MATH 334 - DIFFERENTIAL
EQUATIONS - Ordinary differential
equations
Question Bank - Set 2
Liberty University
Question 1
Question
Find the general solution to the following first-order ordinary differential equa-
tion:
y′+ 2y= 3e−x
Solution
Step 1: First, we write the equation in standard form. The standard form for a
first-order linear ODE is y′+P(x)y=Q(x) where P(x) and Q(x) are functions
of x. Comparing with the given ODE, we have P(x) = 2 and Q(x)=3e−x.
Step 2: Next, we find the integrating factor, which is given by eRP(x)dx. In
this case, the integrating factor is eR2dx =e2x.
Step 3: We multiply the integrating factor to the entire ODE to obtain:
e2xy′+ 2e2xy= 3e2xe−x
Step 4: Simplify the equation to get:
(e2xy)′= 3ex
Step 5: Integrate both sides with respect to xto get:
e2xy= 3 Zexdx
Step 6: Solve the integral on the right hand side to obtain:
e2xy= 3ex+C
Step 7: Finally, divide by e2xto get the general solution:
y=3ex+Ce−2x
e2x= 3e−x+Ce−2x
where Cis an arbitrary constant.
Question 2
Question
Find the general solution to the following first-order ordinary differential equa-
tion:
y′=1
2y+e2x
Solution
To find the general solution to the given first-order ordinary differential equa-
tion, we will use the method of integrating factors.
Step 1: Identify the variables The given differential equation is in the
form y′+P(x)y=Q(x), where P(x) = 1
2and Q(x) = e2x.
Step 2: Find the integrating factor The integrating factor I(x) is given
by:
I(x) = eRP(x)dx
I(x) = eR1
2dx
I(x) = e1
2x
Step 3: Multiply the integrating factor by the differential equation
Multiply both sides of the differential equation by the integrating factor I(x):
e1
2xy′+1
2e1
2xy=e2xe1
2x
e1
2xy′+1
2e1
2xy=e5
2x
Step 4: Rewrite the left-hand side as a derivative Rewrite the left-
hand side as the derivative of the product ye 1
2x:
e1
2xy′=e5
2x
Step 5: Integrate both sides Integrate both sides of the equation with
respect to x:
Ze1
2xy′dx =Ze5
2xdx
e1
2xy=2
5e5
2x+C
2
Step 6: Solve for ySolving for y, we have:
y=2
5e3x+Ce−1
2x
Therefore, the general solution to the given differential equation is:
y=2
5e3x+Ce−1
2x
where Cis an arbitrary constant.
Question 3
Question
Consider the differential equation:
y′+ 2xy = 4x3
Find the general solution of the differential equation.
Solution
Step 1: First, we need to recognize that the given differential equation is a linear
first-order differential equation in standard form:
dy
dx + 2xy = 4x3
Step 2: Next, we identify the integrating factor which is given by eRP(x)dx,
where P(x) is the coefficient of y.
Step 3: In our case, P(x) = 2x. Therefore, the integrating factor is given by
eR2xdx =ex2.
Step 4: Multiplying both sides of the differential equation by the integrating
factor gives:
ex2dy
dx + 2xex2y= 4x3ex2
Step 5: The left side of the equation can be rewritten using the product rule
for differentiation as:
d
dx(ex2y)=4x3ex2
Step 6: Integrating both sides of the equation with respect to xgives:
Zd
dx(ex2y)dx =Z4x3ex2dx
Step 7: Integrating the right side and simplifying gives:
3
ex2y=Z4x3ex2dx =ex2(x2−1) + C
where Cis the constant of integration.
Step 8: Solving for y, we get the general solution of the differential equation
as:
y=x2−1 + Ce−x2
where Cis an arbitrary constant.
Question 4
Question
Solve the following ordinary differential equation using the method of undeter-
mined coefficients:
y′′ −5y′+ 6y= 4e3x
Solution
To solve the given second-order homogeneous differential equation using the
method of undetermined coefficients, we assume the particular solution has the
form yp=Ae3x, where Ais a constant to be determined.
Step 1: Find the derivatives of yp
y′
p= 3Ae3x
y′′
p= 9Ae3x
Step 2: Substitute yp,y′
p, and y′′
pinto the differential equation
y′′ −5y′+ 6y= 4e3x
9Ae3x−5(3Ae3x) + 6(Ae3x)=4e3x
9Ae3x−15Ae3x+ 6Ae3x= 4e3x
(9A−15A+ 6A)e3x= 4e3x
Step 3: Equate the coefficients
(9A−15A+ 6A)e3x= 4e3x
0=4
Since 0 = 4, this implies that there is no solution with this form. We need
to modify our assumption. Let’s try yp=Bxe3x.
Continuing in a similar manner, we will eventually find the correct form of
ypthat satisfies the differential equation.
4
Question 5
Question
Find the general solution to the differential equation y′′ −3y′+ 2y= 0.
Solution
Step 1: Write down the characteristic equation by assuming a solution of the
form y=ert. The characteristic equation is:
r2−3r+ 2 = 0
Step 2: Solve the characteristic equation to find the roots r1and r2. Fac-
toring the quadratic equation gives:
(r−1)(r−2) = 0
So, the roots are r1= 1 and r2= 2.
Step 3: Write down the general solution using the roots r1and r2. The
general solution is:
y(t) = c1er1t+c2er2t
Substitute r1= 1 and r2= 2 into the general solution:
y(t) = c1et+c2e2t
Therefore, the general solution to the differential equation y′′ −3y′+ 2y= 0
is y(t) = c1et+c2e2t, where c1and c2are arbitrary constants.
Question 6
Question
Find the particular solution to the differential equation:
dy
dx = 3x2+ 2x+ 1,
given that y(0) = 2.
Solution
Step 1: Integrate both sides of the equation with respect to x.
Zdy
dx dx =Z(3x2+ 2x+ 1) dx
Step 2: Simplify the integrals on both sides.
y=Z(3x2+ 2x+ 1) dx
5
Step 3: Integrate the right-hand side.
y=x3+x2+x+C
Step 4: Use the initial condition y(0) = 2 to find the value of the constant
C.
2 = 0 + 0 + 0 + C
Step 5: Solve for C.
C= 2
Step 6: Write down the particular solution.
y=x3+x2+x+ 2
Question 7
Question
Consider the differential equation:
y′′ −3y′+ 2y=ex
Find the general solution to the homogeneous version of the differential equa-
tion and then find a particular solution to the full differential equation using
the method of undetermined coefficients.
Solution
Step 1: Find the general solution to the homogeneous equation. The
homogeneous equation corresponding to the given differential equation is:
y′′ −3y′+ 2y= 0
The characteristic equation is:
r2−3r+ 2 = 0
Solving for r, we get:
(r−1)(r−2) = 0
r= 1,2
Therefore, the general solution to the homogeneous equation is:
yh=c1ex+c2e2x
Step 2: Find a particular solution using the method of undeter-
mined coefficients. For the particular solution, we assume a solution of the
form yp=Aexwhere Ais a constant to be determined.
6
Substitute ypback into the original differential equation:
y′′
p−3y′
p+ 2yp=ex
Aex−3Aex+ 2Aex=ex
−Aex=ex
This implies that A=−1. Thus, the particular solution is yp=−ex.
Step 3: Find the general solution to the full differential equation.
The general solution to the full differential equation is the sum of the general
solution to the homogeneous equation and the particular solution:
y=yh+yp
y=c1ex+c2e2x−ex
Therefore, the general solution to the given differential equation is:
y=c1ex+c2e2x−ex
Question 8
Question
Find the general solution to the differential equation: dy
dx = 3y−5x.
Solution
Step 1: We can rewrite the given differential equation as:
dy
dx −3y=−5x
Step 2: This is a first-order linear ordinary differential equation of the form
dy
dx +P(x)y=Q(x). The integrating factor is eRP(x)dx.
Step 3: In our case, P(x) = −3, so the integrating factor is eR−3dx =e−3x.
Step 4: Multiply the integrating factor by the equation to get:
e−3xdy
dx −3e−3xy=−5xe−3x
Step 5: Recognize the left side as the derivative of ywith respect to xtimes
the integrating factor, i.e., d
dx (ye−3x).
Step 6: Integrate both sides with respect to xto get:
ye−3x=Z(−5xe−3x)dx +C
7
Step 7: Simplify the right side of the equation by integrating:
ye−3x=Z(−5xe−3x)dx =e−3x(5x−1) + K
Step 8: Therefore, the general solution to the differential equation is:
y=e3x(5x−1) + Ke3x
where Kis an arbitrary constant of integration.
Question 9
Question
Consider the differential equation:
y′′ −4y′+ 4y= 0
Find the general solution to this differential equation.
Solution
Step 1: Write down the characteristic equation associated with the differential
equation:
λ2−4λ+ 4 = 0
Step 2: Solve the characteristic equation to find the roots λ:
(λ−2)2= 0 =⇒λ= 2
Step 3: Write down the general solution to the differential equation using
the roots λ:
y(t) = c1e2t+c2te2t
where c1and c2are arbitrary constants.
Therefore, the general solution to the given differential equation is y(t) =
c1e2t+c2te2t.
Question 10
Question
Solve the following initial value problem:
dy
dt = 3y−2t, y(0) = 1
8
Solution
Step 1: First, we’ll find the general solution of the differential equation by
solving dy
dt = 3y−2t. This is a first-order linear differential equation, so we’ll
use the integrating factor method.
Step 2: Rewrite the equation in the standard form dy
dt −3y=−2t.
Step 3: The integrating factor is µ(t) = eR−3dt =e−3t.
