MATH 334 - DIFFERENTIAL
EQUATIONS - Ordinary differential
equations
Question Bank - Set 4
Liberty University
Question 1
Question
Consider the following first-order differential equation:
dy
dx =−2x2y
Find the general solution to this differential equation.
Solution
Step 1: Rewrite the equation in a separable form. We have the equation
dy
dx =−2x2y. We can rewrite this as:
dy
y=−2x2dx
Step 2: Integrate both sides. Integrating both sides gives us:
Zdy
y=Z−2x2dx
ln |y|=−2
3x3+C
where Cis the constant of integration.
Step 3: Solve for y. Exponentiating both sides, we get:
|y|=e−2
3x3+C
|y|=eCe−2
3x3
y=±eCe−2
3x3
y=Ke−2
3x3
where K=±eCis a constant.
Step 4: General solution. Therefore, the general solution to the differential
equation is:
y=Ke−2
3x3
where Kis an arbitrary constant.
Question 2
Question
Find the general solution to the differential equation: dy
dx =1
x2+y2.
Solution
Step 1: Rearrange the differential equation as dy
dx =1
x2+y2.
Step 2: Multiply both sides by dx to separate variables: dx = (x2+y2)dy.
Step 3: Integrate both sides with respect to x:Rdx =R(x2+y2)dy.
Step 4: Solving the integrals, we get x=x3
3+y2+C, where Cis the constant
of integration.
Step 5: Rearranging the equation, we find the general solution: y2=x3
3−
x+C.
Therefore, the general solution to the differential equation dy
dx =1
x2+y2is
y2=x3
3−x+C, where Cis a constant.
Question 3
Question
Consider the initial value problem dy
dx = 2x3+ 3y2with y(0) = 1. Find the
particular solution to this initial value problem.
Solution
Step 1: First, let’s rewrite the differential equation as
dy
dx = 2x3+ 3y2.
2
Step 2: Notice that this is a first-order ordinary differential equation that is
separable. We can rewrite the equation as
dy
dx = 2x3+ 3y2=⇒dy
3y2= 2x3dx.
Step 3: Integrating both sides, we get
Z1
3y2dy =Z2x3dx.
Step 4: This gives us
−1
3y=1
2x4+C,
where Cis the constant of integration.
Step 5: Now, we use the initial condition y(0) = 1 to find the particular
solution. Plugging in x= 0 and y= 1 into the equation, we have
−1
3=C.
Step 6: Substituting C=−1
3back into the equation, we have
−1
3y=1
2x4−1
3.
Step 7: Solving for y, we get
y=−3
2x4+ 1.
Therefore, the particular solution to the initial value problem is y=−3
2x4+1 .
Question 4
Question
Consider the following first-order ordinary differential equation:
dy
dx =1−2y
x+y
Use the method of integrating factors to solve this differential equation.
Solution
Step 1: Rewrite the differential equation in the standard form M(x, y)dx +
N(x, y)dy = 0 where Mand Nare functions of xand y.
dy
dx +2y−1
x+y= 0
3
Step 2: Identify Mand N:
M(x, y)=1, N(x, y)=2y−1
Step 3: Compute the integrating factor, denoted by I(x), using the formula:
I(x) = eR(N−My)dx
Step 4: Calculate I(x):
I(x) = eR(2y−1−0)dx =eR(2y−1)dx =e2yx−x
Step 5: Multiply the original differential equation by the integrating factor
I(x):
e2yx−xdy
dx + 2y−1= 0
Step 6: Rewrite the above expression as the derivative of a product:
d
dx(e2yx−xy)=0
Step 7: Integrate both sides with respect to x:
Zd
dx(e2yx−xy)dx =Z0dx
e2yx−xy=C
Step 8: Solve for y:
y=C
e2yx−x
Therefore, the general solution to the given ODE is y=C
e2yx−x, where Cis
an arbitrary constant.
Question 5
Question
Consider the initial value problem:
dy
dx = 2xe−y, y(0) = 1.
Find the solution to the initial value problem.
4
Solution
Step 1: Rewrite the differential equation in the form dy =f(x)dx.
dy
dx = 2xe−y
dy = 2xe−ydx
Step 2: Separate the variables by dividing both sides by e−y.
dy
e−y= 2xdx
Step 3: Integrate both sides with respect to x.
Zeydy =Z2xdx
ey=x2+C1
Step 4: Use the initial condition y(0) = 1 to find the value of the constant
C1.
e1= 02+C1
e=C1
Step 5: Substitute C1=eback into the equation to find the particular
solution.
ey=x2+e
y= lnx2+e
Therefore, the solution to the initial value problem is y= lnx2+e, where
eis Euler’s number.
Question 6
Question
Find the general solution of the differential equation: dy
dx +y=e−x.
Solution
To solve the given first-order linear differential equation, we will use the inte-
grating factor method.
Step 1: Write the given differential equation in standard form dy
dx +P(x)y=
Q(x). Here P(x) = 1 and Q(x) = e−x.
Step 2: Find the integrating factor I(x) = eRP(x)dx.
Integrating factor I(x) = eR1dx =ex
5
Step 3: Multiply both sides of the differential equation by the integrating
factor I(x).
exdy
dx +exy=exe−x
Step 4: Rewrite the left-hand side as the derivative of the product of yand
the integrating factor ex.d
dx(exy) = exe−x
Step 5: Integrate both sides with respect to x.
Zd
dx(exy)dx =Zexe−xdx
exy=Z1dx =x+C
Step 6: Solve for y.
y=x+C
ex
Therefore, the general solution of the given differential equation is y=x+C
ex,
where Cis an arbitrary constant.
Question 7
Question
Find the general solution to the differential equation dy
dx =y2−1
x2+1 .
Solution
Step 1: Rewrite the differential equation in a separable form:
dy
y2−1=dx
x2+ 1
Step 2: Integrate both sides with respect to yand x:
Z1
y2−1dy =Z1
x2+ 1 dx
Step 3: Factor the denominator in the left integral to split it into partial
fractions: Z1
2·1
y−1−1
2·1
y+ 1dy =Z1
x2+ 1 dx
Step 4: Integrate both sides with respect to yand x:
1
2ln |y−1| − 1
2ln |y+ 1|= arctan(x) + C
6
Step 5: Combine the logarithmic terms on the left side using properties of
logarithms:
ln
y−1
√y+ 1
= 2 arctan(x) + C
Step 6: Rewrite the equation in an exponential form:
y−1
√y+ 1
=e2 arctan(x)+C
Step 7: Remove the absolute value on the left side by considering both
positive and negative cases:
y−1
√y+ 1 =±e2 arctan(x)+C
Step 8: Solve for yby squaring both sides and rearranging the terms:
y= 1 + ±e2 arctan(x)+C(y+ 1)
√y+ 1
The general solution to the differential equation is:
y= 1 + ±e2 arctan(x)+C(y+ 1)
√y+ 1
Question 8
Question
Find the general solution to the following differential equation:
y′′ −4y′+ 4y= 0
Solution
To solve the differential equation y′′ −4y′+ 4y= 0, we will first find the
characteristic equation by assuming the form y=ert.
Step 1: Find the characteristic equation Substitute y=ert into the
differential equation:
(r2−4r+ 4)ert = 0
This simplifies to:
(r−2)2= 0
So the characteristic equation is:
r= 2,2
7
Step 2: Find the general solution Since we have repeated real roots,
the general solution is of the form:
y(t) = c1e2t+c2te2t
where c1and c2are arbitrary constants.
Therefore, the general solution to the differential equation y′′ −4y′+ 4y= 0
is:
y(t) = c1e2t+c2te2t
Question 9
Question
Consider the following ordinary differential equation:
y′′ −2y′+y=et
Find the general solution to the differential equation by first solving the
associated homogeneous equation and then finding a particular solution.
Solution
Step 1: First, let’s find the general solution to the associated homogeneous
equation:
y′′ −2y′+y= 0
We assume y=ert for some constant r. Substitute this into the homoge-
neous equation to get:
r2ert −2rert +ert = 0
Divide through by ert:
r2−2r+ 1 = 0
This is a quadratic equation with solutions:
r= 1,1
So, the general solution to the homogeneous equation is:
yh=c1et+c2tet
Step 2: Now, let’s find the particular solution to the nonhomogeneous equa-
tion: Since etis a solution to the homogeneous equation, we try a particular
solution of the form:
yp=Atet
Substitute ypinto the nonhomogeneous equation and solve for A:
(Atet)′′ −2(Atet)′+Atet=et
8
(Aet+ 2Atet−2Aet+Atet) = et
(2Atet) = et
A=1
2
So, the particular solution is:
yp=1
2tet
Step 3: The general solution to the differential equation is the sum of the
general solution to the homogeneous equation and the particular solution:
y=yh+yp=c1et+c2tet+1
2tet
Therefore, the general solution is:
y=c1et+ (c2+1
2)tet
Question 10
Question
Consider the following first-order ordinary differential equation:
dy
dx =e−x−y
Find the particular solution to this differential equation given the initial condi-
tion y(0) = 3.
