MATH 334 - DIFFERENTIAL
EQUATIONS - Ordinary differential
equations
Question Bank - Set 5
Liberty University
Question 1
Question
Consider the initial value problem
dy
dx = 2y, y(0) = 3.
Find the solution y(x) using the method of separation of variables.
Solution
Step 1: Rewrite the differential equation to separate the variables.
dy
dx = 2y⇒dy
y= 2dx
Step 2: Integrate both sides with respect to their respective variables.
Z1
ydy =Z2dx
ln |y|= 2x+C
where Cis the constant of integration.
Step 3: Solve for y.
y=e2x+C=e2xeC=Ce2x
Step 4: Apply the initial condition y(0) = 3 to find the value of C.
3 = Ce0⇒C= 3
Step 5: Write the final solution.
y(x)=3e2x
Therefore, the solution to the initial value problem is y(x) = 3e2x.
Question 2
Question
Find the particular solution of the differential equation y′=2x
ysatisfying the
initial condition y(1) = 3.
Solution
Given the differential equation y′=2x
y, we want to find the particular solution
satisfying the initial condition y(1) = 3.
Step 1: Separate the variables by moving all yterms to one side of the
equation and all xterms to the other side.
dy
dx =2x
y
Step 2: Multiply both sides by yand dx to separate the variables.
y dy = 2x dx
Step 3: Integrate both sides.
Zy dy =Z2x dx
Step 4: Integrate the left side with respect to y.
1
2y2=x2+C
Step 5: Applying the initial condition y(1) = 3 to solve for the constant C.
1
2(3)2= (1)2+C
C=9
2−1
C=7
2
Step 6: Substitute the value of Cback into the general solution.
1
2y2=x2+7
2
Step 7: Solve for y.
y2= 2x2+ 7
y=±p2x2+ 7
Therefore, the particular solution of the differential equation y′=2x
ysatis-
fying the initial condition y(1) = 3 is y=√2x2+ 7.
2
Question 3
Question
Find the general solution to the following differential equation:
y′′ +y= 0
Solution
Step 1: Write the characteristic equation for the given differential equation.
The characteristic equation is obtained by replacing y′′ with r2, where ris the
unknown variable. So, the characteristic equation is:
r2+ 1 = 0
Step 2: Solve the characteristic equation. Setting r2+ 1 = 0 gives us:
r2=−1
r=±i
Therefore, the roots of the characteristic equation are complex conjugates.
Step 3: Write the general solution using the roots of the characteristic equa-
tion. The general solution to a second-order homogeneous differential equation
with complex roots is:
y(t) = eat(c1cos(bt) + c2sin(bt))
where a= 0, b= 1, and c1, c2are arbitrary constants.
Step 4: Substitute the values of aand binto the general solution. Thus,
the general solution to the differential equation y′′ +y= 0 is:
y(t) = c1cos(t) + c2sin(t)
where c1, c2are arbitrary constants.
Question 4
Question
Solve the following differential equation using the method of undetermined co-
efficients: y′′ −4y′+ 4y=e2x.
3
Solution
To solve the given differential equation using the method of undetermined coef-
ficients, we first find the complementary function by solving the characteristic
equation r2−4r+ 4 = 0.This gives us a repeated root at r= 2,so the comple-
mentary function is yc(x) = c1e2x+c2xe2x.
Next, we need to find a particular solution yp(x) for y′′−4y′+4y=e2x.Since
e2xis of the same form as the characteristic equation’s root, we will assume a
particular solution of the form yp(x) = Ax2e2x.
Step 1: Find y′
p(x) and y′′
p(x).
yp(x) = Ax2e2x
y′
p(x) = (2Ax2+ 4Ax)e2x
y′′
p(x) = (4Ax2+ 8Ax + 2A)e2x
Step 2: Substitute yp(x), y′
p(x),and y′′
p(x) into the differential equation and
simplify.
(4Ax2+ 8Ax + 2A)e2x−4(2Ax2+ 4Ax)e2x+ 4Ax2e2x=e2x
(4Ax2+ 8Ax + 2A−8Ax2−16Ax + 4A+ 4Ax2)e2x=e2x
(2A)e2x=e2x
Step 3: Equate the coefficients to find A=1
2.
Step 4: Write down the particular solution.
yp(x) = 1
2x2e2x
Step 5: The general solution is the sum of the complementary function and
the particular solution:
y(x) = yc(x) + yp(x)
y(x) = c1e2x+c2xe2x+1
2x2e2x
Therefore, the solution to the differential equation y′′ −4y′+ 4y=e2xis
y(x) = c1e2x+c2xe2x+1
2x2e2x.
Question 5
Question
Find the general solution to the differential equation: dy
dx =y2−x2
2xy .
4
Solution
To solve this differential equation, we first notice that it is separable. Let’s
rearrange the terms to make it more obvious:
dy
dx =y2−x2
2xy
2y dy = (y2−x2)dx
x
Step 1: Separating variables
2y dy = (y2−x2)dx
x
Step 2: Integrating both sides Integrating both sides gives:
Z2y dy =Z(y2−x2)dx
x
y2=x2
2+C
Step 3: Simplifying the solution Taking the square root of both sides
gives the general solution:
y=±rx2
2+C
So, the general solution to the differential equation is y=±qx2
2+C, where
Cis the constant of integration.
Question 6
Question
Find the general solution to the differential equation:
dy
dx =2x+ 1
y2
Solution
To solve this differential equation, we will first rewrite it in a separable form.
Step 1: Rearrange the terms to separate variables.
y2dy = (2x+ 1) dx
Step 2: Integrate both sides.
Zy2dy =Z(2x+ 1) dx
5
Step 3: Integrate the left side:
1
3y3=x2+x+C
where Cis the constant of integration.
Step 4: Solve for yby taking the cube root of both sides:
y=3
p3x2+ 3x+ 3C
Therefore, the general solution to the differential equation is:
y=3
p3x2+ 3x+ 3C
Question 7
Question
Consider the differential equation dy
dx =x+y
x−y, where y(0) = 1. Find the particu-
lar solution to the initial value problem.
Solution
Step 1: Rewrite the differential equation in a more standard form by multiplying
both sides by (x−y) to separate variables:
dy
dx(x−y) = x+y
xdy
dx −ydy
dx =x+y
Step 2: Rearrange the terms and integrate both sides with respect to x:
xdy
dx −ydy
dx =x+y
Zx dy −Zy dy =Zx+Zy dx
xy −y2
2=x2
2+xy +C
Step 3: Simplify the equation by multiplying through by 2 to clear the
fractions:
2xy −y2=x2+ 2xy + 2C
Step 4: Rearrange the terms to get a quadratic equation in terms of y:
y2−4xy +x2+ 2C= 0
6
Step 5: Apply the initial condition y(0) = 1 to find the particular solution:
1−0+0+2C= 0
2C=−1
C=−1
2
Step 6: Substitute C=−1
2back into the equation to get the particular
solution:
y2−4xy +x2−1=0
Question 8
Question
Solve the following ordinary differential equation:
y′′ + 4y′+ 4y= 0
Solution
Step 1: Write the characteristic equation by assuming y=ert.
−r2+ 4r+ 4 = 0
Step 2: Solve the characteristic equation to find the roots.
r=−4±p42−4(−1)(4)
2(−1)
r=−4±√16
−2
r=−2
Step 3: Write the general solution based on the roots.
y(t) = c1e−2t+c2te−2t
Where c1and c2are constants to be determined.
Step 4: Take the first derivative of y(t).
y′(t) = −2c1e−2t+c2e−2t−2c2te−2t
Step 5: Take the second derivative of y(t).
y′′(t)=4c1e−2t−4c2e−2t+ 4c2te−2t
7
Step 6: Substitute y(t), y′(t), and y′′(t) into the original differential equation
and simplify.
−(4c1e−2t−4c2e−2t+4c2te−2t)+4(−2c1e−2t+c2e−2t−2c2te−2t)+4(c1e−2t+c2te−2t)=0
Step 7: Simplify the equation and find the values of c1and c2that satisfy
the equation. After simplifying, we find c1= 0 and c2is a free variable. So, the
general solution to the differential equation is:
y(t) = c2te−2t
Question 9
Question
Find the general solution to the differential equation
d2y
dx2−5dy
dx + 6y= 0
Solution
Step 1: We first write the characteristic equation for the given differential equa-
tion:
r2−5r+ 6 = 0
Step 2: Solve the characteristic equation to find the roots r1and r2:
The characteristic equation factors to (r−2)(r−3) = 0, so the roots are
r1= 2 and r2= 3.
Step 3: The general solution to the differential equation is then given by
y(x) = c1er1x+c2er2x
where c1and c2are constants to be determined.
Step 4: Substitute the values of r1and r2into the general solution expression:
y(x) = c1e2x+c2e3x
Step 5: Consider the initial conditions which are not given in this problem.
To find the specific solution, additional information such as the initial value of
yand/or its derivative may be needed.
Question 10
Question
Consider the differential equation y′′ + 3y′+ 2y= 0. Find the general solution
of the differential equation.