Step 4: Multiply both sides of the equation by the integrating factor:
e−3tdy
dt −3e−3ty=−2te−3t
Step 5: Recognize the left side as the derivative of the product of e−3tand
yusing the product rule:
d
dt(e−3ty) = −2te−3t
Step 6: Integrate both sides:
Zd
dt(e−3ty)dt =Z−2te−3tdt
Step 7: This simplifies to:
e−3ty=2
9t2+C
Step 8: Now we’ll solve for yby multiplying through by e3t:
y=2
9t2e3t+Ce3t
Step 9: Using the initial condition y(0) = 1, we find:
1 = 2
9(0)2e3·0+Ce3·0=C
Step 10: Therefore, the solution to the initial value problem is:
y=2
9t2e3t+e3t
Question 11
Question
Consider the initial value problem given by y′= sin(x+y), y(0) = 1. Show
that this initial value problem has a unique solution for −∞ <x<∞.
9
Solution
To show the existence and uniqueness of the solution for the initial value prob-
lem, we will use the Picard-Lindel¨of theorem.
Step 1: Existence of the solution
Let f(x, y) = sin(x+y). The function f(x, y) and its partial derivative
∂f
∂y = cos(x+y) are continuous for all (x, y)∈R2. Since the functions are
continuous, the initial value problem has a unique solution in some interval
around x= 0.
Step 2: Local uniqueness
Suppose there are two functions y1(x) and y2(x) that both satisfy the initial
value problem in some interval around x= 0, then the difference z(x) = y1(x)−
y2(x) satisfies the differential equation z′= sin(x+y1)−sin(x+y2) with the
initial condition z(0) = 0.
Step 3: Applying the Mean Value Theorem
By the Mean Value Theorem for integrals, we have
|z(x)|=Zx
0
z′(t)dt
=Zx
0
(sin(t+y1)−sin(t+y2)) dt
=Zx
0
cos(ξt)(y1−y2)dt
for some ξt∈(t, t +y1),(t, t +y2). Since |cos(ξt)| ≤ 1, we have
|z(x)| ≤ |y1−y2|Zx
0
1dt
=|y1−y2| · x
Step 4: Conclusion
Therefore, |z(x)| ≤ |y1−y2| · xwhich implies that |y1(x)−y2(x)|can be
made arbitrarily small by choosing xsufficiently close to 0. Thus, the initial
value problem y′= sin(x+y), y(0) = 1 has a unique solution for −∞ <x<∞.
Question 12
Question
Consider the initial value problem:
dy
dx = (2x+ 3y)2, y(0) = 1.
Find the particular solution to the differential equation.
10
Solution
Step 1: Rewrite the differential equation in standard form.
dy
dx = 4x2+ 12xy + 9y2
Step 2: Recognize that the given differential equation is a Bernoulli differ-
ential equation, which can be converted into a linear differential equation by
substitution. Let v=y−1, then y=v−1and dy
dx =dv
dx (−v−2) = −1
v2
dv
dx .
Step 3: Substitute v=y−1into the differential equation and solve for dv
dx .
−1
v2
dv
dx = 4x2+ 12x−1
v+ 9 1
v2
dv
dx =−4v2x−12 + 9v
Step 4: Recognize that the linear differential equation can be solved using
an integrating factor. The integrating factor is eR−4xdx =e−2x2.
Step 5: Multiply both sides of the linear differential equation by the inte-
grating factor and integrate.
Ze−2x2dv =Z(−4v2x−12 + 9v)e−2x2dx
Step 6: Solve the integrals on both sides and simplify the expression.
v=−2x2v2−12xe−2x2+ 9 Zve−2x2dx
Step 7: Solve for v(x) by integrating the remaining term on the right-hand
side.
v=−2x2v2−12xe−2x2−9 e−2x2
−4+C!
Step 8: Substitute back v=y−1and solve for y(x).
1
y=−2x21
y2
−12xe−2x2+9e−2x2
−4+C
Step 9: Apply the initial condition y(0) = 1 to find the value of the constant
C.
1 = −2·02(12)−12 ·0e0+9
−4+C
1 = −9
4+C
C=13
4
11
Step 10: Substitute Cback into the equation and simplify to find the par-
ticular solution.
1
y=−2x21
y2
−12xe−2x2+9e−2x2
−4+13
4
y=1
−2x2−12xe−2x2+9e−2x2
−4+13
4
Question 13
Question
Consider the differential equation dy
dx =x2+y2
xy .
Find the general solution of the given differential equation.
Solution
Step 1: Let’s start by rewriting the given differential equation in a more conve-
nient form by separating variables:
dy
dx =x2+y2
xy .
Step 2: Rearranging the terms, we get:
dy
dx =1
y·(x+y2
x).
Step 3: We can write this equation as:
y·dy = (x+y2
x)·dx.
Step 4: Integrate both sides:
Zy dy =Z(x+y2
x)dx.
Integrating the left side with respect to ygives:
y2
2.
Step 5: Integrating the right side with respect to xrequires a bit of manip-
ulation. We will first rewrite the integral as:
Zx dx +Zy2
xdx.
12
Integrating xwith respect to xgives:
x2
2.
Step 6: To integrate y2
xwith respect to x, we can use substitution. Let
u=y/x, then y=ux and dy
dx =u+xdu
dx .
Step 7: Substituting back into the expression gives:
Zu2x dx =Z(u+xdu
dx)2x dx.
Step 8: Simplifying gives:
Zu2x dx =Z(u2+ 2uxdu
dx +x2du2
dx2)x dx.
Step 9: This simplifies to:
Zu2x dx =Zu2x dx + 2uZx2dx +Zx3du2
dx2dx.
Step 10: Canceling the terms yields:
2uZx2dx +Zx3du2
dx2dx = 0.
Step 11: Integrating 2uRx2dx gives:
u·x2.
Similarly, integrating Rx3du2
dx2dx gives: u2
2·x2.
Step 12: Therefore, the solution to the differential equation is:
y2
2=x2
2+u·x2+u2
2·x2.
Step 13: Simplifying gives the general solution:
y2=x2+Cx3,
where Cis an arbitrary constant.
Question 14
Question
Consider the first-order ordinary differential equation:
y′+2y
x=x3.
Find the general solution to this differential equation.
13
Solution
To solve the first-order ordinary differential equation, we will use the method
of integrating factors.
Step 1: Write the equation in standard form y′+P(x)y=Q(x).
Comparing the given equation with the standard form, we have P(x) = 2
x
and Q(x) = x3.
Step 2: Find the integrating factor.
The integrating factor I(x) is given by:
I(x) = exp ZP(x)dx.
In this case, P(x) = 2
x, so
I(x) = exp Z2
xdx= exp(2 ln |x|) = |x|2=x2.
Step 3: Multiply the given equation by the integrating factor.
Multiplying both sides of the differential equation by I(x) = x2gives us:
x2y′+ 2xy =x5.
Step 4: Rewrite the left-hand side as the derivative of a product.
The left-hand side can be written as the derivative of the product x2yusing
the product rule:
(x2y)′=x2y′+ 2xy.
Step 5: Integrate both sides.
Integrating both sides of the equation, we get:
Z(x2y)′dx =Zx5dx.
x2y=x6
6+C,
where Cis the constant of integration.
Step 6: Solve for yto find the general solution.
Dividing by x2, we get the general solution:
y=x4
6+C
x2,
where Cis an arbitrary constant.
Question 15
Question
Find the particular solution to the differential equation y′= 3y−5 with initial
condition y(0) = 2.
14
Solution
Step 1: First, we solve the differential equation by separating variables.
dy
dx = 3y−5
dy
3y−5=dx
Step 2: Next, we integrate both sides of the equation.
Z1
3y−5dy =Zdx
1
3ln |3y−5|=x+C
ln |3y−5|= 3x+C
Step 3: Now, we apply the initial condition y(0) = 2 to find the value of the
constant C.
ln |3(2) −5|= 3(0) + C
ln |1|=C
C= 0
Step 4: Substitute the value of Cback into the equation to find the particular
solution.
ln |3y−5|= 3x
|3y−5|=e3x
3y−5 = ±e3x
3y= 5 ±e3x
y=5
3±1
3e3x
Step 5: Finally, apply the initial condition y(0) = 2 to determine the specific
solution.
y(0) = 5
3±1
3
2 = 5
3±1
3
2 = 6
3
2=2
Therefore, the particular solution to the differential equation y′= 3y−5
with initial condition y(0) = 2 is y=5
3+1
3e3x.
15
Question 16
Question
Solve the following first-order ordinary differential equation:
dy
dx = (2x2+ 3y)ex2.
Solution
Step 1: This is a first-order ordinary differential equation of the form dy
dx =
f(x, y). To solve it, we will use the method of separation of variables.
Step 2: Write the given differential equation as
dy
dx = (2x2+ 3y)ex2.
Step 3: Separate the variables by moving all terms involving yto one side
and terms involving xto the other side:
dy
(2x2+ 3y)=ex2dx.
Step 4: Integrate both sides with respect to xand y:
Z1
2x2+ 3ydy =Zex2dx.
Step 5: Integrate the left side with respect to yusing substitution (u=
2x2+ 3y, du = 3dy) to get:
1
3ln |2x2+ 3y|=Zex2dx +C,
where Cis the constant of integration.
Step 6: Integrate the right side with respect to xto get:
1
3ln |2x2+ 3y|=Zex2dx +C.
Step 7: Simplify the expression to solve for y:
ln |2x2+ 3y|= 3 Zex2dx + 3C.
Step 8: Exponentiate both sides to get rid of the natural logarithm:
2x2+ 3y=e3Rex2dx ·e3C.
Step 9: Finally, solve for yto obtain the solution:
y=e3Rex2dx+3C−2x2
3.
