 22
y=1
4x2+ 2x+ 4
So, the solution to the initial value problem is y=1
4x2+ 2x+ 4.
Question 28
Question
Solve the initial value problem:
y′+y= sin(x), y(0) = 1
Solution
Step 1: First, we need to solve the homogeneous differential equation y′+y= 0.
The characteristic equation is r+ 1 = 0, so r=−1. Therefore, the general
solution to the homogeneous equation is yh(x) = c1e−x, where c1is a constant
parameter.
Step 2: Next, we find a particular solution to the non-homogeneous equation
y′+y= sin(x). Since the right-hand side is sin(x), we try a particular solution
of the form yp(x) = Asin(x)+Bcos(x). Taking the derivatives and substituting
back into the equation, we find that A= 0 and B=−1
2. Therefore, a particular
solution is yp(x) = −1
2cos(x).
Step 3: The general solution to the non-homogeneous equation is the sum of
the general solution to the homogeneous equation and the particular solution.
Thus, the general solution is:
y(x) = yh(x) + yp(x) = c1e−x−1
2cos(x)
29
Step 4: Finally, we find the value of the constant c1using the initial condition
y(0) = 1. Substitute x= 0 and y= 1 into the general solution:
1 = c1e0−1
2cos(0)
1 = c1−1
2
c1=3
2
Step 5: Therefore, the solution to the initial value problem is:
y(x) = 3
2e−x−1
2cos(x)
Question 29
Question
Solve the initial value problem:
y′+2
xy=xsin(x), y(1) = 2
Solution
Step 1: The given differential equation is a first-order linear differential equation
in standard form. To solve it, we’ll first find the integrating factor.
The integrating factor µ(x) is given by µ(x) = eR2
xdx.
Step 2: We find R2
xdx.
Z2
xdx = 2 Z1
xdx = 2 ln |x|+C
Step 3: Therefore, the integrating factor is µ(x) = e2 ln |x|=eln(x2)=x2.
Step 4: Now, we multiply the integrating factor by the given differential
equation to get:
x2y′+ 2xy =x3sin(x)
Step 5: We can rewrite this equation as the derivative of a product:
(x2y)′=x3sin(x)
Step 6: Integrating both sides with respect to xgives:
x2y=Zx3sin(x)dx
30
Step 7: Integrating Rx3sin(x)dx by parts:
Zx3sin(x)dx =−x3cos(x)+3Zx2cos(x)dx
Step 8: To evaluate Rx2cos(x)dx, we need to use integration by parts again.
Let u=x2and dv = cos(x)dx.
Step 9: Calculating du and vgives du = 2xdx and v= sin(x).
Step 10: Applying integration by parts:
Zx2cos(x)dx =x2sin(x)−Z2xsin(x)dx
Step 11: Integrating the remaining integral R2xsin(x)dx by parts once more,
we get:
Z2xsin(x)dx = 2xcos(x)−Z2 cos(x)dx
Step 12: Integrating R2 cos(x)dx = 2 sin(x), we substitute this back into the
previous result to get:
Zx2cos(x)dx =x2sin(x)−2xcos(x) + 2 sin(x)
Step 13: Substituting back into our integral equation, we have:
Zx3sin(x)dx =−x3cos(x) + 3(x2sin(x)−2xcos(x) + 2 sin(x))
Step 14: Therefore, the solution to the initial value problem y′+2
xy=
xsin(x), y(1) = 2 is:
y(x) = 1
x2−x3cos(x) + 3(x2sin(x)−2xcos(x) + 2 sin(x))
Question 30
Question
Solve the initial value problem:
dy
dx =2x
1 + y, y(0) = 1
Solution
Step 1: Rewrite the initial value problem and separate variables.
The initial value problem can be rewritten as:
(1 + y)dy = 2x dx
31
Now, we can separate variables:
(1 + y)dy = 2x dx
Step 2: Integrate both sides.
Integrating both sides gives us:
Z(1 + y)dy =Z2x dx
Which simplifies to:
Zdy +Zy dy =Z2x dx
This further simplifies to:
y+y2
2=x2+C
Step 3: Apply initial condition to find the value of the constant.
Using the initial condition y(0) = 1, we have:
1 + 1
2= 02+C=⇒1 + 1
2=C=⇒C=3
2
Step 4: Substitute the value of Cback into the equation and solve for y.
Substitute C=3
2back into the equation gives us:
y+y2
2=x2+3
2
Plugging in the initial condition y(0) = 1, we have:
1 + 1
2= 02+3
2=⇒3
2=3
2
Therefore, the solution to the initial value problem is y=x2+3
2.
Question 31
Question
Consider the initial value problem:
dy
dx =x2+y2, y(0) = 1.
Find the particular solution to the initial value problem on the interval
[−1,1].
32
Solution
Step 1: Note that this is a first-order ordinary differential equation. We can
rewrite it as: dy
dx =x2+y2.
Step 2: This is not a separable equation, so let’s try to use an integrating
factor to solve it. Multiplying both sides by e−y2, we get:
e−y2dy
dx =e−y2x2+ 1.
Step 3: We can rewrite the left-hand side as:
d
dx(e−y2) = e−y2dy
dx −2ye−y2.
Step 4: Substituting this back into the equation and integrating both sides
with respect to x, we get:
Zd
dx(e−y2)dx =Z(e−y2x2+ 1) dx.
Step 5: Integrating both sides gives:
e−y2=1
3x3+x+C,
where Cis the constant of integration.
Step 6: To find the value of C, we use the initial condition y(0) = 1. Sub-
stituting x= 0 and y= 1 into the equation above, we get:
e−1=C.
Step 7: So, the particular solution to the initial value problem is:
e−y2=1
3x3+x+e−1.
Question 32
Question
Consider the initial value problem:
dy
dx =4
x(2 −y), y(1) = 3.
Find the particular solution to the initial value problem.
33
Solution
Step 1: Rewrite the differential equation in the form dy
dx +P(x)y=Q(x) by
expanding and simplifying.
dy
dx =4
x(2 −y)
dy
dx =8
x−4
xy
dy
dx +4
xy=8
x
Step 2: Now, we will solve the differential equation using an integrating
factor µ(x) = eR4
xdx.