Solution
To solve this first-order ordinary differential equation, we can use the method
of integrating factors.
Step 1: Rewrite the equation in the standard form dy
dx +P(x)y=Q(x):
dy
dx +y=e−x
Step 2: Identify P(x) and Q(x) from the standard form:
P(x)=1, Q(x) = e−x
Step 3: Calculate the integrating factor I(x) = eRP(x)dx:
I(x) = eR1dx =ex
9
Step 4: Multiply both sides of the standard form by the integrating factor:
exdy
dx +exy=e2x
Step 5: Rewrite the left side as the derivative of the product exy:
d
dx(exy) = e2x
Step 6: Integrate both sides of the equation:
Zd
dx(exy)dx =Ze2xdx
exy=e2x
2+C
y=1
2+Ce−x
Step 7: Apply the initial condition y(0) = 3 to find the particular solution:
3 = 1
2+C
C=5
2
Step 8: The particular solution to the differential equation is:
y=1
2+5
2e−x
Question 11
Question
Consider the differential equation
dy
dx =2x2+y−3
2y−x
Solve the initial value problem with y(0) = 1.
Solution
To solve this differential equation, we can rewrite it in the form:
(2y−x)dy = (2x2+y−3) dx
10
Step 1: Integrate both sides of the equation.
Z(2y−x)dy =Z(2x2+y−3) dx
Step 2: Integrate the left side with respect to y.
y2−1
2x2=Z(2x2+y−3) dx
Step 3: Integrate the right side with respect to x.
y2−1
2x2=2
3x3+1
2xy −3x+c
Step 4: Simplify the equation by substituting y(0) = 1.
1−0 = 2
3(0)3+1
2(0)(1) −3(0) + c=⇒c= 1
Step 5: Thus, the solution to the initial value problem is:
y2−1
2x2=2
3x3+1
2xy −3x+ 1
Question 12
Question
Find the general solution of the differential equation: dy
dx =2x2+1
y.
Solution
Step 1: Rearrange the differential equation to separate variables.
dy
dx =2x2+ 1
y.
y dy = (2x2+ 1) dx.
Step 2: Integrate both sides of the equation.
Zy dy =Z(2x2+ 1) dx.
Step 3: Solve the integrals.
y2
2=2x3
3+x+C,
where Cis the constant of integration.
11
Step 4: Multiply through by 2 to simplify.
y2=4x3
3+ 2x+ 2C.
Step 5: Take the square root of both sides to find the general solution.
y=±r4x3
3+ 2x+ 2C.
Therefore, the general solution of the differential equation is y=±q4x3
3+ 2x+ 2C.
Question 13
Question
Find the general solution to the following first-order ordinary differential equa-
tion: dy
dx =y+ex
y−ex
Solution
Step 1: Rewrite the given differential equation in a more suitable form by sep-
arating variables. We can rewrite the equation as:
(y−ex)dy = (y+ex)dx
Step 2: Integrate both sides of the equation. The left side can be integrated
as: Z(y−ex)dy =1
2y2−exy+C1
Step 3: Similarly, the right side can be integrated as:
Z(y+ex)dx =y+ex+C2
Step 4: Equating the two integrals, we have:
1
2y2−exy=y+ex+C
Step 5: Rearranging the terms, we get:
1
2y2−y= 3ex+C
Step 6: Finally, the general solution is given by:
y(x) = 1 ±√1+6ex+ 2C
where Cis an arbitrary constant.
12
Question 14
Question
Consider the differential equation y′′ +y= 0. Solve the differential equation
subject to the initial conditions y(0) = 3 and y′(0) = 1.
Solution
Step 1: Write down the characteristic equation of the differential equation
y′′ +y= 0. The characteristic equation is obtained by replacing y′′ with r2,
which gives us r2+ 1 = 0.
Step 2: Solve the characteristic equation r2+ 1 = 0. We have r2=−1, so
r=±√−1 = ±i.
Step 3: General solution of the differential equation. The general solution
of the differential equation is given by y(t) = c1cos(t) + c2sin(t), where c1and
c2are arbitrary constants.
Step 4: Apply the initial conditions y(0) = 3 and y′(0) = 1. Substitute
t= 0 into the general solution and its derivative to get:
y(0) = c1cos(0) + c2sin(0) = c1= 3
y′(0) = −c1sin(0) + c2cos(0) = c2= 1
Step 5: Write down the particular solution. Thus, the particular solution
of the differential equation satisfying the initial conditions is:
y(t) = 3 cos(t) + sin(t)
Question 15
Question
Find the general solution to the differential equation:
y′′ −4y′+ 5y=e2x
Solution
Step 1: First, find the complementary solution by solving the homogeneous
equation y′′ −4y′+ 5y= 0. Let’s assume the solution has the form y=erx.
Substitute y=erx into the homogeneous equation:
r2erx −4rerx + 5erx = 0
Simplify and factor out erx:
erx(r2−4r+ 5) = 0
13
Since erx is never zero, we have:
r2−4r+ 5 = 0
Step 2: Solve the characteristic equation r2−4r+5 = 0. Using the quadratic
formula r=−b±√b2−4ac
2awith a= 1, b=−4, and c= 5:
r=4±p(−4)2−4(1)(5)
2(1)
r=4±√16 −20
2
r=4±√−4
2
r= 2 ±2i
Step 3: Write down the complementary solution. Since the roots are com-
plex, the complementary solution is:
yc=e2x(Acos(2x) + Bsin(2x))
Step 4: Find a particular solution. Since the right-hand side is e2x, let’s try
a particular solution of the form yp=x(Ae2x). Substitute ypinto the original
differential equation and solve for A:
2A−4(2A) + 5(xAe2x) = e2x
2A−8A+ 5xA = 1
5xA −6A= 1
A=−1
6x
Step 5: Write down the full particular solution. The particular solution is:
yp=−1
6xe2x
Step 6: Write down the general solution. The general solution is the sum
of the complementary and particular solutions:
y=yc+yp=e2x(Acos(2x) + Bsin(2x)) −1
6xe2x
Question 16
Question
Solve the following ordinary differential equation:
dy
dx =y2−x2
14
Solution
Step 1: Rewrite the given differential equation in the form dy
dx =f(x, y).
dy
dx =y2−x2
Step 2: Recognize that this is a separable differential equation, so we can
write it as: dy
y2−x2=dx
Step 3: Integrate both sides with respect to y, then solve for y.
Zdy
y2−x2=Zdx
Step 4: To integrate the left side, we need to perform partial fraction de-
composition. Let’s write:
1
y2−x2=A
y+x+B
y−x
Step 5: Solving for Aand Bby multiplying both sides by y2−x2, we get:
1 = A(y−x) + B(y+x)
Step 6: Set y=xto solve for A.
1 = Ax −Bx
A=−B
Step 7: Set y=−xto solve for B.
1 = −Ax −Bx
B=−A= 1
Step 8: Substituting the values of Aand Bback into the partial fraction
decomposition gives us:
1
y2−x2=1
21
y+x−1
y−x
Step 9: Plugging this back into the integral and integrating, we get:
1
2ln
y+x
y−x
=x+C
Step 10: Exponentiate both sides to eliminate the natural logarithm.
y+x
y−x
=e2x+2C
15
Step 11: Simplify the absolute value terms by considering positive and neg-
ative cases. y+x
y−x=±e2x+2C
Step 12: Rearrange to solve for y.
y+x=±e2x+2C(y−x)
y+x=±e2x+2Cy∓e2x+2Cx
y(1 ∓e2x+2C) = x(1 ±e2x+2C)
y=x(1 ±e2x+2C)
1∓e2x+2C
Therefore, the general solution to the ordinary differential equation dy
dx =
y2−x2is:
y=x(1 ±e2x+2C)
1∓e2x+2C
Question 17
Question
Consider the following initial value problem:
y′= 2xy +x, y(0) = 1.
Find the solution y(x) for x≥0.
Solution
Step 1: First, let’s find the general solution to the differential equation.
dy
dx = 2xy +x
dy
dx −2xy =x
This is a first-order linear ordinary differential equation. The integrating factor
is eR−2x dx =e−x2. Multiplying through by the integrating factor, we have:
e−x2dy
dx −2xye−x2=xe−x2
d
dx(ye−x2) = xe−x2
Integrating both sides gives:
ye−x2=Zxe−x2dx
16
ye−x2=−1
2e−x2+C
y=−1
2+Cex2
where Cis an arbitrary constant.
Step 2: Next, we use the initial condition y(0) = 1 to determine the value of
C.
1 = −1
2+C=⇒C=3
2
So, the solution to the initial value problem is:
y(x) = −1
2+3
2ex2=1
2(3ex2−1), x ≥0.