16
Question 17
Question
Solve the following ordinary differential equation:
y′′ + 4y= 0
Solution
Step 1: Write down the characteristic equation associated with the ODE. The
characteristic equation is given by:
r2+ 4 = 0
Step 2: Find the roots of the characteristic equation.
r2=−4
r=±2i
Step 3: Write down the general solution to the ODE using the roots of the
characteristic equation. The general solution is of the form:
y(t) = c1cos(2t) + c2sin(2t)
where c1and c2are arbitrary constants.
Therefore, the general solution to the given ODE is y(t) = c1cos(2t) +
c2sin(2t).
Question 18
Question
Solve the initial value problem:
y′= 2xy, y(0) = 1
Solution
Step 1: We can recognize that this is a separable first-order differential equation.
To solve it, we will separate variables and integrate.
dy
dx = 2xy
dy
y= 2xdx
17
Step 2: Integrate both sides with respect to their respective variables.
Z1
ydy =Z2xdx
ln|y|=x2+C
Step 3: Exponentiate both sides to solve for y.
y=ex2+C
y=Cex2
Step 4: Use the initial condition y(0) = 1 to solve for the constant C.
1 = Ce02
1 = C
Step 5: Thus, the solution to the initial value problem is:
y=ex2
Question 19
Question
Consider the ordinary differential equation:
dy
dx + 2xy =x.
Find the general solution to this differential equation.
Solution
Step 1: First, we identify the integrating factor, which is given by:
µ(x) = eR2x dx =ex2.
Step 2: Multiply both sides of the differential equation by the integrating
factor:
ex2dy
dx + 2xex2y=xex2.
Step 3: Rewrite the left-hand side as the derivative of the product of the
integrating factor and the function y:
d
dx(ex2y) = xex2.
18
Step 4: Integrate both sides with respect to x:
Zd
dx(ex2y)dx =Zxex2dx.
Step 5: Solve the integrals:
ex2y=1
2ex2+C,
where C is the constant of integration.
Step 6: Solve for y by dividing through by the integrating factor:
y=1
2+Ce−x2.
Therefore, the general solution to the given differential equation is:
y=1
2+Ce−x2,
where C is an arbitrary constant.
Question 20
Question
Solve the initial value problem:
dy
dx =2y
x+x3, y(1) = 0
Solution
Step 1: First, rewrite the differential equation in the standard linear form y′+
P(x)y=Q(x). Let P(x) = 2
xand Q(x) = x3. So, the equation becomes:
dy
dx −2
xy=x3
Step 2: Next, identify the integrating factor µ(x) = eRP(x)dx.
µ(x) = eR2
xdx =e2 ln |x|=eln(x2)=x2
Step 3: Multiply the differential equation by the integrating factor µ(x) =
x2.
x2dy
dx −2xy =x5
Step 4: Rewrite the equation using the product rule:
d
dx(x2y) = x5
19
Step 5: Integrate both sides with respect to x:
Zd(x2y) = Zx5dx
x2y=1
6x6+C
Step 6: Solve for y:
y=1
6x4+C
x2
Step 7: Apply the initial condition y(1) = 0 to find the value of C:
0 = 1
6(1)4+C
(1)2
0 = 1
6+C
C=−1
6
Step 8: Substitute C=−1
6back into the general solution:
y=1
6x4−1
6x2
Therefore, the solution to the initial value problem is y=1
6x4−1
6x2.
Question 21
Question
Consider the following first-order ordinary differential equation:
dy
dx =x2−y
x
Solve the initial value problem with the initial condition y(0) = 1.
Solution
To solve the given initial value problem, we can rewrite the differential equation
in the following form:
dy
dx +y
x=x
This is a first order linear differential equation. The integrating factor is
eR1
xdx =eln |x|=|x|.
20
Step 1: Find the integrating factor Multiply both sides of the differen-
tial equation by the integrating factor:
|x|dy
dx +|x|y
x=|x| · x
This simplifies to:
|x|dy
dx +y=x2
Step 2: Solve the differential equation This equation is now in a form
where we can integrate both sides. Integrate both sides with respect to xto
solve for y:
|x|y=1
3x3+C
Step 3: Applying initial condition Given the initial condition y(0) = 1,
we can substitute this in to solve for the constant C:
|0| · 1 = 1
3·03+C
C= 1
Step 4: Final solution Substitute the value of Cback into the equation
to get the final solution:
|x|y=1
3x3+ 1
y=1
3x2+1
|x|
Therefore, the solution to the initial value problem with y(0) = 1 is y=
1
3x2+1
|x|.
Question 22
Question
Consider the initial value problem:
y′=2x
x2+y2, y(0) = 1
Find the interval in which the solution is guaranteed to exist.
21
Solution
Step 1: First, we rewrite the differential equation in terms of yand dy:
dy
dx =2x
x2+y2
Step 2: We notice that the equation is not defined at x2+y2= 0. To avoid
this singularity, we should restrict the domain of the solution.
Step 3: We compute the partial derivative of the right-hand side with respect
to y:
∂
∂y 2x
x2+y2=−4xy
(x2+y2)2
Step 4: Since this derivative is continuous for all real numbers xand y, the
solution to the initial value problem exists on any interval where the right-hand
side of the equation is continuous and satisfies the conditions of the existence
and uniqueness theorem.
Step 5: Therefore, the solution is guaranteed to exist on the largest interval
that does not contain any point (x, y) where x2+y2= 0. In this case, the
solution is guaranteed to exist for all x= 0.
Hence, the interval in which the solution is guaranteed to exist is (−∞,0) ∪(0,∞) .
Question 23
Question
Consider the differential equation:
dy
dx =3x2−y
x
Find the particular solution to the differential equation that satisfies the
initial condition y(1) = 2.
Solution
Step 1: Write the given differential equation in the standard form of a first-order
linear differential equation, dy
dx +P(x)y=Q(x), by moving all terms involving
yto the left side: dy
dx +y
x= 3x2
Step 2: Identify the integrating factor, µ(x), which is defined as µ(x) =
eRP(x)dx. In this case, P(x) = 1
x, so RP(x)dx =R1
xdx = ln |x|. Therefore, the
integrating factor is µ(x) = eln |x|=|x|.
22
Step 3: Multiply both sides of the differential equation by the integrating
factor µ(x):
|x|dy
dx +y
x|x|= 3x2|x|
|x|y′+y= 3x3
Step 4: Rewrite the left side as the derivative of a product:
d
dx(xy)=3x3
Step 5: Integrate both sides with respect to x:
Zd(xy) = Z3x3dx
xy =3
4x4+C
Step 6: Apply the initial condition y(1) = 2 to find the value of the constant
C:
1(2) = 3
4(1)4+C
2 = 3
4+C
C=5
4
Step 7: Substitute the value of Cback into the equation to find the particular
solution:
xy =3
4x4+5
4
2x=3
4x4+5
4
3x4−8x+ 5 = 0
Therefore, the particular solution to the differential equation that satisfies
the initial condition y(1) = 2 is 3x4−8x+ 5 = 0.
Question 24
Question
Solve the following first-order ordinary differential equation:
dy
dx =2y−x
2x−y
23
Solution
To solve the given first-order ordinary differential equation, we first rewrite it
in a more standard form:
(2x−y)dy = (2y−x)dx
Step 1: Let’s expand and simplify the left side of the equation:
2xy −y2= 2y dx −x dx
Step 2: Rearranging the terms, we get:
2xy −2y dx =y2−x dx
Step 3: Now, we divide both sides of the equation by 2xy −2y:
dx
x=dy
y
Step 4: Integrating both sides, we have:
Z1
xdx =Z1
ydy
ln |x|= ln |y|+C
Step 5: Exponentiating both sides, we get:
|x|=|y| · eC
Step 6: Now, we can combine the absolute values into one constant k:
x=y·k
Therefore, the solution to the given differential equation is x=y·k, where
kis a constant.
Question 25
Question
Let y(x) be a solution to the differential equation dy
dx =2x+3y
3x+4ywith initial
condition y(0) = 1. Find the value of y(2).
24
Solution
Step 1: First, let’s rewrite the given differential equation to separate variables:
dy
2x+ 3y=dx
3x+ 4y
Step 2: Integrate both sides with respect to x:
Zdy
2x+ 3y=Zdx
3x+ 4y
Step 3: Simplify the integrals by substitution. Let u= 2x+ 3yand v=
3x+ 4y. Then du = 2dx + 3dy and dv = 3dx + 4dy.
Step 4: Now the integrals become:
1
3Zdu
u=1
4Zdv
v
Step 5: Integrating, we get:
1
3ln |u|=1
4ln |v|+C
Step 6: Substitute back uand v:
1
3ln |2x+ 3y|=1
4ln |3x+ 4y|+C
Step 7: Using the initial condition y(0) = 1, we find C= 0:
1
3ln |2x+ 3y|=1
4ln |3x+ 4y|
Step 8: Now, apply x= 0, y = 1:
1
3ln(3) = 1
4ln(4)
Step 9: Solve for y(2) using x= 2:
1
3ln(7) = 1
4ln(11)
Step 10: Therefore, y(2) = 71/3
111/4.
Question 26
Question
Consider the following first-order ordinary differential equation:
dy
dx =2x+ 3y+ 1
x+y−2
Given the initial condition y(0) = 1, find the particular solution to this differ-
ential equation.