µ(x) = eR4
xdx
µ(x) = e4 ln x
µ(x) = x4
Step 3: Multiply the integrating factor µ(x) to the entire differential equation
and rewrite it in a different form.
x4dy
dx + 4x3y= 8x3
Step 4: Recognize the left-hand side as the derivative of the product of x4y.
d
dx(x4y)=8x3
x4y=Z8x3dx
x4y= 2x4+C
Step 5: Apply the initial condition y(1) = 3 to find the particular solution.
x4y= 2x4+C
14·3=2·14+C
3 = 2 + C
C= 1
So, the particular solution to the initial value problem is y= 2 + 1
x4.
Question 33
Question
Solve the initial value problem:
y′+y=e−t, y(0) = 1
34
Solution
Step 1: First, we find the integrating factor by solving the differential equation
u′(t) = u(t)p(t), where p(t) is the coefficient of y.
u′(t) = u(t)·1
This is a separable differential equation with general solution u(t) = Cet, where
Cis a constant to be determined.
Step 2: Multiplying the given differential equation by the integrating factor
u(t):
ety′+ety=et·e−t
ety′+ety= 1
Step 3: Recognize the left-hand side of the equation as the result of the
product rule for differentiation of yet:
(yet)′= 1
Step 4: Integrate both sides of the equation:
Z(yet)′dt =Z1dt
yet=t+C
Step 5: Apply the initial condition y(0) = 1:
1 = 0 + C
C= 1
Step 6: Therefore, the solution to the initial value problem is:
yet=t+ 1
y=t+ 1
et
Question 34
Question
Consider the initial value problem:
dy
dx =x2+y, y(0) = 1
Find the particular solution to the given initial value problem.
35
Solution
Step 1: First, we rearrange the given differential equation into a separable form.
This involves isolating the variables xand yon opposite sides of the equation.
Thus, we have: dy
dx =x2+y=⇒dy
dx −y=x2
Step 2: Next, we solve this separable first-order linear differential equation
using the integrating factor method. The integrating factor I(x) is given by
eR−1dx =e−x. Multiplying both sides of the equation by the integrating factor,
we get:
e−xdy
dx −e−xy=x2e−x
Step 3: Rewrite the left side of the equation as the derivative of the product
of the integrating factor and yby applying the product rule:
d
dx(e−xy) = x2e−x
Step 4: Integrate both sides with respect to xto solve for y:
Zd
dx(e−xy)dx =Zx2e−xdx
Step 5: Integrating both sides, we have:
e−xy=Zx2e−xdx
Step 6: Solve the integral on the right side:
Zx2e−xdx =−x2e−x−2Z−xe−xdx
Step 7: Continuing to solve the integral on the right side:
Zx2e−xdx =−x2e−x+ 2 Zxe−xdx
Step 8: The integral 2 Rxe−xdx can be solved using integration by parts. We
let u=xand dv =e−xdx, then du =dx and v=−e−x. Applying integration
by parts formula:
Zxe−xdx =−xe−x−Z−e−xdx
Step 9: Simplifying the result, we get:
Zxe−xdx =−xe−x+e−x
36
Step 10: Substitute the computed integral back into the equation and solve
for y:
e−xy=−x2e−x+ 2(−xe−x+e−x) + C
Step 11: Apply the initial condition y(0) = 1 to find the constant C. Plug-
ging in x= 0 and y= 1, we get:
1=0−0 + 2(0 + 1) + C=⇒C=−1
Step 12: Plug the value of Cback into the equation to obtain the particular
solution:
e−xy=−x2e−x+ 2(−xe−x+e−x)−1
Step 13: Simplify the expression to get the final particular solution:
y=−x2+ 2x−1 + ex
Question 35
Question
Solve the initial value problem:
dy
dx =2x−y
x+ 2y, y(0) = 1
Solution
Step 1: Rewrite the initial value problem in the standard form dy
dx =f(x, y):
dy
dx =2x−y
x+ 2y
Step 2: Check for homogeneity by substituting y=vx:
dy
dx =xdv
dx +v, 2x−y
x+ 2y=2−v
1+2v
Step 3: Let v=u+ 1 to get:
du
dx =−3u
1+2u
Step 4: Separate variables and integrate:
Z1+2u
−3udu =Zdx
x
Step 5: Solve the integrals:
ln |u| − 3
2ln |1+2u|= ln |x|+C
37
Step 5: Find the particular solution using the initial condition Applying
the initial condition y(0) = 1, we have:
1
2(1)2= 0 + C
1
2=C
Therefore, the particular solution is:
1
2y2=x2+1
2
y2= 2x2+ 1
y=±p2x2+ 1
Hence, the solution to the initial value problem is y=±√2x2+ 1 with the
initial condition y(0) = 1.
Question 3
Question
Solve the initial value problem:
dy
dx = 2x+ 3y, y(0) = 1
Solution
Step 1: First, we write the differential equation in standard form: dy
dx −3y= 2x
Step 2: The integrating factor is given by eR−3dx =e−3x.
Step 3: Multiplying the integrating factor throughout gives:
e−3xdy
dx −3e−3xy= 2xe−3x
Step 4: Recognizing the left-hand side as a derivative, we can rewrite the
equation as: d
dx e−3xy= 2xe−3x
Step 5: Integrating both sides with respect to x, we get:
Zd
dx e−3xydx =Z2xe−3xdx
Step 6: Integrating the right-hand side gives:
e−3xy=−2
3xe−3x−2
9e−3x+C
3
Step 7: Multiplying through by e3xgives the general solution:
y=−2
3x−2
9+Ce3x
Step 8: Applying the initial condition y(0) = 1, we find C:
1 = −2
9+C⇒C=11
9
Step 9: Therefore, the particular solution to the initial value problem is:
y=−2
3x−2
9+11
9e3x
Question 4
Question
Consider the initial value problem:
dy
dx =e2x−y, y(0) = 1
Find the particular solution to the initial value problem.
Solution
Step 1: First, let’s rewrite the initial value problem in the form of a separable
differential equation: dy
dx +y=e2x
Step 2: Now, we multiply through by the integrating factor eR1dx =ex:
exdy
dx +exy=e3x
Step 3: Notice that the left side can be rewritten as the derivative of the
product exy:d
dx(exy) = e3x
Step 4: Integrating both sides with respect to x, we get:
exy=e3x
3+C
where Cis the constant of integration.
Step 5: Using the initial condition y(0) = 1, we find C:
e0·1 = e0
3+C=⇒1 = 1
3+C=⇒C=2
3
4
Step 6: Substituting back into our equation, we have:
exy=e3x
3+2
3
Step 7: Finally, we solve for y:
y=e3x
3ex+2
3=e2x
3+2
3
Therefore, the particular solution to the initial value problem is y=e2x
3+2
3.