Question 18
Question
Consider the ordinary differential equation:
dy
dx = 2x2y
Find the general solution to this differential equation.
Solution
Step 1: Rewrite the differential equation in separated variables form:
dy
y= 2x2dx
Step 2: Integrate both sides:
Zdy
y=Z2x2dx
Step 3: Evaluate the integrals:
ln |y|=2
3x3+C
Step 4: Solve the equation for yby exponentiating both sides:
|y|=e2
3x3+C
Step 5: Simplify the equation by combining the constants of integration:
|y|=Ke2
3x3where K=eC
Step 6: Consider both cases for the absolute value:
17
y=Ke2
3x3or y=−Ke2
3x3
Therefore, the general solution to the differential equation is:
y=Ke2
3x3or y=−Ke2
3x3
Question 19
Question
Consider the initial value problem:
y′+ 2y=e−x, y(0) = 1.
Find the particular solution to the differential equation that satisfies the initial
condition.
Solution
Step 1: First, we find the integrating factor. The integrating factor is given by
µ(x) = eR2dx =e2x.
Step 2: Multiply both sides of the differential equation by the integrating
factor:
e2xy′+ 2e2xy=e2xe−x.
Step 3: Rewrite the left side as the derivative of the product e2xyusing the
product rule: d
dx(e2xy) = e2xe−x.
Step 4: Integrate both sides with respect to x:
e2xy=−e−x+C,
where Cis the constant of integration.
Step 5: Apply the initial condition y(0) = 1 to solve for the constant C:
e0·1 = −e0+C⇒1 = −1 + C⇒C= 2.
Step 6: Substitute the value of Cback into the equation to find the particular
solution:
e2xy=−e−x+ 2 ⇒y=−e−3x+ 2e−2x.
Therefore, the particular solution to the differential equation that satisfies
the initial condition is y=−e−3x+ 2e−2x.
18
Question 20
Question
Consider the ordinary differential equation
dy
dx =−y
x+x2
with initial condition y(1) = 2. Find the particular solution to the differential
equation.
Solution
Step 1: Let’s first rewrite the differential equation in a more standard form, by
separating variables: dy
dx +y
x=x2
Step 2: This is a first-order linear differential equation, which we can solve
using an integrating factor. In this case, the integrating factor is eR1
xdx =
eln |x|=|x|.
Step 3: Multiplying both sides of the equation by the integrating factor |x|
gives:
|x|dy
dx +y=x3
Step 4: Now we integrate both sides with respect to x:
Z|x|dy
dxdx +Zydx =Zx3dx
Step 5: Integrating both sides gives:
|x|y=1
4x4+C
Step 6: Now we apply the initial condition y(1) = 2 to find the value of the
constant C:
|1| · 2 = 1
4·14+C
2 = 1
4+C
C=7
4
Step 7: Therefore, the particular solution to the differential equation is:
|x|y=1
4x4+7
4
19
Question 21
Question
Solve the following initial value problem:
dy
dx =1
x2+y2, y(1) = 0
Solution
Step 1: Rewrite the differential equation in a more standard form by separating
variables. dy
dx =1
x2+y2=⇒dy
dx =1
x2(1 + (y/x)2)
Step 2: Substituting v=y/x, we have
y=vx =⇒dy
dx =v+xdv
dx
Step 3: Substituting into the differential equation and simplifying, we get
v+xdv
dx =1
x2(1 + v2)=⇒xdv
dx =1−v2
x(1 + v2)
Step 4: Separate variables and integrate both sides:
Z1 + v2
1−v2dv =Z1
xdx
Step 5: Simplify the left-hand side by partial fraction decomposition and
integrate:
Z1 + v2
1−v2dv =Z1
2(1 −v)−1
2(1 + v)dv
Step 6: Integrating both sides gives
1
2ln |1−v| − 1
2ln |1 + v|= ln |x|+C
Step 7: Exponentiate both sides to solve for v:
ln
1−v
1 + v
= 2 ln |x|+C′=⇒1−v
1 + v=Cx2
Step 8: Solve for vin terms of xby manipulating the equation:
v=1−Cx2
1 + Cx2
Step 9: Using the initial condition y(1) = 0, we find C=1
2.
Step 10: Substitute C=1
2back into the equation for vto get the solution
in terms of x:
y(x) = x1−1
2x2
1 + 1
2x2=x(x2−1)
x2+ 2
20
Question 22
Question
Consider the first-order ordinary differential equation:
y′+y2=xy.
Find the general solution of the given differential equation.
Solution
Step 1: Rewrite the equation in standard form:
y′=xy −y2.
Step 2: Factor out yon the right-hand side:
y′=y(x−y).
Step 3: Separate variables by dividing both sides by y(x−y):
dy
y=dx
x−y.
Step 4: Integrate both sides:
Z1
ydy =Z1
x−ydx.
Step 5: Solve the integrals:
ln |y|= ln |x−y|+C,
where Cis the constant of integration.
Step 6: Rewrite the natural logarithms using properties of logarithms:
ln |y|= ln |x−y|+ ln |eC|.
Step 7: Combine the logarithms on the right-hand side:
ln |y|= ln |eC(x−y)|.
Step 8: Use the property of logarithms ln(a) = ln(b) if and only if a=b:
|y|=eC(x−y).
Step 9: Simplify and solve for y:
|y|=Cex−Cy,
where C=±eC.
Step 10: Consider the two cases y > 0 and y < 0: If y > 0, y=Cex. If
y < 0, −y=Cex, then y=−Cex.
Therefore, the general solution to the given differential equation is:
y=Cexor y=−Cex.
21
Question 23
Question
Solve the following first-order ordinary differential equation:
dy
dx =1
x+y
Solution
Step 1: Rearrange the equation by multiplying both sides by (x+y) to separate
variables:
(x+y)dy
dx = 1
Step 2: Integrate both sides with respect to x:
Z(x+y)dy =Zdx
Step 3: Integrate the left-hand side using integration by substitution (u=
x+y,du =dy):
Zu du =Zdx
u2
2=x+C
Step 4: Substitute back u=x+yinto the equation:
(x+y)2
2=x+C
Step 5: Simplify the equation:
x2+ 2xy +y2= 2x+ 2C
Step 6: Rearrange the equation into a more standard form by isolating y:
y2+ 2xy −2x= 2C−x2
Step 7: Use the quadratic formula to solve for y:
y=−2x±p(2x)2−4(1)(−2x+ 2C−x2)
2
y=−2x±√4x2+ 8x−8C+ 4x2
2
y=−2x±√8x2+ 8x−8C
2
22
y=−2x±2√x2+x−C
2
y=−x±px2+x−C
So, the general solution to the differential equation is:
y=−x±px2+x−C
Question 24
Question
Solve the following first-order ordinary differential equation:
dy
dx =3xy
x2+ 1
Solution
Step 1: 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
y=3x
x2+ 1dx
Step 2: Integrate both sides with respect to their respective variables:
Z1
ydy =Z3x
x2+ 1dx
Step 3: Solve the integrals separately:
ln |y|=3
2lnx2+ 1+C
Step 4: Simplify the expression using the properties of logarithms:
ln |y|= lnx2+ 13/2+C
Step 5: Exponentiate both sides to eliminate the natural logarithm:
|y|=eln(x2+1)3/2+C
Step 6: Use the properties of exponents to simplify the expression further:
|y|=eCeln(x2+1)3/2
Step 7: Introduce a new constant k=±eCfor convenience:
|y|=k(x2+ 1)3/2
23
Step 8: Remove the absolute value by considering the positive and negative
cases:
y=±k(x2+ 1)3/2
Therefore, the general solution to the given ordinary differential equation is
y=±k(x2+ 1)3/2, where kis an arbitrary constant.
Question 25
Question
Find the general solution to the differential equation: dy
dx =2y
x+x2.
Solution
Step 1: Rewrite the differential equation in standard form by moving all terms
involving yto the left side: dy
dx −2y
x=x2
Step 2: This is a first-order linear ordinary differential equation of the form
dy
dx +P(x)y=Q(x). Here, P(x) = −2
xand Q(x) = x2.
Step 3: To solve this differential equation, we first find the integrating factor,
denoted by I(x), which is given by:
I(x) = eRP(x)dx
Step 4: Calculate the integrating factor I(x):
I(x) = eR−2
xdx =e−2 ln |x|=eln |x−2|=|x|−2
Step 5: Multiply the integrating factor with the differential equation:
|x|−2dy
dx −2|x|−1y=x2|x|−2
Step 6: Simplify the equation:
d
dx(|x|−2y) = x
Step 7: Integrate both sides with respect to x:
|x|−2y=x2
2+C
Step 8: Multiply through by |x|2to isolate y:
y=x2
2· |x|2+C|x|2
Step 9: Simplify the expression to obtain the general solution:
y=x4
2+Cx2
where Cis an arbitrary constant.