25
Solution
Step 1: Rewrite the given ODE in a more standard form:
(x+y−2)dy = (2x+ 3y+ 1)dx
Step 2: Integrate both sides of the equation:
Z(x+y−2)dy =Z(2x+ 3y+ 1)dx
Step 3: Integrate the left side with respect to y:
Z(y+x−2)dy =1
2y2+xy −2y+C1(where C1is the constant of integration)
Step 4: Integrate the right side with respect to x:
Z(2x+3y+1)dx =x2+3xy+x+C2(where C2is another constant of integration)
Step 5: Equate the two sides of the equation and simplify:
1
2y2+xy −2y+C1=x2+ 3xy +x+C2
Step 6: Rearrange the terms to isolate y:
1
2y2+xy −3xy −2y=x2+x−C1+C2
Step 7: Simplify the equation further:
1
2y2−2xy −2y=x2+x+C
Step 8: Consider the initial condition y(0) = 1 to find the value of the
constant C:
1
2(1)2−2(0)(1) −2(1) = 0 + 0 + C=⇒C=−2
Step 9: Substitute the value of Cback into the equation to find the particular
solution: 1
2y2−2xy −2y=x2+x−2
Therefore, the particular solution to the given differential equation with the
initial condition y(0) = 1 is 1
2y2−2xy −2y=x2+x−2.
Question 27
Question
Solve the following first-order ordinary differential equation:
dy
dx =3x2
2y
26
Solution
Step 1: Rewrite the given differential equation in the form dy =f(x)dx.
dy
dx =3x2
2y
2y dy = 3x2dx
Step 2: Integrate both sides of the equation.
Z2y dy =Z3x2dx
y2=x3+C1
where C1is the constant of integration.
Step 3: Solve for yby taking the square root of both sides.
y=±px3+C1
Therefore, the general solution to the given first-order ordinary differential
equation is y=±√x3+C1.
Question 28
Question
Solve the following initial value problem:
y′=√y, y(0) = 0
Solution
Step 1: Rewrite the differential equation as separable by dividing both sides by
√y:
dy
dτ = 1
Step 2: Integrate both sides with respect to τ:
Zdy =Z1dτ
y=τ+C
Step 3: Use the initial condition y(0) = 0 to find the value of the constant
C:
0 = 0 + C
C= 0
27
Step 4: Substitute C= 0 back into the general solution to get the particular
solution:
y=τ
Step 5: Finally, substitute τback in terms of xto get the final solution:
y=x
Therefore, the solution to the initial value problem is y=x.
Question 29
Question
Solve the initial value problem: y′= 2xy +ex2,y(0) = 1.
Solution
Step 1: First, we identify that this is a first-order linear differential equation in
the form y′+P(x)y=Q(x), where P(x)=2xand Q(x) = ex2.
Step 2: To solve this, we will use an integrating factor µ(x) = eRP(x)dx =ex2.
Step 3: Multiply both sides of the differential equation by the integrating
factor:
ex2y′+ 2xyex2=e2x2y+e2x2
Step 4: Rewrite the left side as the derivative of the product of µ(x) and y:
(ex2y)′=e2x2y+e2x2
Step 5: Integrate both sides with respect to x:
Z(ex2y)′dx =Z(e2x2y+e2x2)dx
ex2y=e2x2y
2+C
Step 6: Solve for y:
y=e2x2y
2ex2+Ce−x2=y
2+Ce−x2
Step 7: Apply the initial condition y(0) = 1:
1 = 1
2+C⇒C=1
2
Step 8: Substitute C=1
2back into the equation:
y=y
2+1
2e−x2
28
Step 9: Solve for y:
y−y
2=1
2e−x2⇒y
2=1
2e−x2⇒y=e−x2
Therefore, the solution to the initial value problem is y=e−x2.
Question 30
Question
Consider the differential equation:
y′′ −2y′+y=ex
Find the general solution to this differential equation.
Solution
Step 1: First, we find the complementary solution by solving the associated
homogeneous equation:
y′′ −2y′+y= 0
Step 2: The characteristic equation is r2−2r+ 1 = 0, which factors as
(r−1)2= 0. Thus, we have a repeated root of r= 1.
Step 3: The complementary solution takes the form:
yc(x) = c1ex+c2xex
Step 4: Next, we find a particular solution to the non-homogeneous equation
by the method of undetermined coefficients. Assume a particular solution in the
form of yp(x) = Aex.
Step 5: Substituting yp(x) into the differential equation, we get:
A−2A+A=ex⇒0 = ex
Step 6: Since the last equation has no solutions, we try an alternative form
for the particular solution. Let’s try yp(x) = Axex.
Step 7: Substituting yp(x) into the differential equation, we get:
2A−2A+Axex=ex
Step 8: This gives us A= 1/2. Therefore, the particular solution is yp(x) =
1
2xex.
Step 9: The general solution to the differential equation is the sum of the
complementary and particular solutions:
y(x) = yc(x) + yp(x) = c1ex+c2xex+1
2xex
Therefore, the general solution to the given differential equation is y(x) =
c1ex+c2xex+1
2xex.
29
Question 31
Question
Solve the following first-order ordinary differential equation:
dy
dx =1
2x+ 3y
Solution
Step 1: Rewrite the equation in a suitable form for separation of variables.
We rewrite the equation as: dy
dx =1
2x+ 3y
2x+ 3ydy
dx = 1
Step 2: Separate the variables.
2x dx + 3y dy =dx
Step 3: Integrate both sides.
Z2x dx +Z3y dy =Zdx
x2+3
2y2=x+C
Step 4: Solve for y.
3
2y2=x−x2+C
y2=2x−2x2+ 2C
3
y=±r2x−2x2+ 2C
3
Thus, the general solution to the differential equation is:
y=±r2x−2x2+ 2C
3
Question 32
Question
Consider the following ordinary differential equation:
dy
dx =e2x−2y
Solve the initial value problem when y(0) = 3.
30
Solution
Step 1: First, let’s rewrite the given differential equation in standard form:
dy
dx + 2y=e2x
This is a first-order linear differential equation.
Step 2: We’ll start by finding the integrating factor, which is given by IF =
eR2dx =e2x.
Step 3: Next, we multiply both sides of the differential equation by the
integrating factor:
e2xdy
dx + 2e2xy=e4x
Step 4: Notice that the left-hand side is the derivative of the product e2xy,
so we rewrite the equation as:
d
dx(e2xy) = e4x
Step 5: We can integrate both sides with respect to xto solve for y:
Zd
dx(e2xy)dx =Ze4xdx
Step 6: Integrating both sides gives:
e2xy=1
4e4x+C
Step 7: Now, we apply the initial condition y(0) = 3 to solve for the constant
C:
e2∗0∗3 = 1
4e4∗0+C
3 = 1
4+C
C=11
4
Step 8: Therefore, the particular solution to the initial value problem is:
e2xy=1
4e4x+11
4
Step 9: Finally, we can solve for y:
y=1
4e2x+11
4e−2x
31
Question 33
Question
Solve the initial value problem:
dy
dx =4x
y3, y(0) = 1
Solution
Step 1: Rewrite the differential equation in separated variables form:
y3dy = 4x dx
Step 2: Integrate both sides:
Zy3dy =Z4x dx
Step 3: Solve the integrals:
1
4y4= 2x2+C
where Cis the constant of integration.
Step 4: Using the initial condition y(0) = 1, we have:
1
4·14= 2 ·02+C=⇒C=1
4
Step 5: Substitute back to find the particular solution:
1
4y4= 2x2+1
4
Step 6: Simplify the equation:
y4= 8x2+ 1
Step 7: Solve for yby taking the fourth root of both sides:
y=4
p8x2+ 1
Therefore, the solution to the initial value problem is y=4
√8x2+ 1 with
the initial condition y(0) = 1.
Question 34
Question
Find the general solution to the differential equation:
dy
dx =1
xy +x2y
32
Solution
To solve this first-order ordinary differential equation, we can rewrite it in the
form y′+P(x)y=Q(x), where P(x) and Q(x) are functions of x. In this case,
we have P(x) = x2and Q(x) = 1
x.
Step 1: We can solve the differential equation using an integrating factor.
The integrating factor µ(x) is defined as µ(x) = eRP(x)dx.
In this case, P(x) = x2, so µ(x) = eRx2dx.
Step 2: Calculating the integral and finding µ(x):
Zx2dx =1
3x3
Thus, µ(x) = e1
3x3.
Step 3: Multiplying the differential equation by the integrating factor:
e1
3x3dy
dx +e1
3x3x2y=e1
3x31
x
Step 4: Rewriting the left-hand side using the product rule:
d
dx e1
3x3y=e1
3x31
x
Step 5: Integrating both sides with respect to x:
Zd
dx e1
3x3ydx =Ze1
3x31
xdx
Step 6: The solution after integrating both sides:
e1
3x3y=Ze1
3x31
xdx +C
Step 7: Simplifying the integral:
e1
3x3y=Ze1
3x31
xdx +C
The general solution is y(x) = e−1
3x3Re1
3x31
xdx +C, where Cis the
constant of integration.
Question 35
Question
Consider the differential equation dy
dx = 2y−x2with initial condition y(0) = 1.
Find the particular solution to the differential equation.
33
Solution
Step 1: First, let’s write the homogeneous equation associated with the given
differential equation: dy
dx = 2y
This is a separable differential equation, so we can rewrite it as:
dy
y= 2dx
Step 2: Integrating both sides gives us:
Z1
ydy =Z2dx
ln |y|= 2x+C1
where C1is the constant of integration.
Step 3: Exponentiating both sides, we get:
|y|=e2x+C1=e2xeC1=Ce2x
where C=eC1.
Step 4: Now, we need to find the particular solution by considering the non-
homogeneous part of the differential equation. Let yp=Ax2+Bx +Cbe the
particular solution.