Question 5
Question
Solve the initial value problem:
y′= 2y+ 3e2x, y(0) = 1
Solution
Step 1: First, we find the general solution to the homogeneous equation y′= 2y.
The homogeneous equation can be solved by separating variables:
dy
dx = 2y
1
ydy = 2dx
Integrating both sides:
ln |y|= 2x+C1
where C1is the constant of integration.
Step 2: Exponentiating both sides to solve for y:
y=e2x+C1=Ce2x
where C=eC1.
Step 3: Now, we find the particular solution to the non-homogeneous equa-
tion y′= 2y+ 3e2xusing the method of undetermined coefficients. Assume
the particular solution has the form yp=Ae2x. Substitute ypinto the non-
homogeneous equation and solve for A:
2Ae2x= 2(Ae2x)+3e2x
A=3
2
5
Step 4: The general solution to the non-homogeneous equation is the sum
of the general solution yh=Ce2xand the particular solution yp=3
2e2x:
y=Ce2x+3
2e2x= (C+3
2)e2x
Step 5: Applying the initial condition y(0) = 1:
1=(C+3
2)e0
1 = C+3
2
C=−1
2
Therefore, the solution to the initial value problem is:
y= (−1
2+3
2)e2x=e2x
Question 6
Question
Consider the initial value problem: y′= 2y+e−x,y(0) = 1. Find the solution
to this initial value problem.
Solution
Step 1: First, we find the integrating factor. The integrating factor µ(x) is given
by µ(x) = eR2dx =e2x.
Step 2: Next, we multiply both sides of the differential equation by the
integrating factor:
e2xy′= 2e2xy+e2xe−x
Step 3: The left side of the equation can be rewritten using the product rule
as: d
dx(e2xy)=2e2xy+ex
Step 4: Integrating both sides with respect to x, we get:
e2xy=Z(2e2xy+ex)dx
Step 5: Solving the integral on the right side, we get:
e2xy=e2xy+ex+C
6
Step 6: Next, we apply the initial condition y(0) = 1:
e0·1 = e0·1 + e0+C
1 = 1 + 1 + C
C=−1
Step 7: Substitute C=−1 back into the equation:
e2xy=e2xy+ex−1
Step 8: Simplifying, we get the solution to the initial value problem:
y=e−x−1
Question 7
Question
Consider the initial value problem:
dy
dx =2x
1 + y2, y(0) = 1
Solve the initial value problem and determine the interval on which the
solution is defined.
Solution
Step 1: Rewrite the initial value problem as:
dy
dx =2x
1 + y2
Step 2: Separate variables:
(1 + y2)dy = 2x dx
Step 3: Integrate both sides:
Z(1 + y2)dy =Z2x dx
Step 4: Simplify the integrals:
y+y3
3=x2+C
Step 5: Apply the initial condition y(0) = 1:
1 + 1
3= 0 + C=⇒C=4
3
7
Step 6: Substitute Cback into the general solution:
y+y3
3=x2+4
3
Step 7: Determine the interval for the solution: Since ymust be real-valued,
1 + y2>0⇒y2>−1⇒y∈R. Therefore, the solution is defined for all real
numbers y.
Step 8: Thus, the solution to the initial value problem is:
y+y3
3=x2+4
3
Question 8
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: First, we rewrite the initial value problem as a separable differential
equation: dy
dx −3y= 2x
Step 2: To solve the separable differential equation, we start by multiplying
both sides by the integrating factor e−3x:
e−3xdy
dx −3e−3xy= 2xe−3x
Step 3: Recognizing that the left side is the derivative of the product e−3xy,
we can rewrite the equation as:
d
dx(e−3xy) = 2xe−3x
Step 4: Integrating both sides with respect to x, we get:
e−3xy=−2
3xe−3x+C
where Cis the constant of integration.
Step 5: To find the particular solution, we use the initial condition y(0) = 1:
e0·1 = −2
3·0·e0+C⇒C= 1
8
Step 6: Substitute C= 1 back into the equation to find the particular
solution:
e−3xy=−2
3xe−3x+ 1
Step 7: Finally, solving for y, we have:
y=−2
3x+e3x
Therefore, the particular solution to the initial value problem is y=−2
3x+
e3x.
Question 9
Question
Solve the initial value problem:
dy
dx + 2y=e−2x, y(0) = 1
Solution
Step 1: First, we find the integrating factor. The integrating factor I(x) is given
by I(x) = eR2dx =e2x.
Step 2: Next, we multiply both sides of the differential equation by the
integrating factor:
e2xdy
dx + 2e2xy=e2xe−2x
Step 3: Rewrite the equation in terms of the product rule:
d
dx ye2x= 1
Step 4: Integrate both sides with respect to x:
Zd
dx ye2xdx =Z1dx
Step 5: Simplify the integrals and constant of integration:
ye2x=x+C
Step 6: To find the value of the constant C, we use the initial condition
y(0) = 1:
1 = 0 + C⇒C= 1
Step 7: Thus, the solution to the initial value problem is:
y=x+ 1
e2x
9
Question 10
Question
Consider the initial value problem:
dy
dx = 2y+x2, y(0) = 1
Solution
To solve this initial value problem, we will use the method of integrating factors.
Step 1: Find the integrating factor The integrating factor is given by
µ(x) = eR2dx =e2x.
Step 2: Multiply the differential equation by the integrating factor
Multiplying both sides of the differential equation by µ(x), we get:
e2xdy
dx = 2e2xy+x2e2x
Step 3: Integrate both sides Integrating both sides with respect to x,
we have: Ze2xdy
dx dx =Z(2e2xy+x2e2x)dx
Step 4: Solve the integrals Integrating the left side with respect to x, we
get:
e2xy= 2 Zxe2xdx +Zx2e2xdx
Step 5: Solve the integrals on the right side Integrating Rxe2xdx by
parts, we find:
Zxe2xdx =1
2xe2x−1
4e2x
Integrating Rx2e2xdx by parts twice, we find:
Zx2e2xdx =1
2x2e2x−xe2x+1
2e2x
Step 6: Substitute back and solve for ySubstitute the results back
into the equation e2xy= 2 Rxe2xdx +Rx2e2xdx and simplify to solve for y(x).