24
Question 26
Question
Find the general solution to the differential equation dy
dx +y=xe−x.
Solution
Step 1: First, we identify the type of differential equation. This is a first-order
linear ordinary differential equation of the form dy
dx +P(x)y=Q(x), where
P(x) = 1 and Q(x) = xe−x.
Step 2: To solve this type of equation, we use an integrating factor defined
by µ(x) = eRP(x)dx. In this case, µ(x) = eR1dx =ex.
Step 3: We multiply the given equation by the integrating factor µ(x):
exdy
dx +exy=xe−xex
Step 4: Simplify the equation:
exdy
dx +exy=x
Step 5: Notice that the left side of the equation is now the derivative of the
product exy. Therefore, we can write the equation as:
d
dx(exy) = x
Step 6: Integrate both sides with respect to x:
Zd
dx(exy)dx =Zx dx
exy=1
2x2+C
Step 7: Solve for yby dividing both sides by ex:
y=1
2xe−x+Ce−x
Step 8: Thus, the general solution to the differential equation is y=1
2xe−x+
Ce−x, where Cis an arbitrary constant.
Question 27
Question
Solve the following ordinary differential equation:
y′=1
1 + x2
25
Solution
We can solve this differential equation by separating variables and then inte-
grating.
Step 1: Write the differential equation in the form dy =f(x)dx.
dy =1
1 + x2dx
Step 2: Integrate both sides.
Zdy =Z1
1 + x2dx
y= arctan(x) + C
Step 3: Therefore, the general solution to the differential equation is y=
arctan(x) + C, where Cis an arbitrary constant.
Question 28
Question
Consider the following first-order ordinary differential equation:
dy
dx =y(2x+ 1)
Find the general solution to this differential equation.
Solution
Step 1: Write the differential equation in standard form:
dy
dx =y(2x+ 1)
Step 2: Separate variables by moving all terms involving yto one side and
all terms involving xto the other side:
dy
y= (2x+ 1)dx
Step 3: Integrate both sides:
Z1
ydy =Z(2x+ 1)dx
Step 4: Solve the integrals:
ln |y|=x2+x+C
26
where Cis the constant of integration.
Step 5: Eliminate the absolute value by taking eof both sides:
|y|=ex2+x+C
y=±ex2+x+C
Step 6: Combine the constants:
y=Aex2+x
where A=±eC.
Step 7: Therefore, the general solution to the differential equation is y=
Aex2+x, where Ais an arbitrary constant.
Question 29
Question
Find the general solution to the following differential equation:
y′′ −4y′+ 4y= 0
Solution
Step 1: First, we assume a solution of the form y=ert, where ris a constant
to be determined. So, the auxiliary equation is:
r2−4r+ 4 = 0
Step 2: Solve the auxiliary equation to find the roots:
(r−2)2= 0
r= 2 (multiplicity 2)
Step 3: Since one root has multiplicity 2, the general solution is:
y=c1e2t+c2te2t
where c1and c2are arbitrary constants.
Question 30
Question
Find the particular solution to the differential equation dy
dx = 2xy given that
y(0) = 3.
27
Solution
Step 1: We start by separating variables.
dy
dx = 2xy
dy
y= 2xdx
Step 2: Integrate both sides.
Z1
ydy =Z2xdx
ln |y|=x2+C
Step 3: Rewrite the equation using the initial condition y(0) = 3.
ln |3|= 0 + C
C= ln 3
Step 4: Solve for y.
ln |y|=x2+ ln 3
y=ex2+ln 3 =ex2·eln 3 = 3ex2
So, the particular solution to the differential equation is y= 3ex2.
Question 31
Question
Solve the following initial value problem:
dy
dx =y(3 −y), y(0) = 1
Solution
Step 1: Rewrite the differential equation in the standard form:
dy
dx = 3y−y2
Step 2: Separate the variables by writing the equation as:
dy
3y−y2=dx
28
Step 3: Perform partial fraction decomposition on the left-hand side:
dy
3y−y2=A
y+B
3−ydy
Step 4: Solve for A and B by equating coefficients:
A(3 −y) + By = 1
Step 5: Solve the system of equations for A and B:
A+B= 0
3A= 1
Thus, A=1
3and B=−1
3.
Step 6: Integrate both sides of the separated equation:
Z1
3y−1
3(3 −y)dy =Zdx
Step 7: Integrate the left-hand side:
1
3ln |y| − 1
3ln |3−y|=x+C
Step 8: Combine the logarithms on the left-hand side:
ln
y
3
√3−y
= 3x+C
Step 9: Exponentiate both sides to get rid of the logarithm:
y
3
√3−y
=e3x+C
Step 10: Rewrite the equation with the absolute value removed:
y
3
√3−y=±e3xeC
Step 11: Let k=±eCfor simplicity:
y=k3
p3−y
Step 12: Plug in the initial condition y(0) = 1 to find the constant k:
1 = k3
√3−1⇒k= 1
Step 13: Substitute k= 1 back into the solution:
y=3
p3−y
29
Step 14: Solve for yby cubing both sides and rearranging:
y3+y−3=0
Step 15: Factor or use numerical methods to solve for y: The solutions are
y≈1.32472, y≈ −1.32472, and y= 1.
Therefore, the solution to the initial value problem is y= 1 for all x, or
y≈1.32472 for x > 0 and y≈ −1.32472 for x < 0.
Question 32
Question
Consider the following first-order linear ordinary differential equation:
dy
dx + (2x−1)y= 4x
Find the general solution to this differential equation.
Solution
Step 1: We first identify the integrating factor, which is given by:
µ(x) = eR(2x−1) dx
µ(x) = ex2−x
Step 2: We multiply both sides of the differential equation by the integrating
factor:
ex2−xdy
dx +ex2−x(2x−1)y=ex2−x4x
Step 3: Recognize the left side of the equation as the result of the product
rule:
d
dx(ex2−xy) = ex2−x4x
Step 4: Integrate both sides of the equation to solve for y:
Zd
dx(ex2−xy)dx =Zex2−x4x dx
ex2−xy=Z4xex2−xdx
Step 5: Integrate the right side of the equation by using substitution (let
u=x2−x):
30
ex2−xy=Z4eudu
ex2−xy= 4 Zeudu
Step 6: Continuing on with the integration, we find:
ex2−xy= 4eu+C
y= 4e−x2+x+Ce−x2+x
Thus, the general solution to the given differential equation is y= 4e−x2+x+
Ce−x2+x, where Cis a constant of integration.
Question 33
Question
Consider the initial value problem:
dy
dx =x+y
x−y, y(0) = 1
Find the particular solution to the differential equation.
Solution
Step 1: Begin by rewriting the given differential equation in a more manageable
form by separating the variables:
dy
dx =x+y
x−y=⇒(x−y)dy = (x+y)dx
Step 2: Integrate both sides:
Z(x−y)dy =Z(x+y)dx
Step 3: Integrate the left side with respect to yand the right side with
respect to x:
Zx dy −Zy dy =Zx dx +Zy dx
Step 4: Solve the integrals:
xy −y2
2=x2
2+xy +C
31
where Cis the constant of integration.
Step 5: Given the initial condition y(0) = 1, plug in x= 0 and y= 1 into
the equation:
0·1−12
2=02
2+ 0 ·1 + C=⇒C=−1
2
Step 6: Substitute C=−1
2back into the equation:
xy −y2
2=x2
2+xy −1
2
Step 7: Simplify the equation to obtain the particular solution:
y2=−x2−1
Therefore, the particular solution to the differential equation is y2=−x2−1.
Question 34
Question
Solve the following initial value problem:
y′+ 2xy =e−x2, y(0) = 1
Solution
Step 1: First, we identify the form of the differential equation as a first-order
linear differential equation. To solve it, we will use an integrating factor. The
integrating factor is given by µ(x) = eR2xdx =ex2.
Step 2: Multiply both sides of the differential equation by the integrating
factor µ(x):
ex2y′+ 2xex2y=ex2e−x2
Step 3: Rewrite the left-hand side as the derivative of the product ex2yusing
the product rule:
(ex2y)′= 1
Step 4: Integrate both sides with respect to x:
Z(ex2y)′dx =Z1dx
Step 5: Solve the integrals on both sides:
ex2y=x+C
where Cis the constant of integration.
32
Step 6: Finally, apply the initial condition y(0) = 1 to find the value of the
constant C:
e0·1 = 0 + C=⇒C= 1
Step 7: Thus, the solution to the initial value problem is:
y(x) = x+ 1
ex2
Question 35
Question
Consider the following differential equation:
x2y′′ −3xy′+ 4y= 0
Find the general solution to this second-order linear homogeneous differential
equation.
Solution
Step 1: Identify the characteristic equation by assuming a solution of the form
y=xr.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute yand its derivatives into the differential equation to
obtain the characteristic equation.
x2r(r−1)xr−2−3xrr−1+ 4xr= 0
r(r−1)xr−3rxr+ 4xr= 0
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
Step 3: Solve for the roots of the characteristic equation.
r= 2
Step 4: Write down the general solution using the roots of the characteristic
equation.
y=c1x2+c2x2ln(x)
where c1and c2are constants.