Step 5: Substitute ypinto the differential equation and differentiate to match
the coefficients of xand constants on both sides. We have:
2(Ax2+Bx +C)−x2= 2Ax2+ 2Bx + 2C−x2
Equating coefficients, we get:
2A= 2,2B= 0,2C= 0
which gives A= 1, B= 0, and C= 0.
Step 6: Therefore, the particular solution is yp=x2.
Step 7: Finally, the general solution is given by the sum of the homogeneous
and particular solutions:
y=Ce2x+x2
Applying the initial condition y(0) = 1, we find C= 1, so the particular solution
to the differential equation is y=e2x+x2.
34
Step 7: Finally, divide by e2xto get the general solution:
y=3ex+Ce−2x
e2x= 3e−x+Ce−2x
where Cis an arbitrary constant.
Question 2
Question
Find the general solution to the following first-order ordinary differential equa-
tion:
y′=1
2y+e2x
Solution
To find the general solution to the given first-order ordinary differential equa-
tion, we will use the method of integrating factors.
Step 1: Identify the variables The given differential equation is in the
form y′+P(x)y=Q(x), where P(x) = 1
2and Q(x) = e2x.
Step 2: Find the integrating factor The integrating factor I(x) is given
by:
I(x) = eRP(x)dx
I(x) = eR1
2dx
I(x) = e1
2x
Step 3: Multiply the integrating factor by the differential equation
Multiply both sides of the differential equation by the integrating factor I(x):
e1
2xy′+1
2e1
2xy=e2xe1
2x
e1
2xy′+1
2e1
2xy=e5
2x
Step 4: Rewrite the left-hand side as a derivative Rewrite the left-
hand side as the derivative of the product ye 1
2x:
e1
2xy′=e5
2x
Step 5: Integrate both sides Integrate both sides of the equation with
respect to x:
Ze1
2xy′dx =Ze5
2xdx
e1
2xy=2
5e5
2x+C
2
Step 6: Solve for ySolving for y, we have:
y=2
5e3x+Ce−1
2x
Therefore, the general solution to the given differential equation is:
y=2
5e3x+Ce−1
2x
where Cis an arbitrary constant.
Question 3
Question
Consider the differential equation:
y′+ 2xy = 4x3
Find the general solution of the differential equation.
Solution
Step 1: First, we need to recognize that the given differential equation is a linear
first-order differential equation in standard form:
dy
dx + 2xy = 4x3
Step 2: Next, we identify the integrating factor which is given by eRP(x)dx,
where P(x) is the coefficient of y.
Step 3: In our case, P(x) = 2x. Therefore, the integrating factor is given by
eR2xdx =ex2.
Step 4: Multiplying both sides of the differential equation by the integrating
factor gives:
ex2dy
dx + 2xex2y= 4x3ex2
Step 5: The left side of the equation can be rewritten using the product rule
for differentiation as:
d
dx(ex2y)=4x3ex2
Step 6: Integrating both sides of the equation with respect to xgives:
Zd
dx(ex2y)dx =Z4x3ex2dx
Step 7: Integrating the right side and simplifying gives:
3
ex2y=Z4x3ex2dx =ex2(x2−1) + C
where Cis the constant of integration.
Step 8: Solving for y, we get the general solution of the differential equation
as:
y=x2−1 + Ce−x2
where Cis an arbitrary constant.
Question 4
Question
Solve the following ordinary differential equation using the method of undeter-
mined coefficients:
y′′ −5y′+ 6y= 4e3x
Solution
To solve the given second-order homogeneous differential equation using the
method of undetermined coefficients, we assume the particular solution has the
form yp=Ae3x, where Ais a constant to be determined.
Step 1: Find the derivatives of yp
y′
p= 3Ae3x
y′′
p= 9Ae3x
Step 2: Substitute yp,y′
p, and y′′
pinto the differential equation
y′′ −5y′+ 6y= 4e3x
9Ae3x−5(3Ae3x) + 6(Ae3x)=4e3x
9Ae3x−15Ae3x+ 6Ae3x= 4e3x
(9A−15A+ 6A)e3x= 4e3x
Step 3: Equate the coefficients
(9A−15A+ 6A)e3x= 4e3x
0=4
Since 0 = 4, this implies that there is no solution with this form. We need
to modify our assumption. Let’s try yp=Bxe3x.
Continuing in a similar manner, we will eventually find the correct form of
ypthat satisfies the differential equation.
4
Question 5
Question
Find the general solution to the differential equation y′′ −3y′+ 2y= 0.
Solution
Step 1: Write down the characteristic equation by assuming a solution of the
form y=ert. The characteristic equation is:
r2−3r+ 2 = 0
Step 2: Solve the characteristic equation to find the roots r1and r2. Fac-
toring the quadratic equation gives:
(r−1)(r−2) = 0
So, the roots are r1= 1 and r2= 2.
Step 3: Write down the general solution using the roots r1and r2. The
general solution is:
y(t) = c1er1t+c2er2t
Substitute r1= 1 and r2= 2 into the general solution:
y(t) = c1et+c2e2t
Therefore, the general solution to the differential equation y′′ −3y′+ 2y= 0
is y(t) = c1et+c2e2t, where c1and c2are arbitrary constants.
Question 6
Question
Find the particular solution to the differential equation:
dy
dx = 3x2+ 2x+ 1,
given that y(0) = 2.
Solution
Step 1: Integrate both sides of the equation with respect to x.
Zdy
dx dx =Z(3x2+ 2x+ 1) dx
Step 2: Simplify the integrals on both sides.
y=Z(3x2+ 2x+ 1) dx
5
Step 3: Integrate the right-hand side.
y=x3+x2+x+C
Step 4: Use the initial condition y(0) = 2 to find the value of the constant
C.
2 = 0 + 0 + 0 + C
Step 5: Solve for C.
C= 2
Step 6: Write down the particular solution.
y=x3+x2+x+ 2
Question 7
Question
Consider the differential equation:
y′′ −3y′+ 2y=ex
Find the general solution to the homogeneous version of the differential equa-
tion and then find a particular solution to the full differential equation using
the method of undetermined coefficients.
Solution
Step 1: Find the general solution to the homogeneous equation. The
homogeneous equation corresponding to the given differential equation is:
y′′ −3y′+ 2y= 0
The characteristic equation is:
r2−3r+ 2 = 0
Solving for r, we get:
(r−1)(r−2) = 0
r= 1,2
Therefore, the general solution to the homogeneous equation is:
yh=c1ex+c2e2x
Step 2: Find a particular solution using the method of undeter-
mined coefficients. For the particular solution, we assume a solution of the
form yp=Aexwhere Ais a constant to be determined.
6
Substitute ypback into the original differential equation:
y′′
p−3y′
p+ 2yp=ex
Aex−3Aex+ 2Aex=ex
−Aex=ex
This implies that A=−1. Thus, the particular solution is yp=−ex.
Step 3: Find the general solution to the full differential equation.
The general solution to the full differential equation is the sum of the general
solution to the homogeneous equation and the particular solution:
y=yh+yp
y=c1ex+c2e2x−ex
Therefore, the general solution to the given differential equation is:
y=c1ex+c2e2x−ex
Question 8
Question
Find the general solution to the differential equation: dy
dx = 3y−5x.
Solution
Step 1: We can rewrite the given differential equation as:
dy
dx −3y=−5x
Step 2: This is a first-order linear ordinary differential equation of the form
dy
dx +P(x)y=Q(x). The integrating factor is eRP(x)dx.
Step 3: In our case, P(x) = −3, so the integrating factor is eR−3dx =e−3x.
Step 4: Multiply the integrating factor by the equation to get:
e−3xdy
dx −3e−3xy=−5xe−3x
Step 5: Recognize the left side as the derivative of ywith respect to xtimes
the integrating factor, i.e., d
dx (ye−3x).
Step 6: Integrate both sides with respect to xto get:
ye−3x=Z(−5xe−3x)dx +C
7
Step 7: Simplify the right side of the equation by integrating:
ye−3x=Z(−5xe−3x)dx =e−3x(5x−1) + K
Step 8: Therefore, the general solution to the differential equation is:
y=e3x(5x−1) + Ke3x
where Kis an arbitrary constant of integration.
Question 9
Question
Consider the differential equation:
y′′ −4y′+ 4y= 0
Find the general solution to this differential equation.
Solution
Step 1: Write down the characteristic equation associated with the differential
equation:
λ2−4λ+ 4 = 0
Step 2: Solve the characteristic equation to find the roots λ:
(λ−2)2= 0 =⇒λ= 2
Step 3: Write down the general solution to the differential equation using
the roots λ:
y(t) = c1e2t+c2te2t
where c1and c2are arbitrary constants.
Therefore, the general solution to the given differential equation is y(t) =
c1e2t+c2te2t.
Question 10
Question
Solve the following initial value problem:
dy
dt = 3y−2t, y(0) = 1
8
Solution
Step 1: First, we’ll find the general solution of the differential equation by
solving dy
dt = 3y−2t. This is a first-order linear differential equation, so we’ll
use the integrating factor method.
Step 2: Rewrite the equation in the standard form dy
dt −3y=−2t.
Step 3: The integrating factor is µ(t) = eR−3dt =e−3t.
Step 4: Multiply both sides of the equation by the integrating factor:
e−3tdy
dt −3e−3ty=−2te−3t
Step 5: Recognize the left side as the derivative of the product of e−3tand
yusing the product rule:
d
dt(e−3ty) = −2te−3t
Step 6: Integrate both sides:
Zd
dt(e−3ty)dt =Z−2te−3tdt
Step 7: This simplifies to:
e−3ty=2
9t2+C
Step 8: Now we’ll solve for yby multiplying through by e3t:
y=2
9t2e3t+Ce3t
Step 9: Using the initial condition y(0) = 1, we find:
1 = 2
9(0)2e3·0+Ce3·0=C
Step 10: Therefore, the solution to the initial value problem is:
y=2
9t2e3t+e3t
Question 11
Question
Consider the initial value problem given by y′= sin(x+y), y(0) = 1. Show
that this initial value problem has a unique solution for −∞ <x<∞.