8
Solution
Step 1: First, we write the characteristic equation by substituting y=ert into
the differential equation y′′ + 3y′+ 2y= 0.
r2+ 3r+ 2 = 0
Step 2: Factorizing the quadratic equation gives us:
(r+ 1)(r+ 2) = 0
Step 3: This leads to the roots r=−1 and r=−2.
Step 4: The general solution of the differential equation is given by:
y(t) = c1e−t+c2e−2t
where c1and c2are arbitrary constants.
Question 11
Question
Find the general solution to the following first-order ordinary differential equa-
tion:
y′+ 2xy =x
Solution
Step 1: This is a first-order linear differential equation of the form
y′+P(x)y=Q(x)
where P(x)=2xand Q(x) = x. To solve this, we will use an integrating factor
µ(x), defined by
µ(x) = eRP(x)dx
Step 2: Compute the integrating factor
µ(x) = eR2xdx =ex2
Step 3: Multiply both sides of the differential equation by the integrating
factor
ex2y′+ 2xex2y=xex2
Step 4: Rewrite the left-hand side as the derivative of the product of yand
the integrating factor
(ex2y)′=xex2
9
Step 5: Integrate both sides with respect to x
Z(ex2y)′dx =Zxex2dx
ex2y=Zxex2dx +C
Step 6: Integrate Rxex2dx using substitution (u=x2)
Zxex2dx =1
2ex2+C′
Step 7: Substitute back into the equation and solve for y
ex2y=1
2ex2+C
y=1
2+Ce−x2
Step 8: Therefore, the general solution to the differential equation is
y=1
2+Ce−x2
Question 12
Question
Solve the following initial value problem:
y′+y=e−x, y(0) = 1
Solution
Step 1: First, we identify this as a first-order linear nonhomogeneous differential
equation of the form y′+p(x)y=q(x).
Step 2: The integrating factor is given by µ(x) = eRp(x)dx. Here, p(x) = 1,
so µ(x) = eR1dx =ex.
Step 3: Multiply both sides of the equation by the integrating factor µ(x):
exy′+exy=exe−x
This simplifies to:
(exy)′= 1
Step 4: Integrate both sides with respect to x:
Z(exy)′dx =Z1dx
10
This gives us:
exy=x+C
Step 5: Apply the initial condition y(0) = 1 to find the value of the constant
C:
e0·1 = 0 + C=⇒C= 1
So, the particular solution to the differential equation is:
exy=x+ 1
Step 6: Solve for y:
y=e−x(x+ 1)
Therefore, the solution to the initial value problem is y=e−x(x+ 1).
Question 13
Question
Consider the differential equation:
y′′ −4y′+ 4y=e2x
Find the general solution of the differential equation.
Solution
Step 1: First, find the complementary function by solving the characteristic
equation:
r2−4r+ 4 = 0
(r−2)2= 0
r= 2(a repeated root)
Therefore, the complementary function is:
yc=c1e2x+c2xe2x
Step 2: To find a particular integral, assume a particular solution of the
form:
yp=Axe2x
Step 3: Calculate the first and second derivatives of yp:
y′
p=Ae2x+ 2Axe2x
y′′
p= 2Ae2x+ 4Axe2x
11
Step 4: Substitute yp, y′
p, y′′
pinto the differential equation:
2Ae2x+ 4Axe2x−4(Ae2x+ 2Axe2x) + 4(Axe2x) = e2x
2Ae2x+ 4Axe2x−4Ae2x−8Axe2x+ 4Axe2x=e2x
2Ae2x−4Ae2x=e2x
−2Ae2x=e2x
A=−1
2
Step 5: The particular solution is:
yp=−1
2xe2x
Step 6: The general solution is the sum of the complementary function and
the particular integral:
y=yc+yp=c1e2x+c2xe2x−1
2xe2x
Question 14
Question
Consider the initial value problem
dy
dx =1
y, y(0) = 1
Find the solution to the differential equation.
Solution
Step 1: Rewrite the differential equation as separate variables.
y dy =dx
Step 2: Integrate both sides of the equation.
Zy dy =Zdx
Step 3: Solve the integrals.
1
2y2=x+C
where Cis the constant of integration.
12
Step 4: Use the initial condition y(0) = 1 to find the value of C.
1
2(1)2= 0 + C
1
2=C
Step 5: Substitute the value of Cback into the equation.
1
2y2=x+1
2
Step 6: Simplify the equation to find the solution in implicit form.
y2= 2x+ 1
Step 7: To find the explicit form of the solution, take the square root of both
sides.
y=±√2x+ 1
Thus, the solution to the differential equation is y=±√2x+ 1.
Question 15
Question
Solve the following first-order ordinary differential equation:
dy
dx =2x+ 3y+ 1
3x+ 2y−2
Solution
Step 1: Rearrange the equation in the form M dx +Ndy = 0 where Mand N
depend only on xand y:
(3x+ 2y−2)dy −(2x+ 3y+ 1)dx = 0
Step 2: Check if the equation is exact by calculating ∂M
∂y and ∂N
∂x :
∂M
∂y = 2 =∂N
∂x =−1
Step 3: Multiply the equation by an integrating factor µ(x, y) to make it
exact. Since µdepends only on x, we have:
1
µ[(3x+ 2y−2)dy −(2x+ 3y+ 1)dx] = 0
13
Step 4: Solve for µ(x) by setting the coefficients of dx and dy equal and
solve for µ:dy
dx =2x+ 3y+ 1
3x+ 2y−2
Therefore, µ(x) becomes:
µ(x) = 1
3x+ 2y−2
Step 5: Multiply the original ODE by the integrating factor µ(x):
1
3x+ 2y−2[(3x+ 2y−2)dy −(2x+ 3y+ 1)dx] = 0
Simplifying, we get:
dy −2x+ 3y+ 1
3x+ 2y−2dx = 0
Step 6: Integrate the equation step by step to find the solution. This will
involve finding a function F(x, y) such that ∂F
∂x =Nand ∂F
∂y =−M.
F(x, y) = Z−2x+ 3y+ 1
3x+ 2y−2dx +g(y)
Step 7: Once F(x, y) is found, set F(x, y) = Cto find the general solution
to the ODE.
Question 16
Question
Solve the following first-order ordinary differential equation:
dy
dx =2x+y+ 4
x−y−2
Solution
To solve the given first-order ordinary differential equation, we will first rear-
range it as follows:
(x−y−2) dy
dx = 2x+y+ 4
Step 1: Let’s expand the left side of the equation using the product rule:
xdy
dx −ydy
dx −2dy
dx = 2x+y+ 4
14
Step 2: Next, let’s rearrange the terms as follows:
xdy
dx −ydy
dx −2dy
dx −y= 2x+ 4
Step 3: Factor out the dy
dx term:
dy
dx(x−y−2) −y= 2x+ 4
Step 4: Now, we can integrate both sides with respect to x:
Zdy
dx(x−y−2) −ydx =Z(2x+ 4)dx
Step 5: Integrating the left side will give:
y(x)−Zydx −Z2dx =Z(2x+ 4)dx
Step 6: Simplify the left side and integrate the right side:
y(x)−Zydx −2x=x2+ 4x+C
Step 7: Finally, the general solution to the differential equation is:
y(x) = x2+ 4x+C+y(x)ex
where Cis an arbitrary constant.
Question 17
Question
Consider the following first-order ordinary differential equation:
dy
dx =y
1 + x2
Find the general solution of the differential equation.
Solution
Step 1: Rewrite the equation as a separable differential equation by multiplying
both sides by (1 + x2):
(1 + x2)dy
dx =y
Step 2: Separate the variables by dividing both sides by yand multiplying
by dx:dy
y=dx
1 + x2
15
Step 3: Integrate both sides with respect to their respective variables:
Z1
ydy =Z1
1 + x2dx
Step 4: Integrate to find the general solution:
ln |y|= arctan(x) + C
where Cis the constant of integration.
Step 5: Exponentiate both sides to solve for y:
y=earctan(x)+C=earctan(x)·eC=C′earctan(x)
where C′=eCis another constant.
Therefore, the general solution to the differential equation is:
y=C′earctan(x)
Question 18
Question
Consider the following first-order ordinary differential equation:
dy
dx =x+y
x−y
Find the general solution to this differential equation.
Solution
Step 1: Rewrite the given differential equation in a more convenient form by
separating the variables:
dy
dx =x+y
x−y=⇒(x−y)dy = (x+y)dx
Step 2: Integrate both sides of the separated variables equation:
Z(x−y)dy =Z(x+y)dx
Step 3: Perform the integrations:
Z(x−y)dy =Z(x+y)dx =⇒xy −y2
2=x2+y2
2+C
Step 4: Rearrange the terms to obtain the general solution in implicit form:
xy −x2= 2C=⇒y=2C+x2
x
So, the general solution to the given initial value problem is y=2C+x2
x,
where Cis an arbitrary constant.