Remember to apply the initial condition y(0) = 1 to find the specific solution.
Question 11
Question
Solve the initial value problem:
dy
dx = 2y+x, y(0) = 1
10
Solution
Step 1: First, let’s solve the differential equation dy
dx = 2y+x. This is a first-order
linear differential equation in standard form dy
dx + (−2y) = x. The integrating
factor µ(x) is given by µ(x) = eR−2dx =e−2x.
Step 2: Multiply both sides of the differential equation by the integrating
factor µ(x):
e−2xdy
dx + (−2e−2xy) = xe−2x
Step 3: Rewrite the left side of the equation as the derivative of (ye−2x):
d
dx(ye−2x) = xe−2x
Step 4: Integrate both sides with respect to x:
Zd
dx(ye−2x)dx =Zxe−2xdx
Step 5: Integrate both sides to get the general solution:
ye−2x=Zxe−2xdx
Step 6: Solve the integral on the right side using integration by parts: Let
u=xand dv =e−2xdx. Then, du =dx and v=−1
2e−2x. The integral
becomes: Zxe−2xdx =−1
2xe−2x+1
2Ze−2xdx
Step 7: Integrate the remaining integral and substitute back into the general
solution: Zxe−2xdx =−1
2xe−2x+1
4e−2x+C
So, the general solution is:
ye−2x=−1
2xe−2x+1
4e−2x+C
Step 8: Apply the initial condition y(0) = 1 to find the value of C:
1 = −1
2(0)e0+1
4e0+C
1 = 1
4+C
C=3
4
Step 9: Substitute C=3
4back into the general solution:
ye−2x=−1
2xe−2x+1
4e−2x+3
4
11
Step 10: Finally, solve for yto get the particular solution:
y=−1
2x+1
4+ 3e2x
Therefore, the solution to the initial value problem is y=−1
2x+1
4+ 3e2x.
Question 12
Question
Solve the initial value problem: dy
dx = 2x−5ywith y(0) = 1.
Solution
Step 1: First, let’s rewrite the differential equation in the standard form of a
first-order linear differential equation:
dy
dx + 5y= 2x.
Step 2: To solve this differential equation, we will use an integrating factor.
The integrating factor is given by µ(x) = eR5dx =e5x.
Step 3: Multiplying both sides of the differential equation by the integrating
factor, we get:
e5xdy
dx + 5e5xy= 2xe5x.
Step 4: Recognizing the left side as the derivative of the product of yand
the integrating factor, we can rewrite the equation as:
d
dx(ye5x)=2xe5x.
Step 5: Integrating both sides with respect to x, we have:
Zd
dx(ye5x)dx =Z2xe5xdx.
Step 6: Simplifying the integrals gives:
ye5x=Z2xe5xdx +C,
where Cis the constant of integration.
Step 7: Solving the integral on the right side, we get:
ye5x= 2 Zxe5xdx +C.
12
Step 8: By using integration by parts, we find:
ye5x= 2 x
5e5x−1
5Ze5xdx+C.
Step 9: Further simplifying gives:
ye5x=2x
5e5x−2
25e5x+C.
Step 10: Finally, we can solve for yby dividing through by e5x:
y=2x
5−2
25 +Ce−5x.
Step 11: To determine the value of the constant C, we use the initial condi-
tion y(0) = 1:
1 = 2(0)
5−2
25 +Ce0.
Step 12: Solving for Cgives:
1 = −2
25 +C⇒C=27
25.
Therefore, the solution to the initial value problem is:
y=2x
5−2
25 +27
25e−5x.
Question 13
Question
Solve the initial value problem:
dy
dx =x2−y, y(0) = 1
Solution
We can solve this first-order ordinary differential equation using the method of
integrating factors.
Step 1: Find the integrating factor The integrating factor for the given
differential equation dy
dx =P(x)y+Q(x) is given by
µ(x) = eRP(x)dx
In this case, P(x) = −1, Q(x) = x2. Thus,
µ(x) = eR−1dx =e−x
13
Step 2: Multiply the given differential equation by the integrating
factor Multiply both sides of the differential equation dy
dx =x2−yby the
integrating factor e−x:
e−xdy
dx =e−x(x2−y)
This simplifies to d
dx(e−xy) = x2e−x
Step 3: Integrate both sides of the equation Integrating both sides
with respect to x gives
Zd
dx(e−xy)dx =Zx2e−xdx
Solving the left-hand side gives
e−xy=Zx2e−xdx
Step 4: Solve the integral The integral on the right-hand side can be
solved using integration by parts or by recognizing it as a product rule for
differentiation. Let’s use integration by parts:
u=x2, dv =e−xdx
du = 2x dx, v =−e−x
Then,
Zx2e−xdx =−x2e−x−Z(−2x)e−xdx =−x2e−x+ 2 Zxe−xdx
This integral can by solved by integrating xe−xby parts again.
Step 5: Finish solving the integral For the integral Rxe−xdx, let u=
x, dv =e−xdx:
du =dx, v =−e−x
Then,
Zxe−xdx =−xe−x−Z(−e−x)dx =−xe−x+Ze−xdx
Thus,
Zx2e−xdx =−x2e−x+ 2(−xe−x−e−x) + C
Step 6: Reintroduce y and solving for C Substitute this back into the
equation µ(x)y=Rx2e−xdx to get
e−xy=−x2e−x+ 2(−xe−x−e−x) + C
14
Applying the initial condition y(0) = 1, we find
e−0·1 = 0 + 2(0 −1) + C=⇒1 = −2 + C=⇒C= 3
Therefore, the solution to the initial value problem is
e−xy=−x2e−x+ 2(−xe−x−e−x)+3
Or equivalently,
y=−x2+ 2(−x−1) + 3ex
Question 14
Question
Solve the initial value problem:
y′+ 2y= 4xwith y(0) = 3
Solution
Step 1: First, we find the integrating factor by computing eR2dx.
Z2dx = 2x
e2x
Step 2: Multiply the given differential equation by the integrating factor
found in Step 1.
e2x(y′+ 2y) = e2x(4x)
(e2xy)′= 4xe2x
Step 3: Integrate both sides of the equation with respect to x.
Z(e2xy)′dx =Z4xe2xdx
e2xy= 2xe2x−Z2e2xdx
Step 4: Solve the integral on the right side.
e2xy= 2xe2x−2e2x+C
y= 2x−2 + Ce−2x
Step 5: Use the initial condition y(0) = 3 to solve for the constant C.