9
Solution
To show the existence and uniqueness of the solution for the initial value prob-
lem, we will use the Picard-Lindel¨of theorem.
Step 1: Existence of the solution
Let f(x, y) = sin(x+y). The function f(x, y) and its partial derivative
∂f
∂y = cos(x+y) are continuous for all (x, y)∈R2. Since the functions are
continuous, the initial value problem has a unique solution in some interval
around x= 0.
Step 2: Local uniqueness
Suppose there are two functions y1(x) and y2(x) that both satisfy the initial
value problem in some interval around x= 0, then the difference z(x) = y1(x)−
y2(x) satisfies the differential equation z′= sin(x+y1)−sin(x+y2) with the
initial condition z(0) = 0.
Step 3: Applying the Mean Value Theorem
By the Mean Value Theorem for integrals, we have
|z(x)|=Zx
0
z′(t)dt
=Zx
0
(sin(t+y1)−sin(t+y2)) dt
=Zx
0
cos(ξt)(y1−y2)dt
for some ξt∈(t, t +y1),(t, t +y2). Since |cos(ξt)| ≤ 1, we have
|z(x)| ≤ |y1−y2|Zx
0
1dt
=|y1−y2| · x
Step 4: Conclusion
Therefore, |z(x)| ≤ |y1−y2| · xwhich implies that |y1(x)−y2(x)|can be
made arbitrarily small by choosing xsufficiently close to 0. Thus, the initial
value problem y′= sin(x+y), y(0) = 1 has a unique solution for −∞ <x<∞.
Question 12
Question
Consider the initial value problem:
dy
dx = (2x+ 3y)2, y(0) = 1.
Find the particular solution to the differential equation.
10
Solution
Step 1: Rewrite the differential equation in standard form.
dy
dx = 4x2+ 12xy + 9y2
Step 2: Recognize that the given differential equation is a Bernoulli differ-
ential equation, which can be converted into a linear differential equation by
substitution. Let v=y−1, then y=v−1and dy
dx =dv
dx (−v−2) = −1
v2
dv
dx .
Step 3: Substitute v=y−1into the differential equation and solve for dv
dx .
−1
v2
dv
dx = 4x2+ 12x−1
v+ 9 1
v2
dv
dx =−4v2x−12 + 9v
Step 4: Recognize that the linear differential equation can be solved using
an integrating factor. The integrating factor is eR−4xdx =e−2x2.
Step 5: Multiply both sides of the linear differential equation by the inte-
grating factor and integrate.
Ze−2x2dv =Z(−4v2x−12 + 9v)e−2x2dx
Step 6: Solve the integrals on both sides and simplify the expression.
v=−2x2v2−12xe−2x2+ 9 Zve−2x2dx
Step 7: Solve for v(x) by integrating the remaining term on the right-hand
side.
v=−2x2v2−12xe−2x2−9 e−2x2
−4+C!
Step 8: Substitute back v=y−1and solve for y(x).
1
y=−2x21
y2
−12xe−2x2+9e−2x2
−4+C
Step 9: Apply the initial condition y(0) = 1 to find the value of the constant
C.
1 = −2·02(12)−12 ·0e0+9
−4+C
1 = −9
4+C
C=13
4
11
Step 10: Substitute Cback into the equation and simplify to find the par-
ticular solution.
1
y=−2x21
y2
−12xe−2x2+9e−2x2
−4+13
4
y=1
−2x2−12xe−2x2+9e−2x2
−4+13
4
Question 13
Question
Consider the differential equation dy
dx =x2+y2
xy .
Find the general solution of the given differential equation.
Solution
Step 1: Let’s start by rewriting the given differential equation in a more conve-
nient form by separating variables:
dy
dx =x2+y2
xy .
Step 2: Rearranging the terms, we get:
dy
dx =1
y·(x+y2
x).
Step 3: We can write this equation as:
y·dy = (x+y2
x)·dx.
Step 4: Integrate both sides:
Zy dy =Z(x+y2
x)dx.
Integrating the left side with respect to ygives:
y2
2.
Step 5: Integrating the right side with respect to xrequires a bit of manip-
ulation. We will first rewrite the integral as:
Zx dx +Zy2
xdx.
12
Integrating xwith respect to xgives:
x2
2.
Step 6: To integrate y2
xwith respect to x, we can use substitution. Let
u=y/x, then y=ux and dy
dx =u+xdu
dx .
Step 7: Substituting back into the expression gives:
Zu2x dx =Z(u+xdu
dx)2x dx.
Step 8: Simplifying gives:
Zu2x dx =Z(u2+ 2uxdu
dx +x2du2
dx2)x dx.
Step 9: This simplifies to:
Zu2x dx =Zu2x dx + 2uZx2dx +Zx3du2
dx2dx.
Step 10: Canceling the terms yields:
2uZx2dx +Zx3du2
dx2dx = 0.
Step 11: Integrating 2uRx2dx gives:
u·x2.
Similarly, integrating Rx3du2
dx2dx gives: u2
2·x2.
Step 12: Therefore, the solution to the differential equation is:
y2
2=x2
2+u·x2+u2
2·x2.
Step 13: Simplifying gives the general solution:
y2=x2+Cx3,
where Cis an arbitrary constant.
Question 14
Question
Consider the first-order ordinary differential equation:
y′+2y
x=x3.
Find the general solution to this differential equation.
13
Solution
To solve the first-order ordinary differential equation, we will use the method
of integrating factors.
Step 1: Write the equation in standard form y′+P(x)y=Q(x).
Comparing the given equation with the standard form, we have P(x) = 2
x
and Q(x) = x3.
Step 2: Find the integrating factor.
The integrating factor I(x) is given by:
I(x) = exp ZP(x)dx.
In this case, P(x) = 2
x, so
I(x) = exp Z2
xdx= exp(2 ln |x|) = |x|2=x2.
Step 3: Multiply the given equation by the integrating factor.
Multiplying both sides of the differential equation by I(x) = x2gives us:
x2y′+ 2xy =x5.
Step 4: Rewrite the left-hand side as the derivative of a product.
The left-hand side can be written as the derivative of the product x2yusing
the product rule:
(x2y)′=x2y′+ 2xy.
Step 5: Integrate both sides.
Integrating both sides of the equation, we get:
Z(x2y)′dx =Zx5dx.
x2y=x6
6+C,
where Cis the constant of integration.
Step 6: Solve for yto find the general solution.
Dividing by x2, we get the general solution:
y=x4
6+C
x2,
where Cis an arbitrary constant.
Question 15
Question
Find the particular solution to the differential equation y′= 3y−5 with initial
condition y(0) = 2.
14
Solution
Step 1: First, we solve the differential equation by separating variables.
dy
dx = 3y−5
dy
3y−5=dx
Step 2: Next, we integrate both sides of the equation.
Z1
3y−5dy =Zdx
1
3ln |3y−5|=x+C
ln |3y−5|= 3x+C
Step 3: Now, we apply the initial condition y(0) = 2 to find the value of the
constant C.
ln |3(2) −5|= 3(0) + C
ln |1|=C
C= 0
Step 4: Substitute the value of Cback into the equation to find the particular
solution.
ln |3y−5|= 3x
|3y−5|=e3x
3y−5 = ±e3x
3y= 5 ±e3x
y=5
3±1
3e3x
Step 5: Finally, apply the initial condition y(0) = 2 to determine the specific
solution.
y(0) = 5
3±1
3
2 = 5
3±1
3
2 = 6
3
2=2
Therefore, the particular solution to the differential equation y′= 3y−5
with initial condition y(0) = 2 is y=5
3+1
3e3x.
15
Question 16
Question
Solve the following first-order ordinary differential equation:
dy
dx = (2x2+ 3y)ex2.
Solution
Step 1: This is a first-order ordinary differential equation of the form dy
dx =
f(x, y). To solve it, we will use the method of separation of variables.
Step 2: Write the given differential equation as
dy
dx = (2x2+ 3y)ex2.
Step 3: Separate the variables by moving all terms involving yto one side
and terms involving xto the other side:
dy
(2x2+ 3y)=ex2dx.
Step 4: Integrate both sides with respect to xand y:
Z1
2x2+ 3ydy =Zex2dx.
Step 5: Integrate the left side with respect to yusing substitution (u=
2x2+ 3y, du = 3dy) to get:
1
3ln |2x2+ 3y|=Zex2dx +C,
where Cis the constant of integration.
Step 6: Integrate the right side with respect to xto get:
1
3ln |2x2+ 3y|=Zex2dx +C.
Step 7: Simplify the expression to solve for y:
ln |2x2+ 3y|= 3 Zex2dx + 3C.
Step 8: Exponentiate both sides to get rid of the natural logarithm:
2x2+ 3y=e3Rex2dx ·e3C.
Step 9: Finally, solve for yto obtain the solution:
y=e3Rex2dx+3C−2x2
3.
16
Question 17
Question
Solve the following ordinary differential equation:
y′′ + 4y= 0
Solution
Step 1: Write down the characteristic equation associated with the ODE. The
characteristic equation is given by:
r2+ 4 = 0
Step 2: Find the roots of the characteristic equation.
r2=−4
r=±2i
Step 3: Write down the general solution to the ODE using the roots of the
characteristic equation. The general solution is of the form:
y(t) = c1cos(2t) + c2sin(2t)
where c1and c2are arbitrary constants.