16
Question 19
Question
Consider the differential equation:
y′+2y
x=x2cos(x)
Find the general solution to the above differential equation.
Solution
Step 1: First, we identify the type of differential equation. This is a first-order
linear ordinary differential equation of the form y′+P(x)y=Q(x).
Step 2: To solve this type of differential equation, we use an integrating
factor which is defined as µ(x) = exp RP(x)dx .
Step 3: Let’s calculate the integrating factor:
µ(x) = exp Z2
xdx
µ(x) = exp(2 ln |x|)
µ(x) = x2
Step 4: Next, we multiply both sides of the differential equation by the
integrating factor µ(x):
x2(y′+2y
x) = x4cos(x)
Step 5: After simplifying, we get:
x2y′+ 2xy =x4cos(x)
Step 6: Now, we can rewrite the equation as the derivative of a product:
(x2y)′=x4cos(x)
Step 7: Integrating both sides with respect to x, we get:
x2y=Zx4cos(x)dx +C
Step 8: Solve the integral on the right hand side to find:
x2y=x4sin(x)−4x3cos(x)−12x2sin(x) + 24xcos(x) + C
Step 9: Finally, the general solution to the differential equation is:
y=x2sin(x)−4x3cos(x)−12x2sin(x) + 24xcos(x) + C
x2
17
Question 20
Question
Solve the following first-order ordinary differential equation:
y′+ 2xy =x2, y(0) = 1
Solution
Step 1: We will first rewrite the given differential equation in standard form,
which is the form y′+P(x)y=Q(x), where P(x) and Q(x) are functions of x.
Given equation: y′+ 2xy =x2We can rewrite this as:
y′=x2−2xy
Step 2: The integrating factor is given by the expression µ(x) = eRP(x)dx.
In this case, P(x)=2x, so
µ(x) = eR2xdx
µ(x) = ex2
Step 3: We multiply both sides of the differential equation by the integrating
factor µ(x):
ex2y′=ex2(x2−2xy)
Step 4: When we apply the product rule to the left side, we get:
d
dx(ex2y) = x2ex2−2xyex2
Step 5: Integrating both sides of the equation with respect to xgives:
ex2y=Z(x2ex2−2xyex2)dx
Step 6: Simplify the integral:
ex2y=1
3x3ex2−yex2+C
where Cis the constant of integration.
Step 7: Let’s now solve for y:
y=1
3x3−y+Ce−x2
Step 8: Applying the initial condition, y(0) = 1, we find:
1=0−1 + C
C= 2
Step 9: Substituting back this value of Cinto the general solution, we get
the final solution:
y=1
3x3−y+ 2e−x2
18
Question 21
Question
Consider the following first-order ordinary differential equation:
dy
dx + 2xy =x
Find the general solution to this differential equation.
Solution
Step 1: We first rewrite the differential equation in standard form:
dy
dx =x−2xy
Step 2: Next, we identify that this is a linear first-order differential equation
of the form: dy
dx +P(x)y=Q(x)
where P(x) = −2xand Q(x) = x.
Step 3: To solve this type of differential equation, we use an integrating
factor µ(x) = eRP(x)dx. Calculating the integrating factor, we have:
µ(x) = eR−2xdx =e−x2
Step 4: We then multiply the integrating factor with the differential equa-
tion:
e−x2dy
dx +e−x2(−2xy) = e−x2x
Step 5: After rewritting the equation, we recognize the left side as the deriva-
tive of the product of yand µ(x):
d
dx(e−x2y) = e−x2x
Step 6: Integrating both sides with respect to x, we get:
e−x2y=Zxe−x2dx
Step 7: We now solve the integral on the right side:
Zxe−x2dx =−1
2e−x2+C
Step 8: Substituting the integral back into our equation from step 6, we
have:
e−x2y=−1
2e−x2+C
19
Step 9: Finally, we solve for yto obtain the general solution:
y=−1
2+Cex2
where Cis the constant of integration.
Question 22
Question
Consider the following first-order ordinary differential equation:
dy
dx =2x+y
x−2y
Find the general solution for the differential equation.
Solution
Step 1: Rewrite the given differential equation in a more suitable form:
(x−2y)dy −(2x+y)dx = 0
Step 2: Check whether the given differential equation is exact by verifying
the equality of mixed partial derivatives:
∂
∂y (x−2y) = −2,∂
∂x (−2x−y) = −2
Since the mixed partial derivatives are equal, the differential equation is exact.
Step 3: Integrate the coefficient of dy with respect to y:
Z(x−2y)dy =xy −y2+g(x)
Step 4: Set the result equal to the function h(x), where h(x) is the an-
tiderivative with respect to xof the coefficient of dx:
xy −y2+g(x) = h(x)
Step 5: Solve for g(x) by comparing terms containing y:
g(x) = h(x)
Step 6: Substitute the solution for g(x) back into the equation:
xy −y2=h(x)
Step 7: Rearrange terms to obtain the general solution:
xy −y2=C
where Cis the constant of integration.
20
Question 23
Question
Consider the initial value problem:
dy
dx = 2xy, y(0) = 1
Find the solution to the initial value problem.
Solution
Step 1: We can rewrite the differential equation as:
dy
dx = 2xy =⇒dy
y= 2x dx
Step 2: Integrating both sides gives us:
Z1
ydy =Z2x dx =⇒ln |y|=x2+C
where Cis the constant of integration.
Step 3: Exponentiating both sides gives us:
y=ex2+C=Cex2
since ex2=eC·ex2.
Step 4: Applying the initial condition y(0) = 1, we find C= 1. Thus, the
solution to the initial value problem is:
y=ex2
Question 24
Question
Find the general solution to the following first-order ordinary differential equa-
tion:
dy
dx + 2xy =x2
Solution
Step 1: This is a first-order linear differential equation in the form dy
dx +P(x)y=
Q(x), where P(x) = 2xand Q(x) = x2. To solve this, we use an integrating
factor, µ(x) = eRP(x)dx.
21
Step 2: Calculate the integrating factor:
µ(x) = eR2xdx =ex2
Step 3: Multiply the differential equation by the integrating factor:
ex2dy
dx + 2xex2y=x2ex2
Step 4: Recognize that the left side is the derivative of the product of µ(x)
and ywith respect to x, so rewrite the equation as:
d
dx(ex2y) = x2ex2
Step 5: Integrate both sides with respect to x:
Zd
dx(ex2y)dx =Zx2ex2dx
Step 6: Solve the integrals:
ex2y=1
2x2ex2+C
Step 7: Finally, solve for yto find the general solution:
y=1
2x2+Ce−x2
Question 25
Question
Find the general solution to the following first-order ordinary differential equa-
tion: dy
dx =y
x+x
Solution
Step 1: Let’s rewrite the given differential equation in a more standard form:
dy
dx −y
x=x
Step 2: This is a linear first-order differential equation in the form dy
dx +
P(x)y=Q(x). Here, P(x) = −1
xand Q(x) = x.
Step 3: To solve this, we first find the integrating factor I(x), which is given
by:
I(x) = eRP(x)dx =eR−1
xdx
22
Step 4: Simplifying the integral to find I(x):
I(x) = e−ln |x|=1
|x|=1
x
Step 5: Multiply both sides of the differential equation by the integrating
factor I(x):
1
x
dy
dx −1
x2y= 1
Step 6: Rewrite the left side as a derivative of a product using the product
rule: d
dx 1
xy= 1
Step 7: Integrating both sides with respect to x:
Zd
dx 1
xydx =Z1dx
1
xy=x+C
Step 8: Multiply through by xto isolate y:
y=x2+Cx
Therefore, the general solution to the differential equation is y=x2+Cx,
where Cis an arbitrary constant.
Question 26
Question
Solve the initial value problem:
y′+2
xy=xln x, y(1) = 1
Solution
Step 1: We first identify the integrating factor for this first-order linear ODE.
The integrating factor µ(x) is given by µ(x) = eR2
xdx.
µ(x) = eR2
xdx
=e2 ln x
=x2
23
Step 2: Next, we multiply both sides of the ODE by the integrating factor
µ(x) = x2.
x2y′+ 2xy =x3ln x
Step 3: We can now rewrite the left-hand side of the ODE as the derivative
of the product of x2and ywith respect to x:
d
dx(x2y) = x3ln x
Step 4: Integrating both sides with respect to x, we have
Zd
dx(x2y)dx =Zx3ln x dx
x2y=1
4x4ln x−1
16x4+C
Step 5: To find the constant of integration C, we apply the initial condition
y(1) = 1.
1 = 1
4−1
16 +C
C= 1 + 1
4−1
16
C=13
16
Step 6: Therefore, the solution to the initial value problem is
y(x) = 1
x21
4x4ln x−1
16x4+13
16
Question 27
Question
Consider the first-order ordinary differential equation:
dy
dx = 2xy +x
Find the general solution to this differential equation.
Solution
Step 1: This is a first-order linear differential equation in the form of:
dy
dx +P(x)y=Q(x)
24
where P(x)=2xand Q(x) = x.
Step 2: To solve this equation, we will use an integrating factor eRP(x)dx.