3=0−2 + Ce0
C= 5
Step 6: Substitute the value of Cback into the general solution to get the
particular solution.
y= 2x−2+5e−2x
Therefore, the solution to the initial value problem is y= 2x−2+5e−2x.
15
Question 15
Question
Solve the initial value problem:
dy
dx = 3x2−2x, y(0) = 4.
Solution
Step 1: First, we find the general solution to the differential equation by inte-
grating both sides with respect to x.
Zdy
dx dx =Z(3x2−2x)dx.
Step 2: Integrating, we get
y=x3−x2+C,
where Cis the constant of integration.
Step 3: To find the specific solution, we use the initial condition y(0) = 4.
4=0−0 + C.
Step 4: Solving for C, we get C= 4.
Step 5: Therefore, the solution to the initial value problem is
y=x3−x2+ 4.
Question 16
Question
Solve the initial value problem:
dy
dx = 3x2−y, y(0) = 1
Solution
Step 1: We notice that the given differential equation is a first-order linear
homogeneous ordinary differential equation. We can rewrite it in standard form
as: dy
dx +y= 3x2
Step 2: To solve this equation, we will use an integrating factor which is
given by eR1dx =ex.
16
Step 3: Multiply both sides of the differential equation by the integrating
factor:
exdy
dx +exy= 3x2ex
Step 4: Notice that the left-hand side can be rewritten using the product
rule as: d
dx(exy)=3x2ex
Step 5: Integrate both sides with respect to x:
Zd
dx(exy)dx =Z3x2exdx
Step 6: Simplify the integrals:
exy=Z3x2exdx
Step 7: Solve the integral on the right-hand side:
exy= 3 Zx2exdx
Step 8: Use integration by parts to find the integral: Let u=x2and dv =
exdx. Then, du = 2xdx and v=ex.
exy= 3(uv −Zvdu)
Step 9: Substitute back the expressions of uand v:
exy= 3(x2ex−Z2xexdx)
Step 10: Solve the remaining integral:
exy= 3(x2ex−2Zxexdx)
Step 11: Use integration by parts again to find the remaining integral: Let
u=xand dv =exdx. Then, du =dx and v=ex.
exy= 3(x2ex−2(uv −Zvdu))
Step 12: Simplify the equation:
exy= 3(x2ex−2(xex−Zexdx))
Step 13: Solve the remaining integral:
exy= 3(x2ex−2(xex−ex) + C)
17
Step 14: Simplify the right-hand side:
exy= 3(x2ex−2xex+ 2ex+C)
Step 15: Divide by exto solve for y:
y= 3x2−2x+2+Ce−x
Step 16: Use the initial condition y(0) = 1 to find the value of C:
1 = 3(02)−2(0) + 2 + C
C=−1
Step 17: Finally, substitute C=−1 back into the general solution to obtain
the particular solution:
y= 3x2−2x+ 1 −e−x
Therefore, the solution to the initial value problem is y= 3x2−2x+ 1 −e−x.
Question 17
Question
Consider the initial value problem:
dy
dx =2xy
x2+y2, y(0) = 1
Solution
Step 1: We rewrite the given differential equation in terms of differentials:
dy
dx =2xy
x2+y2.
Step 2: We separate variables by multiplying both sides by (x2+y2):
(x2+y2)dy = 2xydx.
Step 3: We integrate both sides:
Z(x2+y2)dy =Z2xydx.
Step 4: Integrating the left side with respect to yand the right side with
respect to xgives:
x2y+1
3y3=x2y+C,
where Cis the constant of integration.
18
Step 5: Simplifying, we find:
1
3y3=C.
Step 6: Applying the initial condition y(0) = 1, we find:
1
3=C.
Step 7: Therefore, the particular solution to the initial value problem is
given by: 1
3y3=1
3.
Step 8: Solving for y, we get:
y3= 1.
Thus, the solution to the initial value problem is y= 1.
Question 18
Question
Solve the initial value problem:
y′= 2√y, y(0) = 1.
Solution
Step 1: Rewrite the differential equation in a more manageable form by sepa-
rating variables: dy
dx = 2√y.
Step 2: Divide both sides by √yand multiply both sides by dx to separate
variables: dy
√y= 2dx.
Step 3: Integrate both sides with respect to their respective variables:
Z1
√ydy =Z2dx.
Step 4: Simplify the integrals:
2√y= 2x+C,
where Cis the constant of integration.
19
Step 5: Apply the initial condition y(0) = 1:
2√1 = 2(0) + C.
2 = C.
Step 6: Substitute C= 2 back into the equation from Step 4:
2√y= 2x+ 2.
Step 7: Solve for y:√y=x+ 1.
y= (x+ 1)2.
Therefore, the solution to the initial value problem is y= (x+ 1)2.
Question 19
Question
Consider the initial value problem:
dy
dx =2x+y
2y, y(0) = 1
Solve the initial value problem using the method of separation of variables.
Solution
Step 1: Rewrite the differential equation by separating variables:
2y dy = (2x+y)dx
Step 2: Integrate both sides with respect to their respective variables:
Z2y dy =Z(2x+y)dx
Step 3: Perform the integrations:
y2=x2+xy +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 = 0 + 0 + C=⇒C= 1
Step 5: Substitute the value of Cback into the equation:
y2=x2+xy + 1
20
Step 6: Solve for yin terms of x:
y=−x±√x2+ 4
2
Therefore, the solution to the initial value problem is y=−x±√x2+4
2, where
y(0) = 1.
Question 20
Question
Solve the initial value problem:
dy
dx =2x+y
x+ 2y, y(0) = 1
Solution
Step 1: First, rewrite the given differential equation in a standard form. Let
v=y/x. Then, y=vx, and dy
dx =xdv
dx +v. Substituting these into the
differential equation gives:
xdv
dx +v=2x+vx
x+ 2vx
Step 2: Simplify the equation by multiplying both sides by (x+ 2vx) to get
rid of the denominator:
x(x+ 2vx)dv
dx + (x+ 2vx)v= 2x+vx
Step 3: Expand the left side of the equation and simplify:
x2dv
dx + 2xv2+xv = 2x+vx
Step 4: Rearrange the equation and divide by x:
x2dv
dx + (v−2)v= 2 −v
Step 5: This is now a separable differential equation. Separate the variables
and integrate to solve for v:
dv
(v−2)v=2−v
x2dx
Z1
v−2−1
vdv =Z2−v
x2dx
21
ln |v−2| − ln |v|=−1
x+C
Step 6: Combine the logarithms and simplify:
ln
v−2
v
=−1
x+C
ln |1−2/v|=−1
x+C
1−2
v=Ce−1/x
Step 7: Reintroduce yby substituting v=y/x and solve for y:
1−2x
y=Ce−1/x
y=2x
1−Ce−1/x
Step 8: Apply the initial condition y(0) = 1 to find the value of C:
1 = 2·0
1−C
1=0
Step 9: Since the initial condition leads to a contradiction, there must have
been a mistake made in the solution process. Double-check the steps to find
and correct the error.