33
|y|=eCe−2
3x3
y=±eCe−2
3x3
y=Ke−2
3x3
where K=±eCis a constant.
Step 4: General solution. Therefore, the general solution to the differential
equation is:
y=Ke−2
3x3
where Kis an arbitrary constant.
Question 2
Question
Find the general solution to the differential equation: dy
dx =1
x2+y2.
Solution
Step 1: Rearrange the differential equation as dy
dx =1
x2+y2.
Step 2: Multiply both sides by dx to separate variables: dx = (x2+y2)dy.
Step 3: Integrate both sides with respect to x:Rdx =R(x2+y2)dy.
Step 4: Solving the integrals, we get x=x3
3+y2+C, where Cis the constant
of integration.
Step 5: Rearranging the equation, we find the general solution: y2=x3
3−
x+C.
Therefore, the general solution to the differential equation dy
dx =1
x2+y2is
y2=x3
3−x+C, where Cis a constant.
Question 3
Question
Consider the initial value problem dy
dx = 2x3+ 3y2with y(0) = 1. Find the
particular solution to this initial value problem.
Solution
Step 1: First, let’s rewrite the differential equation as
dy
dx = 2x3+ 3y2.
2
Step 2: Notice that this is a first-order ordinary differential equation that is
separable. We can rewrite the equation as
dy
dx = 2x3+ 3y2=⇒dy
3y2= 2x3dx.
Step 3: Integrating both sides, we get
Z1
3y2dy =Z2x3dx.
Step 4: This gives us
−1
3y=1
2x4+C,
where Cis the constant of integration.
Step 5: Now, we use the initial condition y(0) = 1 to find the particular
solution. Plugging in x= 0 and y= 1 into the equation, we have
−1
3=C.
Step 6: Substituting C=−1
3back into the equation, we have
−1
3y=1
2x4−1
3.
Step 7: Solving for y, we get
y=−3
2x4+ 1.
Therefore, the particular solution to the initial value problem is y=−3
2x4+1 .
Question 4
Question
Consider the following first-order ordinary differential equation:
dy
dx =1−2y
x+y
Use the method of integrating factors to solve this differential equation.
Solution
Step 1: Rewrite the differential equation in the standard form M(x, y)dx +
N(x, y)dy = 0 where Mand Nare functions of xand y.
dy
dx +2y−1
x+y= 0
3
Step 2: Identify Mand N:
M(x, y)=1, N(x, y)=2y−1
Step 3: Compute the integrating factor, denoted by I(x), using the formula:
I(x) = eR(N−My)dx
Step 4: Calculate I(x):
I(x) = eR(2y−1−0)dx =eR(2y−1)dx =e2yx−x
Step 5: Multiply the original differential equation by the integrating factor
I(x):
e2yx−xdy
dx + 2y−1= 0
Step 6: Rewrite the above expression as the derivative of a product:
d
dx(e2yx−xy)=0
Step 7: Integrate both sides with respect to x:
Zd
dx(e2yx−xy)dx =Z0dx
e2yx−xy=C
Step 8: Solve for y:
y=C
e2yx−x
Therefore, the general solution to the given ODE is y=C
e2yx−x, where Cis
an arbitrary constant.
Question 5
Question
Consider the initial value problem:
dy
dx = 2xe−y, y(0) = 1.
Find the solution to the initial value problem.
4
Solution
Step 1: Rewrite the differential equation in the form dy =f(x)dx.
dy
dx = 2xe−y
dy = 2xe−ydx
Step 2: Separate the variables by dividing both sides by e−y.
dy
e−y= 2xdx
Step 3: Integrate both sides with respect to x.
Zeydy =Z2xdx
ey=x2+C1
Step 4: Use the initial condition y(0) = 1 to find the value of the constant
C1.
e1= 02+C1
e=C1
Step 5: Substitute C1=eback into the equation to find the particular
solution.
ey=x2+e
y= lnx2+e
Therefore, the solution to the initial value problem is y= lnx2+e, where
eis Euler’s number.
Question 6
Question
Find the general solution of the differential equation: dy
dx +y=e−x.
Solution
To solve the given first-order linear differential equation, we will use the inte-
grating factor method.
Step 1: Write the given differential equation in standard form dy
dx +P(x)y=
Q(x). Here P(x) = 1 and Q(x) = e−x.
Step 2: Find the integrating factor I(x) = eRP(x)dx.
Integrating factor I(x) = eR1dx =ex
5
Step 3: Multiply both sides of the differential equation by the integrating
factor I(x).
exdy
dx +exy=exe−x
Step 4: Rewrite the left-hand side as the derivative of the product of yand
the integrating factor ex.d
dx(exy) = exe−x
Step 5: Integrate both sides with respect to x.
Zd
dx(exy)dx =Zexe−xdx
exy=Z1dx =x+C
Step 6: Solve for y.
y=x+C
ex
Therefore, the general solution of the given differential equation is y=x+C
ex,
where Cis an arbitrary constant.
Question 7
Question
Find the general solution to the differential equation dy
dx =y2−1
x2+1 .
Solution
Step 1: Rewrite the differential equation in a separable form:
dy
y2−1=dx
x2+ 1
Step 2: Integrate both sides with respect to yand x:
Z1
y2−1dy =Z1
x2+ 1 dx
Step 3: Factor the denominator in the left integral to split it into partial
fractions: Z1
2·1
y−1−1
2·1
y+ 1dy =Z1
x2+ 1 dx
Step 4: Integrate both sides with respect to yand x:
1
2ln |y−1| − 1
2ln |y+ 1|= arctan(x) + C
6
Step 5: Combine the logarithmic terms on the left side using properties of
logarithms:
ln
y−1
√y+ 1
= 2 arctan(x) + C
Step 6: Rewrite the equation in an exponential form:
y−1
√y+ 1
=e2 arctan(x)+C
Step 7: Remove the absolute value on the left side by considering both
positive and negative cases:
y−1
√y+ 1 =±e2 arctan(x)+C
Step 8: Solve for yby squaring both sides and rearranging the terms:
y= 1 + ±e2 arctan(x)+C(y+ 1)
√y+ 1
The general solution to the differential equation is:
y= 1 + ±e2 arctan(x)+C(y+ 1)
√y+ 1
Question 8
Question
Find the general solution to the following differential equation:
y′′ −4y′+ 4y= 0
Solution
To solve the differential equation y′′ −4y′+ 4y= 0, we will first find the
characteristic equation by assuming the form y=ert.
Step 1: Find the characteristic equation Substitute y=ert into the
differential equation:
(r2−4r+ 4)ert = 0
This simplifies to:
(r−2)2= 0
So the characteristic equation is:
r= 2,2
7
Step 2: Find the general solution Since we have repeated real roots,
the general solution is of the form:
y(t) = c1e2t+c2te2t
where c1and c2are arbitrary constants.
Therefore, the general solution to the differential equation y′′ −4y′+ 4y= 0
is:
y(t) = c1e2t+c2te2t
Question 9
Question
Consider the following ordinary differential equation:
y′′ −2y′+y=et
Find the general solution to the differential equation by first solving the
associated homogeneous equation and then finding a particular solution.
Solution
Step 1: First, let’s find the general solution to the associated homogeneous
equation:
y′′ −2y′+y= 0
We assume y=ert for some constant r. Substitute this into the homoge-
neous equation to get:
r2ert −2rert +ert = 0
Divide through by ert:
r2−2r+ 1 = 0
This is a quadratic equation with solutions:
r= 1,1
So, the general solution to the homogeneous equation is:
yh=c1et+c2tet
Step 2: Now, let’s find the particular solution to the nonhomogeneous equa-
tion: Since etis a solution to the homogeneous equation, we try a particular
solution of the form:
yp=Atet
Substitute ypinto the nonhomogeneous equation and solve for A:
(Atet)′′ −2(Atet)′+Atet=et
8
(Aet+ 2Atet−2Aet+Atet) = et
(2Atet) = et
A=1
2
So, the particular solution is:
yp=1
2tet
Step 3: The general solution to the differential equation is the sum of the
general solution to the homogeneous equation and the particular solution:
y=yh+yp=c1et+c2tet+1
2tet
Therefore, the general solution is:
y=c1et+ (c2+1
2)tet
Question 10
Question
Consider the following first-order ordinary differential equation:
dy
dx =e−x−y
Find the particular solution to this differential equation given the initial condi-
tion y(0) = 3.
Solution
To solve this first-order ordinary differential equation, we can use the method
of integrating factors.