The integrating factor is eR2xdx =ex2.
Step 3: Multiply both sides of the differential equation by the integrating
factor:
ex2dy
dx + 2xex2y=ex2x
Step 4: Recognize that the left side is the derivative of the product of yand
the integrating factor: d
dx(ex2y) = ex2x
Step 5: Integrate both sides with respect to x:
Zd
dx(ex2y)dx =Zex2xdx
Step 6: Integrate the right side:
ex2y=Zxex2dx
Step 7: Integrate the right side using substitution u=x2, du = 2xdx:
ex2y=1
2Zeudu =1
2ex2+C
where Cis the constant of integration.
Step 8: Solve for y:
y=1
2+Ce−x2
So, the general solution to the differential equation is y=1
2+Ce−x2, where
Cis an arbitrary constant.
Question 28
Question
Solve the following first-order ordinary differential equation:
dy
dx = (3x2+ 4y)/(2x2−y)
Solution
Step 1: Rewrite the differential equation in a more standard form by multiplying
both sides by 2x2−y:
(2x2−y)dy
dx = 3x2+ 4y
25
Step 2: Notice that the left-hand side resembles the product rule, so we
rewrite it as a derivative with respect to x:
d
dx (2x2−y)y= 3x2+ 4y
Step 3: Integrate both sides with respect to x:
Zd
dx (2x2−y)ydx =Z(3x2+ 4y)dx
Step 4: Simplify and integrate:
(2x2−y)y=x3+ 4xy +C
Step 5: Rearrange the equation to express y explicitly:
y2−2x2y=x3+ 4xy +C
Step 6: Complete the square on the left-hand side to make it a perfect square
trinomial:
(y−x)2=x3+ 4xy +C+x2
Step 7: Take the square root of both sides and solve for y:
y−x=±px3+ 4xy +C+x2
y=x±px3+ 4xy +C+x2
Therefore, the solution to the differential equation is y=x±px3+ 4xy +C+x2.
Question 29
Question
Consider the initial value problem:
y′=2x+y+ 1
x−y−1, y(0) = 1
Determine the interval on which the solution exists and show that the solution
can be explicitly written.
Solution
Step 1: Rewrite the given differential equation in standard form.
(x−y−1) dy
dx = 2x+y+ 1
26
Step 2: Rewrite the equation in a more familiar form by expanding the left
side of the equation.
xdy
dx −ydy
dx −dy
dx = 2x+y+ 1
Step 3: Rearrange the terms to isolate dy
dx .
dy
dx +y=x+ 1
Step 4: Find the integrating factor by taking the exponential of the integral
of the coefficient of y.
Let µ(x) = eR1dx =ex
Step 5: Use the integrating factor to solve for y.
exdy
dx +exy=ex(x+ 1)
Step 6: Recognize the left side as the product rule of the integrating factor
with y.d
dx(exy) = ex(x+ 1)
Step 7: Integrate both sides to find the general solution.
exy=Zex(x+ 1) dx =ex(x−1) + C
Step 8: Solve for y in the general solution.
y=x−1 + Ce−x
Step 9: Use the initial condition y(0) = 1 to find the specific solution.
1=0−1 + Ce0=⇒C= 2
Step 10: Substitute C= 2 back into the general solution to get the particular
solution.
y=x−1+2e−x
Therefore, the solution to the initial value problem is y=x−1 + 2e−xfor
x > 0.
Question 30
Question
Consider the differential equation y′′ −4y′+ 4y=e2x. Find the general solution
to this differential equation.
27
Solution
Step 1: First, find the characteristic equation by solving the auxiliary equation
r2−4r+4 = 0. The characteristic equation is (r−2)2= 0, which has a repeated
root r= 2.
Step 2: The general solution to the homogeneous equation y′′ −4y′+ 4y= 0
is then yh(x) = c1e2x+c2xe2x, where c1and c2are arbitrary constants.
Step 3: To find a particular solution to the non-homogeneous equation y′′ −
4y′+ 4y=e2x, try a particular solution of the form yp(x) = ae2x.
Step 4: Substitute yp(x) into the differential equation to find a.
(2a)e2x−4(2a)e2x+ 4ae2x=e2x
−6ae2x+ 4ae2x=e2x
−2ae2x=e2x
a=−1
2
Step 5: Therefore, the particular solution to the non-homogeneous equation
is yp(x) = −1
2e2x.
Step 6: The general solution to the differential equation is then the sum of
the homogeneous and particular solutions:
y(x) = yh(x) + yp(x) = c1e2x+c2xe2x−1
2e2x
where c1and c2are arbitrary constants.
Question 31
Question
Consider the differential equation dy
dx =√ywhere y(0) = 1. Find the particular
solution to this initial value problem.
Solution
Step 1: Rewrite the differential equation as dy
√y=dx.
Step 2: Integrate both sides with respect to x.
Z1
√ydy =Zdx
Step 3: Simplify the integrals.
2√y=x+C
Step 4: Use the initial condition y(0) = 1 to find the value of the constant
C.
2√1 = 0 + C
28
C= 2
Step 5: Substitute the value of Cback into the general solution.
2√y=x+ 2
Step 6: Solve for y.√y=x
2+ 1
y=x
2+ 12
Therefore, the particular solution to the initial value problem is y=x
2+ 12.
Question 32
Question
Solve the following first-order ordinary differential equation:
dy
dx +y=e−x
Solution
Step 1: First, we identify that this is a linear first-order ODE of the form
dy
dx +P(x)y=Q(x), where P(x) = 1 and Q(x) = e−x.
Step 2: To solve this differential equation, we first find the integrating factor
I(x), which is given by I(x) = eRP(x)dx =eR1dx =ex.
Step 3: We multiply both sides of the ODE by the integrating factor:
exdy
dx +exy=exe−x
Step 4: Notice that the left-hand side is now the result of applying the
product rule for differentiation, specifically d
dx (exy). So, we can rewrite the
equation as: d
dx(exy)=1
Step 5: Integrating both sides with respect to x, we get:
exy=x+C
Step 6: Solving for y, we have:
y=e−x(x+C)
Therefore, the general solution to the differential equation dy
dx +y=e−xis
y=e−x(x+C), where Cis an arbitrary constant.
29
Question 33
Question
Solve the following first-order ordinary differential equation:
dy
dx =3x2+ 2y
2x
Solution
Step 1: Rewrite the given ODE in a more suitable form. Let’s rewrite the ODE
as: dy
dx =3x2
2x+2y
2x
dy
dx =3
2x+y
x
Step 2: This is now a separable differential equation. We can rewrite it as:
dy
dx −y
x=3
2x
Step 3: Now, we can use an integrating factor to solve this linear ODE. The
integrating factor is given by eR−1
xdx =e−ln |x|=1
|x|.
Step 4: Multiply the integrating factor by the entire equation to get:
1
|x|
dy
dx −1
|x|
y
x=3
2
Step 5: Rewrite the equation in its integrated form:
d
dx(y
|x|) = 3
2
Step 6: Integrate both sides with respect to x:
Zd
dx(y
|x|)dx =Z3
2dx
Step 7: The left-hand side simplifies to y
|x|=3
2x+C, where Cis the constant
of integration.
Step 8: Finally, we can solve for y:
y=3
2x|x|+C|x|
y=3
2x2+Cx
Thus, the solution to the given first-order ODE is y=3
2x2+Cx, where C
is a constant.
30
Question 34
Question
Solve the following first-order ordinary differential equation:
dy
dx =y2−x2
Solution
Step 1: First, we rewrite the given differential equation in the standard form of
a separable differential equation:
dy
dx =y2−x2
Step 2: Rearrange the equation to separate the variables xand y:
dy
y2−x2=dx
Step 3: Factor the denominator on the left side using the difference of
squares: dy
(y−x)(y+x)=dx
Step 4: Use partial fraction decomposition to split the expression on the left
side into two separate fractions:
dy
(y−x)(y+x)=A
y−x+B
y+x
Step 5: Multiply both sides by the denominator to solve for Aand B:
dy =A(y+x)dx +B(y−x)dx
Step 6: Integrate both sides with respect to yon the left side and xon the
right side to solve for y:
Zdy =ZA(y+x)dx +ZB(y−x)dx
Step 7: Integrate all terms separately:
y=Axy +A
2x2+By −B
2x2+C
Step 8: Rearrange the equation by combining the terms with yand those
without y:
y(1 −A) = A−B
2x2+C
31
Step 9: To solve for y, we set 1 −A= 0 and A−B= 0 (since xcan be any
value):
A= 1, B = 1
Step 10: Substitute A= 1 and B= 1 back into the equation:
y=x2+C
So, the general solution to the differential equation is y=x2+C, where C
is an arbitrary constant.
Question 35
Question
Consider the differential equation:
y′+ 2xy =x2.
Find the particular solution to the differential equation that satisfies the initial
condition y(0) = −1.
Solution
Step 1: First, we rearrange the given differential equation into the standard
form for a first-order linear differential equation:
y′+ 2xy =x2.