Therefore, the general solution to the given ODE is y(t) = c1cos(2t) +
c2sin(2t).
Question 18
Question
Solve the initial value problem:
y′= 2xy, y(0) = 1
Solution
Step 1: We can recognize that this is a separable first-order differential equation.
To solve it, we will separate variables and integrate.
dy
dx = 2xy
dy
y= 2xdx
17
Step 2: Integrate both sides with respect to their respective variables.
Z1
ydy =Z2xdx
ln|y|=x2+C
Step 3: Exponentiate both sides to solve for y.
y=ex2+C
y=Cex2
Step 4: Use the initial condition y(0) = 1 to solve for the constant C.
1 = Ce02
1 = C
Step 5: Thus, the solution to the initial value problem is:
y=ex2
Question 19
Question
Consider the ordinary differential equation:
dy
dx + 2xy =x.
Find the general solution to this differential equation.
Solution
Step 1: First, we identify the integrating factor, which is given by:
µ(x) = eR2x dx =ex2.
Step 2: Multiply both sides of the differential equation by the integrating
factor:
ex2dy
dx + 2xex2y=xex2.
Step 3: Rewrite the left-hand side as the derivative of the product of the
integrating factor and the function y:
d
dx(ex2y) = xex2.
18
Step 4: Integrate both sides with respect to x:
Zd
dx(ex2y)dx =Zxex2dx.
Step 5: Solve the integrals:
ex2y=1
2ex2+C,
where C is the constant of integration.
Step 6: Solve for y by dividing through by the integrating factor:
y=1
2+Ce−x2.
Therefore, the general solution to the given differential equation is:
y=1
2+Ce−x2,
where C is an arbitrary constant.
Question 20
Question
Solve the initial value problem:
dy
dx =2y
x+x3, y(1) = 0
Solution
Step 1: First, rewrite the differential equation in the standard linear form y′+
P(x)y=Q(x). Let P(x) = 2
xand Q(x) = x3. So, the equation becomes:
dy
dx −2
xy=x3
Step 2: Next, identify the integrating factor µ(x) = eRP(x)dx.
µ(x) = eR2
xdx =e2 ln |x|=eln(x2)=x2
Step 3: Multiply the differential equation by the integrating factor µ(x) =
x2.
x2dy
dx −2xy =x5
Step 4: Rewrite the equation using the product rule:
d
dx(x2y) = x5
19
Step 5: Integrate both sides with respect to x:
Zd(x2y) = Zx5dx
x2y=1
6x6+C
Step 6: Solve for y:
y=1
6x4+C
x2
Step 7: Apply the initial condition y(1) = 0 to find the value of C:
0 = 1
6(1)4+C
(1)2
0 = 1
6+C
C=−1
6
Step 8: Substitute C=−1
6back into the general solution:
y=1
6x4−1
6x2
Therefore, the solution to the initial value problem is y=1
6x4−1
6x2.
Question 21
Question
Consider the following first-order ordinary differential equation:
dy
dx =x2−y
x
Solve the initial value problem with the initial condition y(0) = 1.
Solution
To solve the given initial value problem, we can rewrite the differential equation
in the following form:
dy
dx +y
x=x
This is a first order linear differential equation. The integrating factor is
eR1
xdx =eln |x|=|x|.
20
Step 1: Find the integrating factor Multiply both sides of the differen-
tial equation by the integrating factor:
|x|dy
dx +|x|y
x=|x| · x
This simplifies to:
|x|dy
dx +y=x2
Step 2: Solve the differential equation This equation is now in a form
where we can integrate both sides. Integrate both sides with respect to xto
solve for y:
|x|y=1
3x3+C
Step 3: Applying initial condition Given the initial condition y(0) = 1,
we can substitute this in to solve for the constant C:
|0| · 1 = 1
3·03+C
C= 1
Step 4: Final solution Substitute the value of Cback into the equation
to get the final solution:
|x|y=1
3x3+ 1
y=1
3x2+1
|x|
Therefore, the solution to the initial value problem with y(0) = 1 is y=
1
3x2+1
|x|.
Question 22
Question
Consider the initial value problem:
y′=2x
x2+y2, y(0) = 1
Find the interval in which the solution is guaranteed to exist.
21
Solution
Step 1: First, we rewrite the differential equation in terms of yand dy:
dy
dx =2x
x2+y2
Step 2: We notice that the equation is not defined at x2+y2= 0. To avoid
this singularity, we should restrict the domain of the solution.
Step 3: We compute the partial derivative of the right-hand side with respect
to y:
∂
∂y 2x
x2+y2=−4xy
(x2+y2)2
Step 4: Since this derivative is continuous for all real numbers xand y, the
solution to the initial value problem exists on any interval where the right-hand
side of the equation is continuous and satisfies the conditions of the existence
and uniqueness theorem.
Step 5: Therefore, the solution is guaranteed to exist on the largest interval
that does not contain any point (x, y) where x2+y2= 0. In this case, the
solution is guaranteed to exist for all x= 0.
Hence, the interval in which the solution is guaranteed to exist is (−∞,0) ∪(0,∞) .
Question 23
Question
Consider the differential equation:
dy
dx =3x2−y
x
Find the particular solution to the differential equation that satisfies the
initial condition y(1) = 2.
Solution
Step 1: Write the given differential equation in the standard form of a first-order
linear differential equation, dy
dx +P(x)y=Q(x), by moving all terms involving
yto the left side: dy
dx +y
x= 3x2
Step 2: Identify the integrating factor, µ(x), which is defined as µ(x) =
eRP(x)dx. In this case, P(x) = 1
x, so RP(x)dx =R1
xdx = ln |x|. Therefore, the
integrating factor is µ(x) = eln |x|=|x|.
22
Step 3: Multiply both sides of the differential equation by the integrating
factor µ(x):
|x|dy
dx +y
x|x|= 3x2|x|
|x|y′+y= 3x3
Step 4: Rewrite the left side as the derivative of a product:
d
dx(xy)=3x3
Step 5: Integrate both sides with respect to x:
Zd(xy) = Z3x3dx
xy =3
4x4+C
Step 6: Apply the initial condition y(1) = 2 to find the value of the constant
C:
1(2) = 3
4(1)4+C
2 = 3
4+C
C=5
4
Step 7: Substitute the value of Cback into the equation to find the particular
solution:
xy =3
4x4+5
4
2x=3
4x4+5
4
3x4−8x+ 5 = 0
Therefore, the particular solution to the differential equation that satisfies
the initial condition y(1) = 2 is 3x4−8x+ 5 = 0.
Question 24
Question
Solve the following first-order ordinary differential equation:
dy
dx =2y−x
2x−y
23
Solution
To solve the given first-order ordinary differential equation, we first rewrite it
in a more standard form:
(2x−y)dy = (2y−x)dx
Step 1: Let’s expand and simplify the left side of the equation:
2xy −y2= 2y dx −x dx
Step 2: Rearranging the terms, we get:
2xy −2y dx =y2−x dx
Step 3: Now, we divide both sides of the equation by 2xy −2y:
dx
x=dy
y
Step 4: Integrating both sides, we have:
Z1
xdx =Z1
ydy
ln |x|= ln |y|+C
Step 5: Exponentiating both sides, we get:
|x|=|y| · eC
Step 6: Now, we can combine the absolute values into one constant k:
x=y·k
Therefore, the solution to the given differential equation is x=y·k, where
kis a constant.
Question 25
Question
Let y(x) be a solution to the differential equation dy
dx =2x+3y
3x+4ywith initial
condition y(0) = 1. Find the value of y(2).
24
Solution
Step 1: First, let’s rewrite the given differential equation to separate variables:
dy
2x+ 3y=dx
3x+ 4y
Step 2: Integrate both sides with respect to x:
Zdy
2x+ 3y=Zdx
3x+ 4y
Step 3: Simplify the integrals by substitution. Let u= 2x+ 3yand v=
3x+ 4y. Then du = 2dx + 3dy and dv = 3dx + 4dy.
Step 4: Now the integrals become:
1
3Zdu
u=1
4Zdv
v
Step 5: Integrating, we get:
1
3ln |u|=1
4ln |v|+C
Step 6: Substitute back uand v:
1
3ln |2x+ 3y|=1
4ln |3x+ 4y|+C
Step 7: Using the initial condition y(0) = 1, we find C= 0:
1
3ln |2x+ 3y|=1
4ln |3x+ 4y|
Step 8: Now, apply x= 0, y = 1:
1
3ln(3) = 1
4ln(4)
Step 9: Solve for y(2) using x= 2:
1
3ln(7) = 1
4ln(11)
Step 10: Therefore, y(2) = 71/3
111/4.
Question 26
Question
Consider the following first-order ordinary differential equation:
dy
dx =2x+ 3y+ 1
x+y−2
Given the initial condition y(0) = 1, find the particular solution to this differ-
ential equation.
25
Solution
Step 1: Rewrite the given ODE in a more standard form:
(x+y−2)dy = (2x+ 3y+ 1)dx
Step 2: Integrate both sides of the equation:
Z(x+y−2)dy =Z(2x+ 3y+ 1)dx
Step 3: Integrate the left side with respect to y:
Z(y+x−2)dy =1
2y2+xy −2y+C1(where C1is the constant of integration)
Step 4: Integrate the right side with respect to x:
Z(2x+3y+1)dx =x2+3xy+x+C2(where C2is another constant of integration)
Step 5: Equate the two sides of the equation and simplify:
1
2y2+xy −2y+C1=x2+ 3xy +x+C2
Step 6: Rearrange the terms to isolate y:
1
2y2+xy −3xy −2y=x2+x−C1+C2
Step 7: Simplify the equation further:
1
2y2−2xy −2y=x2+x+C
Step 8: Consider the initial condition y(0) = 1 to find the value of the
constant C:
1
2(1)2−2(0)(1) −2(1) = 0 + 0 + C=⇒C=−2
Step 9: Substitute the value of Cback into the equation to find the particular
solution: 1
2y2−2xy −2y=x2+x−2
Therefore, the particular solution to the given differential equation with the
initial condition y(0) = 1 is 1
2y2−2xy −2y=x2+x−2.