Question 21
Question
Consider the initial value problem
dy
dx = 3y+ex, y(0) = 1.
Find the solution to the initial value problem.
Solution
Step 1: First, we find the general solution to the corresponding homogeneous
equation dy
dx = 3y. Since this is a separable differential equation, we can rewrite
it as dy
y= 3dx and integrate both sides:
Z1
ydy =Z3dx.
22
This gives us ln |y|= 3x+C1, where C1is the constant of integration. Expo-
nentiating both sides, we obtain |y|=e3x+C1=Ce3x, where C=±eC1.
Step 2: Now, we consider the non-homogeneous part of the equation. Be-
cause y=Ce3xis also a solution to dy
dx = 3y, we seek a particular solution to
the non-homogeneous equation in the form yp=Axex, where Ais a constant
we need to find. Computing y′
pand plugging it into the differential equation,
we get
Axex= 3(Axex) + ex.
This simplifies to A=1
3.
Step 3: The general solution to the non-homogeneous equation is the sum of
the general solution to the homogeneous equation and the particular solution:
y=Ce3x+1
3xex.
Now we use the initial condition y(0) = 1 to find C:
1 = C+ 0 ⇒C= 1.
Therefore, the solution to the initial value problem is
y=e3x+1
3xex.
Question 22
Question
Solve the initial value problem: dy
dx = 2y−x, with y(0) = 3.
Solution
Step 1: First, let’s rewrite the differential equation as dy
dx −2y=−x.
Step 2: Next, we find the integrating factor by computing eR−2dx =e−2x.
Step 3: Multiply the integrating factor by both sides of the differential equa-
tion to get:
e−2xdy
dx −2e−2xy=−xe−2x
.
Step 4: Recognize that the left side is the derivative of the product e−2xy,
which simplifies to: d
dx(e−2xy) = −xe−2x
.
Step 5: Integrate the equation with respect to xto get:
e−2xy=Z−xe−2xdx
23
.
Step 6: Solve the integral to obtain:
e−2xy=1
4xe−2x+C
.
Step 7: Now, apply the initial condition y(0) = 3 to find the value of the
constant C:
e0·3 = 1
4·0 + C
.
3 = C
.
Step 8: Substitute C= 3 back into the equation to find the particular
solution:
e−2xy=1
4xe−2x+ 3
.
y=1
4x+ 3e2x
.
Hence, the solution to the initial value problem is y=1
4x+ 3e2x.
Question 23
Question
Solve the initial value problem:
dy
dx + 2y=e−x, y(0) = 1
Solution
Step 1: First, we find the integrating factor by solving the auxiliary equation
v′+ 2v= 0.
r+ 2 = 0 =⇒r=−2
So, the integrating factor is v(x) = e−2x.
Step 2: We multiply both sides of the given differential equation by the
integrating factor v(x).
e−2xdy
dx + 2e−2xy=e−3x
Step 3: Rewrite the equation in terms of the product rule for differentiation.
d
dx(e−2xy) = e−3x
24
Step 4: Integrate both sides with respect to x.
Zd
dx(e−2xy)dx =Ze−3xdx
e−2xy=−1
3e−3x+C
where Cis the constant of integration.
Step 5: Solve for yby multiplying through by e2x.
y=−1
3e−x+Ce2x
Step 6: Use the initial condition y(0) = 1 to solve for C.
1 = −1
3−C=⇒C=4
3
Step 7: Substitute C=4
3back into the general solution to get the particular
solution.
y=−1
3e−x+4
3e2x
Therefore, the solution to the initial value problem is y=−1
3e−x+4
3e2x.
Question 24
Question
Solve the initial value problem
y′= sin(2x) + y2, y(0) = 1.
Solution
To solve the initial value problem, we will use the method of integrating factors.
Step 1: Rewrite the initial value problem in the form y′+P(x)y=Q(x).
The given initial value problem can be rewritten as
y′−y2= sin(2x).
Step 2: Find the integrating factor µ(x) = eRP(x)dx. Here, P(x) = −1.
Thus, the integrating factor is
µ(x) = eR−1dx =e−x.
Step 3: Multiply the integrating factor by the differential equation. Multi-
plying both sides of the differential equation by µ(x), we get
e−xy′−e−xy2=e−xsin(2x).
25
Step 4: Rewrite the left side as the derivative of a product. The left side
can be rewritten using the product rule as
(e−xy)′=e−xsin(2x).
Step 5: Integrate both sides with respect to x. Integrating both sides, we
have Z(e−xy)′dx =Ze−xsin(2x)dx.
Integrating, we get
e−xy=Ze−xsin(2x)dx.
Step 6: Solve for y. To find y, we solve the integral on the right side and
then solve for y.
e−xy=−1
5cos(2x) + C,
where Cis the constant of integration.
Thus, the general solution is
y(x) = −1
5excos(2x) + Cex.
Step 7: Use the initial condition to find the value of C. Applying the initial
condition y(0) = 1, we get
1 = −1
5cos(0) + C.
This simplifies to 1 = −1
5+C, so C=6
5.
Step 8: Write the final solution. Therefore, the solution to the initial value
problem is
y(x) = −1
5excos(2x) + 6
5ex.
Question 25
Question
Solve the initial value problem:
dy
dx =y
x+ex, y(1) = 2
26
Solution
Step 1: Rewrite the differential equation in standard form.
dy
dx −y
x=ex
Step 2: Identify the integrating factor µ(x), which is given by µ(x) = eR−1
xdx.
µ(x) = eR−1
xdx =e−ln |x|=1
x
Step 3: Multiply the integrating factor µ(x) throughout the standard form
equation. 1
x
dy
dx −1
x2y=ex·1
x
Step 4: Rewrite the left-hand side of the equation using the product rule,
and simplify the right-hand side.
d
dx y
x=ex·1
x
Step 5: Integrate both sides with respect to x.