Step 1: Rewrite the equation in the standard form dy
dx +P(x)y=Q(x):
dy
dx +y=e−x
Step 2: Identify P(x) and Q(x) from the standard form:
P(x)=1, Q(x) = e−x
Step 3: Calculate the integrating factor I(x) = eRP(x)dx:
I(x) = eR1dx =ex
9
Step 4: Multiply both sides of the standard form by the integrating factor:
exdy
dx +exy=e2x
Step 5: Rewrite the left side as the derivative of the product exy:
d
dx(exy) = e2x
Step 6: Integrate both sides of the equation:
Zd
dx(exy)dx =Ze2xdx
exy=e2x
2+C
y=1
2+Ce−x
Step 7: Apply the initial condition y(0) = 3 to find the particular solution:
3 = 1
2+C
C=5
2
Step 8: The particular solution to the differential equation is:
y=1
2+5
2e−x
Question 11
Question
Consider the differential equation
dy
dx =2x2+y−3
2y−x
Solve the initial value problem with y(0) = 1.
Solution
To solve this differential equation, we can rewrite it in the form:
(2y−x)dy = (2x2+y−3) dx
10
Step 1: Integrate both sides of the equation.
Z(2y−x)dy =Z(2x2+y−3) dx
Step 2: Integrate the left side with respect to y.
y2−1
2x2=Z(2x2+y−3) dx
Step 3: Integrate the right side with respect to x.
y2−1
2x2=2
3x3+1
2xy −3x+c
Step 4: Simplify the equation by substituting y(0) = 1.
1−0 = 2
3(0)3+1
2(0)(1) −3(0) + c=⇒c= 1
Step 5: Thus, the solution to the initial value problem is:
y2−1
2x2=2
3x3+1
2xy −3x+ 1
Question 12
Question
Find the general solution of the differential equation: dy
dx =2x2+1
y.
Solution
Step 1: Rearrange the differential equation to separate variables.
dy
dx =2x2+ 1
y.
y dy = (2x2+ 1) dx.
Step 2: Integrate both sides of the equation.
Zy dy =Z(2x2+ 1) dx.
Step 3: Solve the integrals.
y2
2=2x3
3+x+C,
where Cis the constant of integration.
11
Step 4: Multiply through by 2 to simplify.
y2=4x3
3+ 2x+ 2C.
Step 5: Take the square root of both sides to find the general solution.
y=±r4x3
3+ 2x+ 2C.
Therefore, the general solution of the differential equation is y=±q4x3
3+ 2x+ 2C.
Question 13
Question
Find the general solution to the following first-order ordinary differential equa-
tion: dy
dx =y+ex
y−ex
Solution
Step 1: Rewrite the given differential equation in a more suitable form by sep-
arating variables. We can rewrite the equation as:
(y−ex)dy = (y+ex)dx
Step 2: Integrate both sides of the equation. The left side can be integrated
as: Z(y−ex)dy =1
2y2−exy+C1
Step 3: Similarly, the right side can be integrated as:
Z(y+ex)dx =y+ex+C2
Step 4: Equating the two integrals, we have:
1
2y2−exy=y+ex+C
Step 5: Rearranging the terms, we get:
1
2y2−y= 3ex+C
Step 6: Finally, the general solution is given by:
y(x) = 1 ±√1+6ex+ 2C
where Cis an arbitrary constant.
12
Question 14
Question
Consider the differential equation y′′ +y= 0. Solve the differential equation
subject to the initial conditions y(0) = 3 and y′(0) = 1.
Solution
Step 1: Write down the characteristic equation of the differential equation
y′′ +y= 0. The characteristic equation is obtained by replacing y′′ with r2,
which gives us r2+ 1 = 0.
Step 2: Solve the characteristic equation r2+ 1 = 0. We have r2=−1, so
r=±√−1 = ±i.
Step 3: General solution of the differential equation. The general solution
of the differential equation is given by y(t) = c1cos(t) + c2sin(t), where c1and
c2are arbitrary constants.
Step 4: Apply the initial conditions y(0) = 3 and y′(0) = 1. Substitute
t= 0 into the general solution and its derivative to get:
y(0) = c1cos(0) + c2sin(0) = c1= 3
y′(0) = −c1sin(0) + c2cos(0) = c2= 1
Step 5: Write down the particular solution. Thus, the particular solution
of the differential equation satisfying the initial conditions is:
y(t) = 3 cos(t) + sin(t)
Question 15
Question
Find the general solution to the differential equation:
y′′ −4y′+ 5y=e2x
Solution
Step 1: First, find the complementary solution by solving the homogeneous
equation y′′ −4y′+ 5y= 0. Let’s assume the solution has the form y=erx.
Substitute y=erx into the homogeneous equation:
r2erx −4rerx + 5erx = 0
Simplify and factor out erx:
erx(r2−4r+ 5) = 0
13
Since erx is never zero, we have:
r2−4r+ 5 = 0
Step 2: Solve the characteristic equation r2−4r+5 = 0. Using the quadratic
formula r=−b±√b2−4ac
2awith a= 1, b=−4, and c= 5:
r=4±p(−4)2−4(1)(5)
2(1)
r=4±√16 −20
2
r=4±√−4
2
r= 2 ±2i
Step 3: Write down the complementary solution. Since the roots are com-
plex, the complementary solution is:
yc=e2x(Acos(2x) + Bsin(2x))
Step 4: Find a particular solution. Since the right-hand side is e2x, let’s try
a particular solution of the form yp=x(Ae2x). Substitute ypinto the original
differential equation and solve for A:
2A−4(2A) + 5(xAe2x) = e2x
2A−8A+ 5xA = 1
5xA −6A= 1
A=−1
6x
Step 5: Write down the full particular solution. The particular solution is:
yp=−1
6xe2x
Step 6: Write down the general solution. The general solution is the sum
of the complementary and particular solutions:
y=yc+yp=e2x(Acos(2x) + Bsin(2x)) −1
6xe2x
Question 16
Question
Solve the following ordinary differential equation:
dy
dx =y2−x2
14
Solution
Step 1: Rewrite the given differential equation in the form dy
dx =f(x, y).
dy
dx =y2−x2
Step 2: Recognize that this is a separable differential equation, so we can
write it as: dy
y2−x2=dx
Step 3: Integrate both sides with respect to y, then solve for y.
Zdy
y2−x2=Zdx
Step 4: To integrate the left side, we need to perform partial fraction de-
composition. Let’s write:
1
y2−x2=A
y+x+B
y−x
Step 5: Solving for Aand Bby multiplying both sides by y2−x2, we get:
1 = A(y−x) + B(y+x)
Step 6: Set y=xto solve for A.
1 = Ax −Bx
A=−B
Step 7: Set y=−xto solve for B.
1 = −Ax −Bx
B=−A= 1
Step 8: Substituting the values of Aand Bback into the partial fraction
decomposition gives us:
1
y2−x2=1
21
y+x−1
y−x
Step 9: Plugging this back into the integral and integrating, we get:
1
2ln
y+x
y−x
=x+C
Step 10: Exponentiate both sides to eliminate the natural logarithm.
y+x
y−x
=e2x+2C
15
Step 11: Simplify the absolute value terms by considering positive and neg-
ative cases. y+x
y−x=±e2x+2C
Step 12: Rearrange to solve for y.
y+x=±e2x+2C(y−x)
y+x=±e2x+2Cy∓e2x+2Cx
y(1 ∓e2x+2C) = x(1 ±e2x+2C)
y=x(1 ±e2x+2C)
1∓e2x+2C
Therefore, the general solution to the ordinary differential equation dy
dx =
y2−x2is:
y=x(1 ±e2x+2C)
1∓e2x+2C
Question 17
Question
Consider the following initial value problem:
y′= 2xy +x, y(0) = 1.
Find the solution y(x) for x≥0.
Solution
Step 1: First, let’s find the general solution to the differential equation.
dy
dx = 2xy +x
dy
dx −2xy =x
This is a first-order linear ordinary differential equation. The integrating factor
is eR−2x dx =e−x2. Multiplying through by the integrating factor, we have:
e−x2dy
dx −2xye−x2=xe−x2
d
dx(ye−x2) = xe−x2
Integrating both sides gives:
ye−x2=Zxe−x2dx
16
ye−x2=−1
2e−x2+C
y=−1
2+Cex2
where Cis an arbitrary constant.
Step 2: Next, we use the initial condition y(0) = 1 to determine the value of
C.
1 = −1
2+C=⇒C=3
2
So, the solution to the initial value problem is:
y(x) = −1
2+3
2ex2=1
2(3ex2−1), x ≥0.
Question 18
Question
Consider the ordinary differential equation:
dy
dx = 2x2y
Find the general solution to this differential equation.