Question 27
Question
Solve the following first-order ordinary differential equation:
dy
dx =3x2
2y
26
Solution
Step 1: Rewrite the given differential equation in the form dy =f(x)dx.
dy
dx =3x2
2y
2y dy = 3x2dx
Step 2: Integrate both sides of the equation.
Z2y dy =Z3x2dx
y2=x3+C1
where C1is the constant of integration.
Step 3: Solve for yby taking the square root of both sides.
y=±px3+C1
Therefore, the general solution to the given first-order ordinary differential
equation is y=±√x3+C1.
Question 28
Question
Solve the following initial value problem:
y′=√y, y(0) = 0
Solution
Step 1: Rewrite the differential equation as separable by dividing both sides by
√y:
dy
dτ = 1
Step 2: Integrate both sides with respect to τ:
Zdy =Z1dτ
y=τ+C
Step 3: Use the initial condition y(0) = 0 to find the value of the constant
C:
0 = 0 + C
C= 0
27
Step 4: Substitute C= 0 back into the general solution to get the particular
solution:
y=τ
Step 5: Finally, substitute τback in terms of xto get the final solution:
y=x
Therefore, the solution to the initial value problem is y=x.
Question 29
Question
Solve the initial value problem: y′= 2xy +ex2,y(0) = 1.
Solution
Step 1: First, we identify that this is a first-order linear differential equation in
the form y′+P(x)y=Q(x), where P(x)=2xand Q(x) = ex2.
Step 2: To solve this, we will use an integrating factor µ(x) = eRP(x)dx =ex2.
Step 3: Multiply both sides of the differential equation by the integrating
factor:
ex2y′+ 2xyex2=e2x2y+e2x2
Step 4: Rewrite the left side as the derivative of the product of µ(x) and y:
(ex2y)′=e2x2y+e2x2
Step 5: Integrate both sides with respect to x:
Z(ex2y)′dx =Z(e2x2y+e2x2)dx
ex2y=e2x2y
2+C
Step 6: Solve for y:
y=e2x2y
2ex2+Ce−x2=y
2+Ce−x2
Step 7: Apply the initial condition y(0) = 1:
1 = 1
2+C⇒C=1
2
Step 8: Substitute C=1
2back into the equation:
y=y
2+1
2e−x2
28
Step 9: Solve for y:
y−y
2=1
2e−x2⇒y
2=1
2e−x2⇒y=e−x2
Therefore, the solution to the initial value problem is y=e−x2.
Question 30
Question
Consider the differential equation:
y′′ −2y′+y=ex
Find the general solution to this differential equation.
Solution
Step 1: First, we find the complementary solution by solving the associated
homogeneous equation:
y′′ −2y′+y= 0
Step 2: The characteristic equation is r2−2r+ 1 = 0, which factors as
(r−1)2= 0. Thus, we have a repeated root of r= 1.
Step 3: The complementary solution takes the form:
yc(x) = c1ex+c2xex
Step 4: Next, we find a particular solution to the non-homogeneous equation
by the method of undetermined coefficients. Assume a particular solution in the
form of yp(x) = Aex.
Step 5: Substituting yp(x) into the differential equation, we get:
A−2A+A=ex⇒0 = ex
Step 6: Since the last equation has no solutions, we try an alternative form
for the particular solution. Let’s try yp(x) = Axex.
Step 7: Substituting yp(x) into the differential equation, we get:
2A−2A+Axex=ex
Step 8: This gives us A= 1/2. Therefore, the particular solution is yp(x) =
1
2xex.
Step 9: The general solution to the differential equation is the sum of the
complementary and particular solutions:
y(x) = yc(x) + yp(x) = c1ex+c2xex+1
2xex
Therefore, the general solution to the given differential equation is y(x) =
c1ex+c2xex+1
2xex.
29
Question 31
Question
Solve the following first-order ordinary differential equation:
dy
dx =1
2x+ 3y
Solution
Step 1: Rewrite the equation in a suitable form for separation of variables.
We rewrite the equation as: dy
dx =1
2x+ 3y
2x+ 3ydy
dx = 1
Step 2: Separate the variables.
2x dx + 3y dy =dx
Step 3: Integrate both sides.
Z2x dx +Z3y dy =Zdx
x2+3
2y2=x+C
Step 4: Solve for y.
3
2y2=x−x2+C
y2=2x−2x2+ 2C
3
y=±r2x−2x2+ 2C
3
Thus, the general solution to the differential equation is:
y=±r2x−2x2+ 2C
3
Question 32
Question
Consider the following ordinary differential equation:
dy
dx =e2x−2y
Solve the initial value problem when y(0) = 3.
30
Solution
Step 1: First, let’s rewrite the given differential equation in standard form:
dy
dx + 2y=e2x
This is a first-order linear differential equation.
Step 2: We’ll start by finding the integrating factor, which is given by IF =
eR2dx =e2x.
Step 3: Next, we multiply both sides of the differential equation by the
integrating factor:
e2xdy
dx + 2e2xy=e4x
Step 4: Notice that the left-hand side is the derivative of the product e2xy,
so we rewrite the equation as:
d
dx(e2xy) = e4x
Step 5: We can integrate both sides with respect to xto solve for y:
Zd
dx(e2xy)dx =Ze4xdx
Step 6: Integrating both sides gives:
e2xy=1
4e4x+C
Step 7: Now, we apply the initial condition y(0) = 3 to solve for the constant
C:
e2∗0∗3 = 1
4e4∗0+C
3 = 1
4+C
C=11
4
Step 8: Therefore, the particular solution to the initial value problem is:
e2xy=1
4e4x+11
4
Step 9: Finally, we can solve for y:
y=1
4e2x+11
4e−2x
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Question 33
Question
Solve the initial value problem:
dy
dx =4x
y3, y(0) = 1
Solution
Step 1: Rewrite the differential equation in separated variables form:
y3dy = 4x dx
Step 2: Integrate both sides:
Zy3dy =Z4x dx
Step 3: Solve the integrals:
1
4y4= 2x2+C
where Cis the constant of integration.
Step 4: Using the initial condition y(0) = 1, we have:
1
4·14= 2 ·02+C=⇒C=1
4
Step 5: Substitute back to find the particular solution:
1
4y4= 2x2+1
4
Step 6: Simplify the equation:
y4= 8x2+ 1
Step 7: Solve for yby taking the fourth root of both sides:
y=4
p8x2+ 1
Therefore, the solution to the initial value problem is y=4
√8x2+ 1 with
the initial condition y(0) = 1.
Question 34
Question
Find the general solution to the differential equation:
dy
dx =1
xy +x2y
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Solution
To solve this first-order ordinary differential equation, we can rewrite it in the
form y′+P(x)y=Q(x), where P(x) and Q(x) are functions of x. In this case,
we have P(x) = x2and Q(x) = 1
x.
Step 1: We can solve the differential equation using an integrating factor.
The integrating factor µ(x) is defined as µ(x) = eRP(x)dx.
In this case, P(x) = x2, so µ(x) = eRx2dx.
Step 2: Calculating the integral and finding µ(x):
Zx2dx =1
3x3
Thus, µ(x) = e1
3x3.
Step 3: Multiplying the differential equation by the integrating factor:
e1
3x3dy
dx +e1
3x3x2y=e1
3x31
x
Step 4: Rewriting the left-hand side using the product rule:
d
dx e1
3x3y=e1
3x31
x
Step 5: Integrating both sides with respect to x:
Zd
dx e1
3x3ydx =Ze1
3x31
xdx
Step 6: The solution after integrating both sides:
e1
3x3y=Ze1
3x31
xdx +C
Step 7: Simplifying the integral:
e1
3x3y=Ze1
3x31
xdx +C
The general solution is y(x) = e−1
3x3Re1
3x31
xdx +C, where Cis the
constant of integration.
Question 35
Question
Consider the differential equation dy
dx = 2y−x2with initial condition y(0) = 1.
Find the particular solution to the differential equation.
33
Solution
Step 1: First, let’s write the homogeneous equation associated with the given
differential equation: dy
dx = 2y
This is a separable differential equation, so we can rewrite it as:
dy
y= 2dx
Step 2: Integrating both sides gives us:
Z1
ydy =Z2dx
ln |y|= 2x+C1
where C1is the constant of integration.
Step 3: Exponentiating both sides, we get:
|y|=e2x+C1=e2xeC1=Ce2x
where C=eC1.
Step 4: Now, we need to find the particular solution by considering the non-
homogeneous part of the differential equation. Let yp=Ax2+Bx +Cbe the
particular solution.
Step 5: Substitute ypinto the differential equation and differentiate to match
the coefficients of xand constants on both sides. We have:
2(Ax2+Bx +C)−x2= 2Ax2+ 2Bx + 2C−x2
Equating coefficients, we get:
2A= 2,2B= 0,2C= 0
which gives A= 1, B= 0, and C= 0.
Step 6: Therefore, the particular solution is yp=x2.
Step 7: Finally, the general solution is given by the sum of the homogeneous
and particular solutions:
y=Ce2x+x2
Applying the initial condition y(0) = 1, we find C= 1, so the particular solution
to the differential equation is y=e2x+x2.
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