Step 2: To solve this equation, we will use an integrating factor µ(x) =
eR2x dx =ex2.
Step 3: Multiplying both sides of the differential equation by the integrating
factor µ(x), we get:
ex2y′+ 2xex2y=x2ex2.
Step 4: We can rewrite the left side of the equation as the derivative of the
product ex2yusing the product rule:
(ex2y)′=x2ex2.
Step 5: Integrating both sides, we have:
ex2y=Zx2ex2dx +C,
where Cis the constant of integration.
Step 6: Solving the integral on the right-hand side by substitution (u=
x2, du = 2xdx), we get:
Zx2ex2dx =1
2ex2+D,
32
Question 2
Question
Find the particular solution of the differential equation y′=2x
ysatisfying the
initial condition y(1) = 3.
Solution
Given the differential equation y′=2x
y, we want to find the particular solution
satisfying the initial condition y(1) = 3.
Step 1: Separate the variables by moving all yterms to one side of the
equation and all xterms to the other side.
dy
dx =2x
y
Step 2: Multiply both sides by yand dx to separate the variables.
y dy = 2x dx
Step 3: Integrate both sides.
Zy dy =Z2x dx
Step 4: Integrate the left side with respect to y.
1
2y2=x2+C
Step 5: Applying the initial condition y(1) = 3 to solve for the constant C.
1
2(3)2= (1)2+C
C=9
2−1
C=7
2
Step 6: Substitute the value of Cback into the general solution.
1
2y2=x2+7
2
Step 7: Solve for y.
y2= 2x2+ 7
y=±p2x2+ 7
Therefore, the particular solution of the differential equation y′=2x
ysatis-
fying the initial condition y(1) = 3 is y=√2x2+ 7.
2
Question 3
Question
Find the general solution to the following differential equation:
y′′ +y= 0
Solution
Step 1: Write the characteristic equation for the given differential equation.
The characteristic equation is obtained by replacing y′′ with r2, where ris the
unknown variable. So, the characteristic equation is:
r2+ 1 = 0
Step 2: Solve the characteristic equation. Setting r2+ 1 = 0 gives us:
r2=−1
r=±i
Therefore, the roots of the characteristic equation are complex conjugates.
Step 3: Write the general solution using the roots of the characteristic equa-
tion. The general solution to a second-order homogeneous differential equation
with complex roots is:
y(t) = eat(c1cos(bt) + c2sin(bt))
where a= 0, b= 1, and c1, c2are arbitrary constants.
Step 4: Substitute the values of aand binto the general solution. Thus,
the general solution to the differential equation y′′ +y= 0 is:
y(t) = c1cos(t) + c2sin(t)
where c1, c2are arbitrary constants.
Question 4
Question
Solve the following differential equation using the method of undetermined co-
efficients: y′′ −4y′+ 4y=e2x.
3
Solution
To solve the given differential equation using the method of undetermined coef-
ficients, we first find the complementary function by solving the characteristic
equation r2−4r+ 4 = 0.This gives us a repeated root at r= 2,so the comple-
mentary function is yc(x) = c1e2x+c2xe2x.
Next, we need to find a particular solution yp(x) for y′′−4y′+4y=e2x.Since
e2xis of the same form as the characteristic equation’s root, we will assume a
particular solution of the form yp(x) = Ax2e2x.
Step 1: Find y′
p(x) and y′′
p(x).
yp(x) = Ax2e2x
y′
p(x) = (2Ax2+ 4Ax)e2x
y′′
p(x) = (4Ax2+ 8Ax + 2A)e2x
Step 2: Substitute yp(x), y′
p(x),and y′′
p(x) into the differential equation and
simplify.
(4Ax2+ 8Ax + 2A)e2x−4(2Ax2+ 4Ax)e2x+ 4Ax2e2x=e2x
(4Ax2+ 8Ax + 2A−8Ax2−16Ax + 4A+ 4Ax2)e2x=e2x
(2A)e2x=e2x
Step 3: Equate the coefficients to find A=1
2.
Step 4: Write down the particular solution.
yp(x) = 1
2x2e2x
Step 5: The general solution is the sum of the complementary function and
the particular solution:
y(x) = yc(x) + yp(x)
y(x) = c1e2x+c2xe2x+1
2x2e2x
Therefore, the solution to the differential equation y′′ −4y′+ 4y=e2xis
y(x) = c1e2x+c2xe2x+1
2x2e2x.
Question 5
Question
Find the general solution to the differential equation: dy
dx =y2−x2
2xy .
4
Solution
To solve this differential equation, we first notice that it is separable. Let’s
rearrange the terms to make it more obvious:
dy
dx =y2−x2
2xy
2y dy = (y2−x2)dx
x
Step 1: Separating variables
2y dy = (y2−x2)dx
x
Step 2: Integrating both sides Integrating both sides gives:
Z2y dy =Z(y2−x2)dx
x
y2=x2
2+C
Step 3: Simplifying the solution Taking the square root of both sides
gives the general solution:
y=±rx2
2+C
So, the general solution to the differential equation is y=±qx2
2+C, where
Cis the constant of integration.
Question 6
Question
Find the general solution to the differential equation:
dy
dx =2x+ 1
y2
Solution
To solve this differential equation, we will first rewrite it in a separable form.
Step 1: Rearrange the terms to separate variables.
y2dy = (2x+ 1) dx
Step 2: Integrate both sides.
Zy2dy =Z(2x+ 1) dx
5
Step 3: Integrate the left side:
1
3y3=x2+x+C
where Cis the constant of integration.
Step 4: Solve for yby taking the cube root of both sides:
y=3
p3x2+ 3x+ 3C
Therefore, the general solution to the differential equation is:
y=3
p3x2+ 3x+ 3C
Question 7
Question
Consider the differential equation dy
dx =x+y
x−y, where y(0) = 1. Find the particu-
lar solution to the initial value problem.
Solution
Step 1: Rewrite the differential equation in a more standard form by multiplying
both sides by (x−y) to separate variables:
dy
dx(x−y) = x+y
xdy
dx −ydy
dx =x+y
Step 2: Rearrange the terms and integrate both sides with respect to x:
xdy
dx −ydy
dx =x+y
Zx dy −Zy dy =Zx+Zy dx
xy −y2
2=x2
2+xy +C
Step 3: Simplify the equation by multiplying through by 2 to clear the
fractions:
2xy −y2=x2+ 2xy + 2C
Step 4: Rearrange the terms to get a quadratic equation in terms of y:
y2−4xy +x2+ 2C= 0
6
Step 5: Apply the initial condition y(0) = 1 to find the particular solution:
1−0+0+2C= 0
2C=−1
C=−1
2
Step 6: Substitute C=−1
2back into the equation to get the particular
solution:
y2−4xy +x2−1=0
Question 8
Question
Solve the following ordinary differential equation:
y′′ + 4y′+ 4y= 0
Solution
Step 1: Write the characteristic equation by assuming y=ert.
−r2+ 4r+ 4 = 0
Step 2: Solve the characteristic equation to find the roots.
r=−4±p42−4(−1)(4)
2(−1)
r=−4±√16
−2
r=−2
Step 3: Write the general solution based on the roots.
y(t) = c1e−2t+c2te−2t
Where c1and c2are constants to be determined.
Step 4: Take the first derivative of y(t).
y′(t) = −2c1e−2t+c2e−2t−2c2te−2t
Step 5: Take the second derivative of y(t).
y′′(t)=4c1e−2t−4c2e−2t+ 4c2te−2t
7
Step 6: Substitute y(t), y′(t), and y′′(t) into the original differential equation
and simplify.
−(4c1e−2t−4c2e−2t+4c2te−2t)+4(−2c1e−2t+c2e−2t−2c2te−2t)+4(c1e−2t+c2te−2t)=0
Step 7: Simplify the equation and find the values of c1and c2that satisfy
the equation. After simplifying, we find c1= 0 and c2is a free variable. So, the
general solution to the differential equation is:
y(t) = c2te−2t
Question 9
Question
Find the general solution to the differential equation
d2y
dx2−5dy
dx + 6y= 0
Solution
Step 1: We first write the characteristic equation for the given differential equa-
tion:
r2−5r+ 6 = 0
Step 2: Solve the characteristic equation to find the roots r1and r2:
The characteristic equation factors to (r−2)(r−3) = 0, so the roots are
r1= 2 and r2= 3.
Step 3: The general solution to the differential equation is then given by
y(x) = c1er1x+c2er2x
where c1and c2are constants to be determined.
Step 4: Substitute the values of r1and r2into the general solution expression:
y(x) = c1e2x+c2e3x
Step 5: Consider the initial conditions which are not given in this problem.
To find the specific solution, additional information such as the initial value of
yand/or its derivative may be needed.
Question 10
Question
Consider the differential equation y′′ + 3y′+ 2y= 0. Find the general solution
of the differential equation.