Zd
dx y
xdx =Zex·1
xdx
Step 6: Solve the integrals.
y
x=Zex·1
xdx =Z1
xexdx =ex+C
Step 7: Multiply by xto solve for y.
y=x(ex+C)
Step 8: Use the initial condition y(1) = 2 to solve for C.
2 = 1(e1+C) =⇒C= 2 −e
Step 9: Substitute Cback into the equation to obtain the final solution.
y=x(ex+ 2 −e)
Question 26
Question
Solve the initial value problem:
y′=y
x+xy, y(1) = 1.
27
Solution
Step 1: This is a first-order linear differential equation in the form y′−P(x)y=
Q(x). We can rewrite the given equation as:
y′−1
xy=x2.
Step 2: To solve the homogeneous part of the equation, we first find the
integrating factor I(x) = eR−1
xdx =e−ln x=1
x.
Step 3: Multiply the homogeneous equation by the integrating factor:
1
xy′−1
x2y=x
d
dx 1
xy=x
Zd
dx 1
xydx =Zxdx
1
xy=1
2x2+C,
where Cis an arbitrary constant.
Step 4: Now, we add the particular solution to the general solution:
y=x2+Cx
y(1) = 1 + C= 1
C= 0.
Therefore, the solution to the initial value problem is y=x2.
Question 27
Question
Solve the initial value problem: dy
dx =√y, with y(0) = 4.
Solution
Step 1: Rewrite the differential equation in a separable form:
dy
dx =√y
dy
√y=dx
Step 2: Integrate both sides:
Z1
√ydy =Zdx
28
2√y=x+C1
Step 3: Solve for yusing the initial condition y(0) = 4:
2√4 = 0 + C1
C1= 4
Step 4: Substitute C1= 4 back into the general solution:
2√y=x+ 4
√y=1
2x+ 2
Step 5: Solve for y:
y=1
2x+ 22
y=1
4x2+ 2x+ 4
So, the solution to the initial value problem is y=1
4x2+ 2x+ 4.
Question 28
Question
Solve the initial value problem:
y′+y= sin(x), y(0) = 1
Solution
Step 1: First, we need to solve the homogeneous differential equation y′+y= 0.
The characteristic equation is r+ 1 = 0, so r=−1. Therefore, the general
solution to the homogeneous equation is yh(x) = c1e−x, where c1is a constant
parameter.
Step 2: Next, we find a particular solution to the non-homogeneous equation
y′+y= sin(x). Since the right-hand side is sin(x), we try a particular solution
of the form yp(x) = Asin(x)+Bcos(x). Taking the derivatives and substituting
back into the equation, we find that A= 0 and B=−1
2. Therefore, a particular
solution is yp(x) = −1
2cos(x).
Step 3: The general solution to the non-homogeneous equation is the sum of
the general solution to the homogeneous equation and the particular solution.
Thus, the general solution is:
y(x) = yh(x) + yp(x) = c1e−x−1
2cos(x)
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Step 4: Finally, we find the value of the constant c1using the initial condition
y(0) = 1. Substitute x= 0 and y= 1 into the general solution:
1 = c1e0−1
2cos(0)
1 = c1−1
2
c1=3
2
Step 5: Therefore, the solution to the initial value problem is:
y(x) = 3
2e−x−1
2cos(x)
Question 29
Question
Solve the initial value problem:
y′+2
xy=xsin(x), y(1) = 2
Solution
Step 1: The given differential equation is a first-order linear differential equation
in standard form. To solve it, we’ll first find the integrating factor.
The integrating factor µ(x) is given by µ(x) = eR2
xdx.
Step 2: We find R2
xdx.
Z2
xdx = 2 Z1
xdx = 2 ln |x|+C
Step 3: Therefore, the integrating factor is µ(x) = e2 ln |x|=eln(x2)=x2.
Step 4: Now, we multiply the integrating factor by the given differential
equation to get:
x2y′+ 2xy =x3sin(x)
Step 5: We can rewrite this equation as the derivative of a product:
(x2y)′=x3sin(x)
Step 6: Integrating both sides with respect to xgives:
x2y=Zx3sin(x)dx
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Step 7: Integrating Rx3sin(x)dx by parts:
Zx3sin(x)dx =−x3cos(x)+3Zx2cos(x)dx
Step 8: To evaluate Rx2cos(x)dx, we need to use integration by parts again.
Let u=x2and dv = cos(x)dx.
Step 9: Calculating du and vgives du = 2xdx and v= sin(x).
Step 10: Applying integration by parts:
Zx2cos(x)dx =x2sin(x)−Z2xsin(x)dx
Step 11: Integrating the remaining integral R2xsin(x)dx by parts once more,
we get:
Z2xsin(x)dx = 2xcos(x)−Z2 cos(x)dx
Step 12: Integrating R2 cos(x)dx = 2 sin(x), we substitute this back into the
previous result to get:
Zx2cos(x)dx =x2sin(x)−2xcos(x) + 2 sin(x)
Step 13: Substituting back into our integral equation, we have:
Zx3sin(x)dx =−x3cos(x) + 3(x2sin(x)−2xcos(x) + 2 sin(x))
Step 14: Therefore, the solution to the initial value problem y′+2
xy=
xsin(x), y(1) = 2 is:
y(x) = 1
x2−x3cos(x) + 3(x2sin(x)−2xcos(x) + 2 sin(x))
Question 30
Question
Solve the initial value problem:
dy
dx =2x
1 + y, y(0) = 1
Solution
Step 1: Rewrite the initial value problem and separate variables.
The initial value problem can be rewritten as:
(1 + y)dy = 2x dx
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Now, we can separate variables:
(1 + y)dy = 2x dx
Step 2: Integrate both sides.
Integrating both sides gives us:
Z(1 + y)dy =Z2x dx
Which simplifies to:
Zdy +Zy dy =Z2x dx
This further simplifies to:
y+y2
2=x2+C
Step 3: Apply initial condition to find the value of the constant.
Using the initial condition y(0) = 1, we have:
1 + 1
2= 02+C=⇒1 + 1
2=C=⇒C=3
2
Step 4: Substitute the value of Cback into the equation and solve for y.