Solution
Step 1: Rewrite the differential equation in separated variables form:
dy
y= 2x2dx
Step 2: Integrate both sides:
Zdy
y=Z2x2dx
Step 3: Evaluate the integrals:
ln |y|=2
3x3+C
Step 4: Solve the equation for yby exponentiating both sides:
|y|=e2
3x3+C
Step 5: Simplify the equation by combining the constants of integration:
|y|=Ke2
3x3where K=eC
Step 6: Consider both cases for the absolute value:
17
y=Ke2
3x3or y=−Ke2
3x3
Therefore, the general solution to the differential equation is:
y=Ke2
3x3or y=−Ke2
3x3
Question 19
Question
Consider the initial value problem:
y′+ 2y=e−x, y(0) = 1.
Find the particular solution to the differential equation that satisfies the initial
condition.
Solution
Step 1: First, we find the integrating factor. The integrating factor is given by
µ(x) = eR2dx =e2x.
Step 2: Multiply both sides of the differential equation by the integrating
factor:
e2xy′+ 2e2xy=e2xe−x.
Step 3: Rewrite the left side as the derivative of the product e2xyusing the
product rule: d
dx(e2xy) = e2xe−x.
Step 4: Integrate both sides with respect to x:
e2xy=−e−x+C,
where Cis the constant of integration.
Step 5: Apply the initial condition y(0) = 1 to solve for the constant C:
e0·1 = −e0+C⇒1 = −1 + C⇒C= 2.
Step 6: Substitute the value of Cback into the equation to find the particular
solution:
e2xy=−e−x+ 2 ⇒y=−e−3x+ 2e−2x.
Therefore, the particular solution to the differential equation that satisfies
the initial condition is y=−e−3x+ 2e−2x.
18
Question 20
Question
Consider the ordinary differential equation
dy
dx =−y
x+x2
with initial condition y(1) = 2. Find the particular solution to the differential
equation.
Solution
Step 1: Let’s first rewrite the differential equation in a more standard form, by
separating variables: dy
dx +y
x=x2
Step 2: This is a first-order linear differential equation, which we can solve
using an integrating factor. In this case, the integrating factor is eR1
xdx =
eln |x|=|x|.
Step 3: Multiplying both sides of the equation by the integrating factor |x|
gives:
|x|dy
dx +y=x3
Step 4: Now we integrate both sides with respect to x:
Z|x|dy
dxdx +Zydx =Zx3dx
Step 5: Integrating both sides gives:
|x|y=1
4x4+C
Step 6: Now we apply the initial condition y(1) = 2 to find the value of the
constant C:
|1| · 2 = 1
4·14+C
2 = 1
4+C
C=7
4
Step 7: Therefore, the particular solution to the differential equation is:
|x|y=1
4x4+7
4
19
Question 21
Question
Solve the following initial value problem:
dy
dx =1
x2+y2, y(1) = 0
Solution
Step 1: Rewrite the differential equation in a more standard form by separating
variables. dy
dx =1
x2+y2=⇒dy
dx =1
x2(1 + (y/x)2)
Step 2: Substituting v=y/x, we have
y=vx =⇒dy
dx =v+xdv
dx
Step 3: Substituting into the differential equation and simplifying, we get
v+xdv
dx =1
x2(1 + v2)=⇒xdv
dx =1−v2
x(1 + v2)
Step 4: Separate variables and integrate both sides:
Z1 + v2
1−v2dv =Z1
xdx
Step 5: Simplify the left-hand side by partial fraction decomposition and
integrate:
Z1 + v2
1−v2dv =Z1
2(1 −v)−1
2(1 + v)dv
Step 6: Integrating both sides gives
1
2ln |1−v| − 1
2ln |1 + v|= ln |x|+C
Step 7: Exponentiate both sides to solve for v:
ln
1−v
1 + v
= 2 ln |x|+C′=⇒1−v
1 + v=Cx2
Step 8: Solve for vin terms of xby manipulating the equation:
v=1−Cx2
1 + Cx2
Step 9: Using the initial condition y(1) = 0, we find C=1
2.
Step 10: Substitute C=1
2back into the equation for vto get the solution
in terms of x:
y(x) = x1−1
2x2
1 + 1
2x2=x(x2−1)
x2+ 2
20
Question 22
Question
Consider the first-order ordinary differential equation:
y′+y2=xy.
Find the general solution of the given differential equation.
Solution
Step 1: Rewrite the equation in standard form:
y′=xy −y2.
Step 2: Factor out yon the right-hand side:
y′=y(x−y).
Step 3: Separate variables by dividing both sides by y(x−y):
dy
y=dx
x−y.
Step 4: Integrate both sides:
Z1
ydy =Z1
x−ydx.
Step 5: Solve the integrals:
ln |y|= ln |x−y|+C,
where Cis the constant of integration.
Step 6: Rewrite the natural logarithms using properties of logarithms:
ln |y|= ln |x−y|+ ln |eC|.
Step 7: Combine the logarithms on the right-hand side:
ln |y|= ln |eC(x−y)|.
Step 8: Use the property of logarithms ln(a) = ln(b) if and only if a=b:
|y|=eC(x−y).
Step 9: Simplify and solve for y:
|y|=Cex−Cy,
where C=±eC.
Step 10: Consider the two cases y > 0 and y < 0: If y > 0, y=Cex. If
y < 0, −y=Cex, then y=−Cex.
Therefore, the general solution to the given differential equation is:
y=Cexor y=−Cex.
21
Question 23
Question
Solve the following first-order ordinary differential equation:
dy
dx =1
x+y
Solution
Step 1: Rearrange the equation by multiplying both sides by (x+y) to separate
variables:
(x+y)dy
dx = 1
Step 2: Integrate both sides with respect to x:
Z(x+y)dy =Zdx
Step 3: Integrate the left-hand side using integration by substitution (u=
x+y,du =dy):
Zu du =Zdx
u2
2=x+C
Step 4: Substitute back u=x+yinto the equation:
(x+y)2
2=x+C
Step 5: Simplify the equation:
x2+ 2xy +y2= 2x+ 2C
Step 6: Rearrange the equation into a more standard form by isolating y:
y2+ 2xy −2x= 2C−x2
Step 7: Use the quadratic formula to solve for y:
y=−2x±p(2x)2−4(1)(−2x+ 2C−x2)
2
y=−2x±√4x2+ 8x−8C+ 4x2
2
y=−2x±√8x2+ 8x−8C
2
22
y=−2x±2√x2+x−C
2
y=−x±px2+x−C
So, the general solution to the differential equation is:
y=−x±px2+x−C
Question 24
Question
Solve the following first-order ordinary differential equation:
dy
dx =3xy
x2+ 1
Solution
Step 1: 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
y=3x
x2+ 1dx
Step 2: Integrate both sides with respect to their respective variables:
Z1
ydy =Z3x
x2+ 1dx
Step 3: Solve the integrals separately:
ln |y|=3
2lnx2+ 1+C
Step 4: Simplify the expression using the properties of logarithms:
ln |y|= lnx2+ 13/2+C
Step 5: Exponentiate both sides to eliminate the natural logarithm:
|y|=eln(x2+1)3/2+C
Step 6: Use the properties of exponents to simplify the expression further:
|y|=eCeln(x2+1)3/2
Step 7: Introduce a new constant k=±eCfor convenience:
|y|=k(x2+ 1)3/2
23
Step 8: Remove the absolute value by considering the positive and negative
cases:
y=±k(x2+ 1)3/2
Therefore, the general solution to the given ordinary differential equation is
y=±k(x2+ 1)3/2, where kis an arbitrary constant.
Question 25
Question
Find the general solution to the differential equation: dy
dx =2y
x+x2.
Solution
Step 1: Rewrite the differential equation in standard form by moving all terms
involving yto the left side: dy
dx −2y
x=x2
Step 2: This is a first-order linear ordinary differential equation of the form
dy
dx +P(x)y=Q(x). Here, P(x) = −2
xand Q(x) = x2.
Step 3: To solve this differential equation, we first find the integrating factor,
denoted by I(x), which is given by:
I(x) = eRP(x)dx
Step 4: Calculate the integrating factor I(x):
I(x) = eR−2
xdx =e−2 ln |x|=eln |x−2|=|x|−2
Step 5: Multiply the integrating factor with the differential equation:
|x|−2dy
dx −2|x|−1y=x2|x|−2
Step 6: Simplify the equation:
d
dx(|x|−2y) = x
Step 7: Integrate both sides with respect to x:
|x|−2y=x2
2+C
Step 8: Multiply through by |x|2to isolate y:
y=x2
2· |x|2+C|x|2
Step 9: Simplify the expression to obtain the general solution:
y=x4
2+Cx2
where Cis an arbitrary constant.
24
Question 26
Question
Find the general solution to the differential equation dy
dx +y=xe−x.
Solution
Step 1: First, we identify the type of differential equation. This is a first-order
linear ordinary differential equation of the form dy
dx +P(x)y=Q(x), where
P(x) = 1 and Q(x) = xe−x.
Step 2: To solve this type of equation, we use an integrating factor defined
by µ(x) = eRP(x)dx. In this case, µ(x) = eR1dx =ex.