8
Solution
Step 1: First, we write the characteristic equation by substituting y=ert into
the differential equation y′′ + 3y′+ 2y= 0.
r2+ 3r+ 2 = 0
Step 2: Factorizing the quadratic equation gives us:
(r+ 1)(r+ 2) = 0
Step 3: This leads to the roots r=−1 and r=−2.
Step 4: The general solution of the differential equation is given by:
y(t) = c1e−t+c2e−2t
where c1and c2are arbitrary constants.
Question 11
Question
Find the general solution to the following first-order ordinary differential equa-
tion:
y′+ 2xy =x
Solution
Step 1: This is a first-order linear differential equation of the form
y′+P(x)y=Q(x)
where P(x)=2xand Q(x) = x. To solve this, we will use an integrating factor
µ(x), defined by
µ(x) = eRP(x)dx
Step 2: Compute the integrating factor
µ(x) = eR2xdx =ex2
Step 3: Multiply both sides of the differential equation by the integrating
factor
ex2y′+ 2xex2y=xex2
Step 4: Rewrite the left-hand side as the derivative of the product of yand
the integrating factor
(ex2y)′=xex2
9
Step 5: Integrate both sides with respect to x
Z(ex2y)′dx =Zxex2dx
ex2y=Zxex2dx +C
Step 6: Integrate Rxex2dx using substitution (u=x2)
Zxex2dx =1
2ex2+C′
Step 7: Substitute back into the equation and solve for y
ex2y=1
2ex2+C
y=1
2+Ce−x2
Step 8: Therefore, the general solution to the differential equation is
y=1
2+Ce−x2
Question 12
Question
Solve the following initial value problem:
y′+y=e−x, y(0) = 1
Solution
Step 1: First, we identify this as a first-order linear nonhomogeneous differential
equation of the form y′+p(x)y=q(x).
Step 2: The integrating factor is given by µ(x) = eRp(x)dx. Here, p(x) = 1,
so µ(x) = eR1dx =ex.
Step 3: Multiply both sides of the equation by the integrating factor µ(x):
exy′+exy=exe−x
This simplifies to:
(exy)′= 1
Step 4: Integrate both sides with respect to x:
Z(exy)′dx =Z1dx
10
This gives us:
exy=x+C
Step 5: Apply the initial condition y(0) = 1 to find the value of the constant
C:
e0·1 = 0 + C=⇒C= 1
So, the particular solution to the differential equation is:
exy=x+ 1
Step 6: Solve for y:
y=e−x(x+ 1)
Therefore, the solution to the initial value problem is y=e−x(x+ 1).
Question 13
Question
Consider the differential equation:
y′′ −4y′+ 4y=e2x
Find the general solution of the differential equation.
Solution
Step 1: First, find the complementary function by solving the characteristic
equation:
r2−4r+ 4 = 0
(r−2)2= 0
r= 2(a repeated root)
Therefore, the complementary function is:
yc=c1e2x+c2xe2x
Step 2: To find a particular integral, assume a particular solution of the
form:
yp=Axe2x
Step 3: Calculate the first and second derivatives of yp:
y′
p=Ae2x+ 2Axe2x
y′′
p= 2Ae2x+ 4Axe2x
11
Step 4: Substitute yp, y′
p, y′′
pinto the differential equation:
2Ae2x+ 4Axe2x−4(Ae2x+ 2Axe2x) + 4(Axe2x) = e2x
2Ae2x+ 4Axe2x−4Ae2x−8Axe2x+ 4Axe2x=e2x
2Ae2x−4Ae2x=e2x
−2Ae2x=e2x
A=−1
2
Step 5: The particular solution is:
yp=−1
2xe2x
Step 6: The general solution is the sum of the complementary function and
the particular integral:
y=yc+yp=c1e2x+c2xe2x−1
2xe2x
Question 14
Question
Consider the initial value problem
dy
dx =1
y, y(0) = 1
Find the solution to the differential equation.
Solution
Step 1: Rewrite the differential equation as separate variables.
y dy =dx
Step 2: Integrate both sides of the equation.
Zy dy =Zdx
Step 3: Solve the integrals.
1
2y2=x+C
where Cis the constant of integration.
12
Step 4: Use the initial condition y(0) = 1 to find the value of C.
1
2(1)2= 0 + C
1
2=C
Step 5: Substitute the value of Cback into the equation.
1
2y2=x+1
2
Step 6: Simplify the equation to find the solution in implicit form.
y2= 2x+ 1
Step 7: To find the explicit form of the solution, take the square root of both
sides.
y=±√2x+ 1
Thus, the solution to the differential equation is y=±√2x+ 1.
Question 15
Question
Solve the following first-order ordinary differential equation:
dy
dx =2x+ 3y+ 1
3x+ 2y−2
Solution
Step 1: Rearrange the equation in the form M dx +Ndy = 0 where Mand N
depend only on xand y:
(3x+ 2y−2)dy −(2x+ 3y+ 1)dx = 0
Step 2: Check if the equation is exact by calculating ∂M
∂y and ∂N
∂x :
∂M
∂y = 2 =∂N
∂x =−1
Step 3: Multiply the equation by an integrating factor µ(x, y) to make it
exact. Since µdepends only on x, we have:
1
µ[(3x+ 2y−2)dy −(2x+ 3y+ 1)dx] = 0
13
Step 4: Solve for µ(x) by setting the coefficients of dx and dy equal and
solve for µ:dy
dx =2x+ 3y+ 1
3x+ 2y−2
Therefore, µ(x) becomes:
µ(x) = 1
3x+ 2y−2
Step 5: Multiply the original ODE by the integrating factor µ(x):
1
3x+ 2y−2[(3x+ 2y−2)dy −(2x+ 3y+ 1)dx] = 0
Simplifying, we get:
dy −2x+ 3y+ 1
3x+ 2y−2dx = 0
Step 6: Integrate the equation step by step to find the solution. This will
involve finding a function F(x, y) such that ∂F
∂x =Nand ∂F
∂y =−M.
F(x, y) = Z−2x+ 3y+ 1
3x+ 2y−2dx +g(y)
Step 7: Once F(x, y) is found, set F(x, y) = Cto find the general solution
to the ODE.
Question 16
Question
Solve the following first-order ordinary differential equation:
dy
dx =2x+y+ 4
x−y−2
Solution
To solve the given first-order ordinary differential equation, we will first rear-
range it as follows:
(x−y−2) dy
dx = 2x+y+ 4
Step 1: Let’s expand the left side of the equation using the product rule:
xdy
dx −ydy
dx −2dy
dx = 2x+y+ 4
14
Step 2: Next, let’s rearrange the terms as follows:
xdy
dx −ydy
dx −2dy
dx −y= 2x+ 4
Step 3: Factor out the dy
dx term:
dy
dx(x−y−2) −y= 2x+ 4
Step 4: Now, we can integrate both sides with respect to x:
Zdy
dx(x−y−2) −ydx =Z(2x+ 4)dx
Step 5: Integrating the left side will give:
y(x)−Zydx −Z2dx =Z(2x+ 4)dx
Step 6: Simplify the left side and integrate the right side:
y(x)−Zydx −2x=x2+ 4x+C
Step 7: Finally, the general solution to the differential equation is:
y(x) = x2+ 4x+C+y(x)ex
where Cis an arbitrary constant.
Question 17
Question
Consider the following first-order ordinary differential equation:
dy
dx =y
1 + x2
Find the general solution of the differential equation.
Solution
Step 1: Rewrite the equation as a separable differential equation by multiplying
both sides by (1 + x2):
(1 + x2)dy
dx =y
Step 2: Separate the variables by dividing both sides by yand multiplying
by dx:dy
y=dx
1 + x2
15
Step 3: Integrate both sides with respect to their respective variables:
Z1
ydy =Z1
1 + x2dx
Step 4: Integrate to find the general solution:
ln |y|= arctan(x) + C
where Cis the constant of integration.
Step 5: Exponentiate both sides to solve for y:
y=earctan(x)+C=earctan(x)·eC=C′earctan(x)
where C′=eCis another constant.
Therefore, the general solution to the differential equation is:
y=C′earctan(x)
Question 18
Question
Consider the following first-order ordinary differential equation:
dy
dx =x+y
x−y
Find the general solution to this differential equation.
Solution
Step 1: Rewrite the given differential equation in a more convenient form by
separating the variables:
dy
dx =x+y
x−y=⇒(x−y)dy = (x+y)dx
Step 2: Integrate both sides of the separated variables equation:
Z(x−y)dy =Z(x+y)dx
Step 3: Perform the integrations:
Z(x−y)dy =Z(x+y)dx =⇒xy −y2
2=x2+y2
2+C
Step 4: Rearrange the terms to obtain the general solution in implicit form:
xy −x2= 2C=⇒y=2C+x2
x
So, the general solution to the given initial value problem is y=2C+x2
x,
where Cis an arbitrary constant.
16
Question 19
Question
Consider the differential equation:
y′+2y
x=x2cos(x)
Find the general solution to the above differential equation.
Solution
Step 1: First, we identify the type of differential equation. This is a first-order
linear ordinary differential equation of the form y′+P(x)y=Q(x).