Substitute C=3
2back into the equation gives us:
y+y2
2=x2+3
2
Plugging in the initial condition y(0) = 1, we have:
1 + 1
2= 02+3
2=⇒3
2=3
2
Therefore, the solution to the initial value problem is y=x2+3
2.
Question 31
Question
Consider the initial value problem:
dy
dx =x2+y2, y(0) = 1.
Find the particular solution to the initial value problem on the interval
[−1,1].
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Solution
Step 1: Note that this is a first-order ordinary differential equation. We can
rewrite it as: dy
dx =x2+y2.
Step 2: This is not a separable equation, so let’s try to use an integrating
factor to solve it. Multiplying both sides by e−y2, we get:
e−y2dy
dx =e−y2x2+ 1.
Step 3: We can rewrite the left-hand side as:
d
dx(e−y2) = e−y2dy
dx −2ye−y2.
Step 4: Substituting this back into the equation and integrating both sides
with respect to x, we get:
Zd
dx(e−y2)dx =Z(e−y2x2+ 1) dx.
Step 5: Integrating both sides gives:
e−y2=1
3x3+x+C,
where Cis the constant of integration.
Step 6: To find the value of C, we use the initial condition y(0) = 1. Sub-
stituting x= 0 and y= 1 into the equation above, we get:
e−1=C.
Step 7: So, the particular solution to the initial value problem is:
e−y2=1
3x3+x+e−1.
Question 32
Question
Consider the initial value problem:
dy
dx =4
x(2 −y), y(1) = 3.
Find the particular solution to the initial value problem.
33
Solution
Step 1: Rewrite the differential equation in the form dy
dx +P(x)y=Q(x) by
expanding and simplifying.
dy
dx =4
x(2 −y)
dy
dx =8
x−4
xy
dy
dx +4
xy=8
x
Step 2: Now, we will solve the differential equation using an integrating
factor µ(x) = eR4
xdx.
µ(x) = eR4
xdx
µ(x) = e4 ln x
µ(x) = x4
Step 3: Multiply the integrating factor µ(x) to the entire differential equation
and rewrite it in a different form.
x4dy
dx + 4x3y= 8x3
Step 4: Recognize the left-hand side as the derivative of the product of x4y.
d
dx(x4y)=8x3
x4y=Z8x3dx
x4y= 2x4+C
Step 5: Apply the initial condition y(1) = 3 to find the particular solution.
x4y= 2x4+C
14·3=2·14+C
3 = 2 + C
C= 1
So, the particular solution to the initial value problem is y= 2 + 1
x4.
Question 33
Question
Solve the initial value problem:
y′+y=e−t, y(0) = 1
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Solution
Step 1: First, we find the integrating factor by solving the differential equation
u′(t) = u(t)p(t), where p(t) is the coefficient of y.
u′(t) = u(t)·1
This is a separable differential equation with general solution u(t) = Cet, where
Cis a constant to be determined.
Step 2: Multiplying the given differential equation by the integrating factor
u(t):
ety′+ety=et·e−t
ety′+ety= 1
Step 3: Recognize the left-hand side of the equation as the result of the
product rule for differentiation of yet:
(yet)′= 1
Step 4: Integrate both sides of the equation:
Z(yet)′dt =Z1dt
yet=t+C
Step 5: Apply the initial condition y(0) = 1:
1 = 0 + C
C= 1
Step 6: Therefore, the solution to the initial value problem is:
yet=t+ 1
y=t+ 1
et
Question 34
Question
Consider the initial value problem:
dy
dx =x2+y, y(0) = 1
Find the particular solution to the given initial value problem.
35
Solution
Step 1: First, we rearrange the given differential equation into a separable form.
This involves isolating the variables xand yon opposite sides of the equation.
Thus, we have: dy
dx =x2+y=⇒dy
dx −y=x2
Step 2: Next, we solve this separable first-order linear differential equation
using the integrating factor method. The integrating factor I(x) is given by
eR−1dx =e−x. Multiplying both sides of the equation by the integrating factor,
we get:
e−xdy
dx −e−xy=x2e−x
Step 3: Rewrite the left side of the equation as the derivative of the product
of the integrating factor and yby applying the product rule:
d
dx(e−xy) = x2e−x
Step 4: Integrate both sides with respect to xto solve for y:
Zd
dx(e−xy)dx =Zx2e−xdx
Step 5: Integrating both sides, we have:
e−xy=Zx2e−xdx
Step 6: Solve the integral on the right side:
Zx2e−xdx =−x2e−x−2Z−xe−xdx
Step 7: Continuing to solve the integral on the right side:
Zx2e−xdx =−x2e−x+ 2 Zxe−xdx
Step 8: The integral 2 Rxe−xdx can be solved using integration by parts. We
let u=xand dv =e−xdx, then du =dx and v=−e−x. Applying integration
by parts formula:
Zxe−xdx =−xe−x−Z−e−xdx
Step 9: Simplifying the result, we get:
Zxe−xdx =−xe−x+e−x
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Step 10: Substitute the computed integral back into the equation and solve
for y:
e−xy=−x2e−x+ 2(−xe−x+e−x) + C
Step 11: Apply the initial condition y(0) = 1 to find the constant C. Plug-
ging in x= 0 and y= 1, we get:
1=0−0 + 2(0 + 1) + C=⇒C=−1
Step 12: Plug the value of Cback into the equation to obtain the particular
solution:
e−xy=−x2e−x+ 2(−xe−x+e−x)−1
Step 13: Simplify the expression to get the final particular solution:
y=−x2+ 2x−1 + ex
Question 35
Question
Solve the initial value problem:
dy
dx =2x−y
x+ 2y, y(0) = 1
Solution
Step 1: Rewrite the initial value problem in the standard form dy
dx =f(x, y):
dy
dx =2x−y
x+ 2y
Step 2: Check for homogeneity by substituting y=vx:
dy
dx =xdv
dx +v, 2x−y
x+ 2y=2−v
1+2v
Step 3: Let v=u+ 1 to get:
du
dx =−3u
1+2u
Step 4: Separate variables and integrate:
Z1+2u
−3udu =Zdx
x
Step 5: Solve the integrals:
ln |u| − 3
2ln |1+2u|= ln |x|+C
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Step 6: Substitute back u=y
x−1:
ln
y
x−1
−3
2ln
2y
x+ 1
= ln |x|+C
Step 7: Apply the initial condition y(0) = 1 to find the constant C:
C= ln 1 = 0
Step 8: Simplify the equation further if necessary.
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