Step 3: We multiply the given equation by the integrating factor µ(x):
exdy
dx +exy=xe−xex
Step 4: Simplify the equation:
exdy
dx +exy=x
Step 5: Notice that the left side of the equation is now the derivative of the
product exy. Therefore, we can write the equation as:
d
dx(exy) = x
Step 6: Integrate both sides with respect to x:
Zd
dx(exy)dx =Zx dx
exy=1
2x2+C
Step 7: Solve for yby dividing both sides by ex:
y=1
2xe−x+Ce−x
Step 8: Thus, the general solution to the differential equation is y=1
2xe−x+
Ce−x, where Cis an arbitrary constant.
Question 27
Question
Solve the following ordinary differential equation:
y′=1
1 + x2
25
Solution
We can solve this differential equation by separating variables and then inte-
grating.
Step 1: Write the differential equation in the form dy =f(x)dx.
dy =1
1 + x2dx
Step 2: Integrate both sides.
Zdy =Z1
1 + x2dx
y= arctan(x) + C
Step 3: Therefore, the general solution to the differential equation is y=
arctan(x) + C, where Cis an arbitrary constant.
Question 28
Question
Consider the following first-order ordinary differential equation:
dy
dx =y(2x+ 1)
Find the general solution to this differential equation.
Solution
Step 1: Write the differential equation in standard form:
dy
dx =y(2x+ 1)
Step 2: Separate variables by moving all terms involving yto one side and
all terms involving xto the other side:
dy
y= (2x+ 1)dx
Step 3: Integrate both sides:
Z1
ydy =Z(2x+ 1)dx
Step 4: Solve the integrals:
ln |y|=x2+x+C
26
where Cis the constant of integration.
Step 5: Eliminate the absolute value by taking eof both sides:
|y|=ex2+x+C
y=±ex2+x+C
Step 6: Combine the constants:
y=Aex2+x
where A=±eC.
Step 7: Therefore, the general solution to the differential equation is y=
Aex2+x, where Ais an arbitrary constant.
Question 29
Question
Find the general solution to the following differential equation:
y′′ −4y′+ 4y= 0
Solution
Step 1: First, we assume a solution of the form y=ert, where ris a constant
to be determined. So, the auxiliary equation is:
r2−4r+ 4 = 0
Step 2: Solve the auxiliary equation to find the roots:
(r−2)2= 0
r= 2 (multiplicity 2)
Step 3: Since one root has multiplicity 2, the general solution is:
y=c1e2t+c2te2t
where c1and c2are arbitrary constants.
Question 30
Question
Find the particular solution to the differential equation dy
dx = 2xy given that
y(0) = 3.
27
Solution
Step 1: We start by separating variables.
dy
dx = 2xy
dy
y= 2xdx
Step 2: Integrate both sides.
Z1
ydy =Z2xdx
ln |y|=x2+C
Step 3: Rewrite the equation using the initial condition y(0) = 3.
ln |3|= 0 + C
C= ln 3
Step 4: Solve for y.
ln |y|=x2+ ln 3
y=ex2+ln 3 =ex2·eln 3 = 3ex2
So, the particular solution to the differential equation is y= 3ex2.
Question 31
Question
Solve the following initial value problem:
dy
dx =y(3 −y), y(0) = 1
Solution
Step 1: Rewrite the differential equation in the standard form:
dy
dx = 3y−y2
Step 2: Separate the variables by writing the equation as:
dy
3y−y2=dx
28
Step 3: Perform partial fraction decomposition on the left-hand side:
dy
3y−y2=A
y+B
3−ydy
Step 4: Solve for A and B by equating coefficients:
A(3 −y) + By = 1
Step 5: Solve the system of equations for A and B:
A+B= 0
3A= 1
Thus, A=1
3and B=−1
3.
Step 6: Integrate both sides of the separated equation:
Z1
3y−1
3(3 −y)dy =Zdx
Step 7: Integrate the left-hand side:
1
3ln |y| − 1
3ln |3−y|=x+C
Step 8: Combine the logarithms on the left-hand side:
ln
y
3
√3−y
= 3x+C
Step 9: Exponentiate both sides to get rid of the logarithm:
y
3
√3−y
=e3x+C
Step 10: Rewrite the equation with the absolute value removed:
y
3
√3−y=±e3xeC
Step 11: Let k=±eCfor simplicity:
y=k3
p3−y
Step 12: Plug in the initial condition y(0) = 1 to find the constant k:
1 = k3
√3−1⇒k= 1
Step 13: Substitute k= 1 back into the solution:
y=3
p3−y
29
Step 14: Solve for yby cubing both sides and rearranging:
y3+y−3=0
Step 15: Factor or use numerical methods to solve for y: The solutions are
y≈1.32472, y≈ −1.32472, and y= 1.
Therefore, the solution to the initial value problem is y= 1 for all x, or
y≈1.32472 for x > 0 and y≈ −1.32472 for x < 0.
Question 32
Question
Consider the following first-order linear ordinary differential equation:
dy
dx + (2x−1)y= 4x
Find the general solution to this differential equation.
Solution
Step 1: We first identify the integrating factor, which is given by:
µ(x) = eR(2x−1) dx
µ(x) = ex2−x
Step 2: We multiply both sides of the differential equation by the integrating
factor:
ex2−xdy
dx +ex2−x(2x−1)y=ex2−x4x
Step 3: Recognize the left side of the equation as the result of the product
rule:
d
dx(ex2−xy) = ex2−x4x
Step 4: Integrate both sides of the equation to solve for y:
Zd
dx(ex2−xy)dx =Zex2−x4x dx
ex2−xy=Z4xex2−xdx
Step 5: Integrate the right side of the equation by using substitution (let
u=x2−x):
30
ex2−xy=Z4eudu
ex2−xy= 4 Zeudu
Step 6: Continuing on with the integration, we find:
ex2−xy= 4eu+C
y= 4e−x2+x+Ce−x2+x
Thus, the general solution to the given differential equation is y= 4e−x2+x+
Ce−x2+x, where Cis a constant of integration.
Question 33
Question
Consider the initial value problem:
dy
dx =x+y
x−y, y(0) = 1
Find the particular solution to the differential equation.
Solution
Step 1: Begin by rewriting the given differential equation in a more manageable
form by separating the variables:
dy
dx =x+y
x−y=⇒(x−y)dy = (x+y)dx
Step 2: Integrate both sides:
Z(x−y)dy =Z(x+y)dx
Step 3: Integrate the left side with respect to yand the right side with
respect to x:
Zx dy −Zy dy =Zx dx +Zy dx
Step 4: Solve the integrals:
xy −y2
2=x2
2+xy +C
31
where Cis the constant of integration.
Step 5: Given the initial condition y(0) = 1, plug in x= 0 and y= 1 into
the equation:
0·1−12
2=02
2+ 0 ·1 + C=⇒C=−1
2
Step 6: Substitute C=−1
2back into the equation:
xy −y2
2=x2
2+xy −1
2
Step 7: Simplify the equation to obtain the particular solution:
y2=−x2−1
Therefore, the particular solution to the differential equation is y2=−x2−1.
Question 34
Question
Solve the following initial value problem:
y′+ 2xy =e−x2, y(0) = 1
Solution
Step 1: First, we identify the form of the differential equation as a first-order
linear differential equation. To solve it, we will use an integrating factor. The
integrating factor is given by µ(x) = eR2xdx =ex2.
Step 2: Multiply both sides of the differential equation by the integrating
factor µ(x):
ex2y′+ 2xex2y=ex2e−x2
Step 3: Rewrite the left-hand side as the derivative of the product ex2yusing
the product rule:
(ex2y)′= 1
Step 4: Integrate both sides with respect to x:
Z(ex2y)′dx =Z1dx
Step 5: Solve the integrals on both sides:
ex2y=x+C
where Cis the constant of integration.
32
Step 6: Finally, apply the initial condition y(0) = 1 to find the value of the
constant C:
e0·1 = 0 + C=⇒C= 1
Step 7: Thus, the solution to the initial value problem is:
y(x) = x+ 1
ex2
Question 35
Question
Consider the following differential equation:
x2y′′ −3xy′+ 4y= 0
Find the general solution to this second-order linear homogeneous differential
equation.
Solution
Step 1: Identify the characteristic equation by assuming a solution of the form
y=xr.
y=xr
y′=rxr−1
y′′ =r(r−1)xr−2
Step 2: Substitute yand its derivatives into the differential equation to
obtain the characteristic equation.
x2r(r−1)xr−2−3xrr−1+ 4xr= 0
r(r−1)xr−3rxr+ 4xr= 0
r2−r−3r+ 4 = 0
r2−4r+ 4 = 0
(r−2)2= 0
Step 3: Solve for the roots of the characteristic equation.
r= 2
Step 4: Write down the general solution using the roots of the characteristic
equation.
y=c1x2+c2x2ln(x)
where c1and c2are constants.
33