Step 2: To solve this type of differential equation, we use an integrating
factor which is defined as µ(x) = exp RP(x)dx .
Step 3: Let’s calculate the integrating factor:
µ(x) = exp Z2
xdx
µ(x) = exp(2 ln |x|)
µ(x) = x2
Step 4: Next, we multiply both sides of the differential equation by the
integrating factor µ(x):
x2(y′+2y
x) = x4cos(x)
Step 5: After simplifying, we get:
x2y′+ 2xy =x4cos(x)
Step 6: Now, we can rewrite the equation as the derivative of a product:
(x2y)′=x4cos(x)
Step 7: Integrating both sides with respect to x, we get:
x2y=Zx4cos(x)dx +C
Step 8: Solve the integral on the right hand side to find:
x2y=x4sin(x)−4x3cos(x)−12x2sin(x) + 24xcos(x) + C
Step 9: Finally, the general solution to the differential equation is:
y=x2sin(x)−4x3cos(x)−12x2sin(x) + 24xcos(x) + C
x2
17
Question 20
Question
Solve the following first-order ordinary differential equation:
y′+ 2xy =x2, y(0) = 1
Solution
Step 1: We will first rewrite the given differential equation in standard form,
which is the form y′+P(x)y=Q(x), where P(x) and Q(x) are functions of x.
Given equation: y′+ 2xy =x2We can rewrite this as:
y′=x2−2xy
Step 2: The integrating factor is given by the expression µ(x) = eRP(x)dx.
In this case, P(x)=2x, so
µ(x) = eR2xdx
µ(x) = ex2
Step 3: We multiply both sides of the differential equation by the integrating
factor µ(x):
ex2y′=ex2(x2−2xy)
Step 4: When we apply the product rule to the left side, we get:
d
dx(ex2y) = x2ex2−2xyex2
Step 5: Integrating both sides of the equation with respect to xgives:
ex2y=Z(x2ex2−2xyex2)dx
Step 6: Simplify the integral:
ex2y=1
3x3ex2−yex2+C
where Cis the constant of integration.
Step 7: Let’s now solve for y:
y=1
3x3−y+Ce−x2
Step 8: Applying the initial condition, y(0) = 1, we find:
1=0−1 + C
C= 2
Step 9: Substituting back this value of Cinto the general solution, we get
the final solution:
y=1
3x3−y+ 2e−x2
18
Question 21
Question
Consider the following first-order ordinary differential equation:
dy
dx + 2xy =x
Find the general solution to this differential equation.
Solution
Step 1: We first rewrite the differential equation in standard form:
dy
dx =x−2xy
Step 2: Next, we identify that this is a linear first-order differential equation
of the form: dy
dx +P(x)y=Q(x)
where P(x) = −2xand Q(x) = x.
Step 3: To solve this type of differential equation, we use an integrating
factor µ(x) = eRP(x)dx. Calculating the integrating factor, we have:
µ(x) = eR−2xdx =e−x2
Step 4: We then multiply the integrating factor with the differential equa-
tion:
e−x2dy
dx +e−x2(−2xy) = e−x2x
Step 5: After rewritting the equation, we recognize the left side as the deriva-
tive of the product of yand µ(x):
d
dx(e−x2y) = e−x2x
Step 6: Integrating both sides with respect to x, we get:
e−x2y=Zxe−x2dx
Step 7: We now solve the integral on the right side:
Zxe−x2dx =−1
2e−x2+C
Step 8: Substituting the integral back into our equation from step 6, we
have:
e−x2y=−1
2e−x2+C
19
Step 9: Finally, we solve for yto obtain the general solution:
y=−1
2+Cex2
where Cis the constant of integration.
Question 22
Question
Consider the following first-order ordinary differential equation:
dy
dx =2x+y
x−2y
Find the general solution for the differential equation.
Solution
Step 1: Rewrite the given differential equation in a more suitable form:
(x−2y)dy −(2x+y)dx = 0
Step 2: Check whether the given differential equation is exact by verifying
the equality of mixed partial derivatives:
∂
∂y (x−2y) = −2,∂
∂x (−2x−y) = −2
Since the mixed partial derivatives are equal, the differential equation is exact.
Step 3: Integrate the coefficient of dy with respect to y:
Z(x−2y)dy =xy −y2+g(x)
Step 4: Set the result equal to the function h(x), where h(x) is the an-
tiderivative with respect to xof the coefficient of dx:
xy −y2+g(x) = h(x)
Step 5: Solve for g(x) by comparing terms containing y:
g(x) = h(x)
Step 6: Substitute the solution for g(x) back into the equation:
xy −y2=h(x)
Step 7: Rearrange terms to obtain the general solution:
xy −y2=C
where Cis the constant of integration.
20
Question 23
Question
Consider the initial value problem:
dy
dx = 2xy, y(0) = 1
Find the solution to the initial value problem.
Solution
Step 1: We can rewrite the differential equation as:
dy
dx = 2xy =⇒dy
y= 2x dx
Step 2: Integrating both sides gives us:
Z1
ydy =Z2x dx =⇒ln |y|=x2+C
where Cis the constant of integration.
Step 3: Exponentiating both sides gives us:
y=ex2+C=Cex2
since ex2=eC·ex2.
Step 4: Applying the initial condition y(0) = 1, we find C= 1. Thus, the
solution to the initial value problem is:
y=ex2
Question 24
Question
Find the general solution to the following first-order ordinary differential equa-
tion:
dy
dx + 2xy =x2
Solution
Step 1: This is a first-order linear differential equation in the form dy
dx +P(x)y=
Q(x), where P(x) = 2xand Q(x) = x2. To solve this, we use an integrating
factor, µ(x) = eRP(x)dx.
21
Step 2: Calculate the integrating factor:
µ(x) = eR2xdx =ex2
Step 3: Multiply the differential equation by the integrating factor:
ex2dy
dx + 2xex2y=x2ex2
Step 4: Recognize that the left side is the derivative of the product of µ(x)
and ywith respect to x, so rewrite the equation as:
d
dx(ex2y) = x2ex2
Step 5: Integrate both sides with respect to x:
Zd
dx(ex2y)dx =Zx2ex2dx
Step 6: Solve the integrals:
ex2y=1
2x2ex2+C
Step 7: Finally, solve for yto find the general solution:
y=1
2x2+Ce−x2
Question 25
Question
Find the general solution to the following first-order ordinary differential equa-
tion: dy
dx =y
x+x
Solution
Step 1: Let’s rewrite the given differential equation in a more standard form:
dy
dx −y
x=x
Step 2: This is a linear first-order differential equation in the form dy
dx +
P(x)y=Q(x). Here, P(x) = −1
xand Q(x) = x.
Step 3: To solve this, we first find the integrating factor I(x), which is given
by:
I(x) = eRP(x)dx =eR−1
xdx
22
Step 4: Simplifying the integral to find I(x):
I(x) = e−ln |x|=1
|x|=1
x
Step 5: Multiply both sides of the differential equation by the integrating
factor I(x):
1
x
dy
dx −1
x2y= 1
Step 6: Rewrite the left side as a derivative of a product using the product
rule: d
dx 1
xy= 1
Step 7: Integrating both sides with respect to x:
Zd
dx 1
xydx =Z1dx
1
xy=x+C
Step 8: Multiply through by xto isolate y:
y=x2+Cx
Therefore, the general solution to the differential equation is y=x2+Cx,
where Cis an arbitrary constant.
Question 26
Question
Solve the initial value problem:
y′+2
xy=xln x, y(1) = 1
Solution
Step 1: We first identify the integrating factor for this first-order linear ODE.
The integrating factor µ(x) is given by µ(x) = eR2
xdx.
µ(x) = eR2
xdx
=e2 ln x
=x2
23
Step 2: Next, we multiply both sides of the ODE by the integrating factor
µ(x) = x2.
x2y′+ 2xy =x3ln x
Step 3: We can now rewrite the left-hand side of the ODE as the derivative
of the product of x2and ywith respect to x:
d
dx(x2y) = x3ln x
Step 4: Integrating both sides with respect to x, we have
Zd
dx(x2y)dx =Zx3ln x dx
x2y=1
4x4ln x−1
16x4+C
Step 5: To find the constant of integration C, we apply the initial condition
y(1) = 1.
1 = 1
4−1
16 +C
C= 1 + 1
4−1
16
C=13
16
Step 6: Therefore, the solution to the initial value problem is
y(x) = 1
x21
4x4ln x−1
16x4+13
16
Question 27
Question
Consider the first-order ordinary differential equation:
dy
dx = 2xy +x
Find the general solution to this differential equation.
Solution
Step 1: This is a first-order linear differential equation in the form of:
dy
dx +P(x)y=Q(x)
24
where P(x)=2xand Q(x) = x.
Step 2: To solve this equation, we will use an integrating factor eRP(x)dx.
The integrating factor is eR2xdx =ex2.
Step 3: Multiply both sides of the differential equation by the integrating
factor:
ex2dy
dx + 2xex2y=ex2x
Step 4: Recognize that the left side is the derivative of the product of yand
the integrating factor: d
dx(ex2y) = ex2x
Step 5: Integrate both sides with respect to x:
Zd
dx(ex2y)dx =Zex2xdx
Step 6: Integrate the right side:
ex2y=Zxex2dx
Step 7: Integrate the right side using substitution u=x2, du = 2xdx:
ex2y=1
2Zeudu =1
2ex2+C
where Cis the constant of integration.
Step 8: Solve for y:
y=1
2+Ce−x2
So, the general solution to the differential equation is y=1
2+Ce−x2, where
Cis an arbitrary constant.
Question 28
Question
Solve the following first-order ordinary differential equation:
dy
dx = (3x2+ 4y)/(2x2−y)
Solution
Step 1: Rewrite the differential equation in a more standard form by multiplying
both sides by 2x2−y:
(2x2−y)dy
dx = 3x2+ 4y
25
Step 2: Notice that the left-hand side resembles the product rule, so we
rewrite it as a derivative with respect to x:
d
dx (2x2−y)y= 3x2+ 4y
Step 3: Integrate both sides with respect to x:
Zd
dx (2x2−y)ydx =Z(3x2+ 4y)dx
Step 4: Simplify and integrate:
(2x2−y)y=x3+ 4xy +C
Step 5: Rearrange the equation to express y explicitly:
y2−2x2y=x3+ 4xy +C
Step 6: Complete the square on the left-hand side to make it a perfect square
trinomial:
(y−x)2=x3+ 4xy +C+x2
Step 7: Take the square root of both sides and solve for y:
y−x=±px3+ 4xy +C+x2
y=x±px3+ 4xy +C+x2
Therefore, the solution to the differential equation is y=x±px3+ 4xy +C+x2.
Question 29
Question
Consider the initial value problem:
y′=2x+y+ 1
x−y−1, y(0) = 1
Determine the interval on which the solution exists and show that the solution
can be explicitly written.
Solution
Step 1: Rewrite the given differential equation in standard form.
(x−y−1) dy
dx = 2x+y+ 1
26
Step 2: Rewrite the equation in a more familiar form by expanding the left
side of the equation.
xdy
dx −ydy
dx −dy
dx = 2x+y+ 1
Step 3: Rearrange the terms to isolate dy
dx .
dy
dx +y=x+ 1
Step 4: Find the integrating factor by taking the exponential of the integral
of the coefficient of y.
Let µ(x) = eR1dx =ex
Step 5: Use the integrating factor to solve for y.
exdy
dx +exy=ex(x+ 1)
Step 6: Recognize the left side as the product rule of the integrating factor
with y.d
dx(exy) = ex(x+ 1)
Step 7: Integrate both sides to find the general solution.
exy=Zex(x+ 1) dx =ex(x−1) + C
Step 8: Solve for y in the general solution.
y=x−1 + Ce−x
Step 9: Use the initial condition y(0) = 1 to find the specific solution.
1=0−1 + Ce0=⇒C= 2
Step 10: Substitute C= 2 back into the general solution to get the particular
solution.
y=x−1+2e−x
Therefore, the solution to the initial value problem is y=x−1 + 2e−xfor
x > 0.
Question 30
Question
Consider the differential equation y′′ −4y′+ 4y=e2x. Find the general solution
to this differential equation.
27
Solution
Step 1: First, find the characteristic equation by solving the auxiliary equation
r2−4r+4 = 0. The characteristic equation is (r−2)2= 0, which has a repeated
root r= 2.
Step 2: The general solution to the homogeneous equation y′′ −4y′+ 4y= 0
is then yh(x) = c1e2x+c2xe2x, where c1and c2are arbitrary constants.
Step 3: To find a particular solution to the non-homogeneous equation y′′ −
4y′+ 4y=e2x, try a particular solution of the form yp(x) = ae2x.
Step 4: Substitute yp(x) into the differential equation to find a.
(2a)e2x−4(2a)e2x+ 4ae2x=e2x
−6ae2x+ 4ae2x=e2x
−2ae2x=e2x
a=−1
2
Step 5: Therefore, the particular solution to the non-homogeneous equation
is yp(x) = −1
2e2x.
Step 6: The general solution to the differential equation is then the sum of
the homogeneous and particular solutions:
y(x) = yh(x) + yp(x) = c1e2x+c2xe2x−1
2e2x
where c1and c2are arbitrary constants.
Question 31
Question
Consider the differential equation dy
dx =√ywhere y(0) = 1. Find the particular
solution to this initial value problem.
Solution
Step 1: Rewrite the differential equation as dy
√y=dx.
Step 2: Integrate both sides with respect to x.
Z1
√ydy =Zdx
Step 3: Simplify the integrals.
2√y=x+C
Step 4: Use the initial condition y(0) = 1 to find the value of the constant
C.
2√1 = 0 + C
28
C= 2
Step 5: Substitute the value of Cback into the general solution.
2√y=x+ 2
Step 6: Solve for y.√y=x
2+ 1
y=x
2+ 12
Therefore, the particular solution to the initial value problem is y=x
2+ 12.
Question 32
Question
Solve the following first-order ordinary differential equation:
dy
dx +y=e−x
Solution
Step 1: First, we identify that this is a linear first-order ODE of the form
dy
dx +P(x)y=Q(x), where P(x) = 1 and Q(x) = e−x.
Step 2: To solve this differential equation, we first find the integrating factor
I(x), which is given by I(x) = eRP(x)dx =eR1dx =ex.
Step 3: We multiply both sides of the ODE by the integrating factor:
exdy
dx +exy=exe−x
Step 4: Notice that the left-hand side is now the result of applying the
product rule for differentiation, specifically d
dx (exy). So, we can rewrite the
equation as: d
dx(exy)=1
Step 5: Integrating both sides with respect to x, we get:
exy=x+C
Step 6: Solving for y, we have:
y=e−x(x+C)
Therefore, the general solution to the differential equation dy
dx +y=e−xis
y=e−x(x+C), where Cis an arbitrary constant.
29
Question 33
Question
Solve the following first-order ordinary differential equation:
dy
dx =3x2+ 2y
2x
Solution
Step 1: Rewrite the given ODE in a more suitable form. Let’s rewrite the ODE
as: dy
dx =3x2
2x+2y
2x
dy
dx =3
2x+y
x
Step 2: This is now a separable differential equation. We can rewrite it as:
dy
dx −y
x=3
2x
Step 3: Now, we can use an integrating factor to solve this linear ODE. The
integrating factor is given by eR−1
xdx =e−ln |x|=1
|x|.
Step 4: Multiply the integrating factor by the entire equation to get:
1
|x|
dy
dx −1
|x|
y
x=3
2
Step 5: Rewrite the equation in its integrated form:
d
dx(y
|x|) = 3
2
Step 6: Integrate both sides with respect to x:
Zd
dx(y
|x|)dx =Z3
2dx
Step 7: The left-hand side simplifies to y
|x|=3
2x+C, where Cis the constant
of integration.
Step 8: Finally, we can solve for y:
y=3
2x|x|+C|x|
y=3
2x2+Cx
Thus, the solution to the given first-order ODE is y=3
2x2+Cx, where C
is a constant.
30
Question 34
Question
Solve the following first-order ordinary differential equation:
dy
dx =y2−x2
Solution
Step 1: First, we rewrite the given differential equation in the standard form of
a separable differential equation:
dy
dx =y2−x2
Step 2: Rearrange the equation to separate the variables xand y:
dy
y2−x2=dx
Step 3: Factor the denominator on the left side using the difference of
squares: dy
(y−x)(y+x)=dx
Step 4: Use partial fraction decomposition to split the expression on the left
side into two separate fractions:
dy
(y−x)(y+x)=A
y−x+B
y+x
Step 5: Multiply both sides by the denominator to solve for Aand B:
dy =A(y+x)dx +B(y−x)dx
Step 6: Integrate both sides with respect to yon the left side and xon the
right side to solve for y:
Zdy =ZA(y+x)dx +ZB(y−x)dx
Step 7: Integrate all terms separately:
y=Axy +A
2x2+By −B
2x2+C
Step 8: Rearrange the equation by combining the terms with yand those
without y:
y(1 −A) = A−B
2x2+C
31
Step 9: To solve for y, we set 1 −A= 0 and A−B= 0 (since xcan be any
value):
A= 1, B = 1
Step 10: Substitute A= 1 and B= 1 back into the equation:
y=x2+C
So, the general solution to the differential equation is y=x2+C, where C
is an arbitrary constant.
Question 35
Question
Consider the differential equation:
y′+ 2xy =x2.
Find the particular solution to the differential equation that satisfies the initial
condition y(0) = −1.
Solution
Step 1: First, we rearrange the given differential equation into the standard
form for a first-order linear differential equation:
y′+ 2xy =x2.
Step 2: To solve this equation, we will use an integrating factor µ(x) =
eR2x dx =ex2.
Step 3: Multiplying both sides of the differential equation by the integrating
factor µ(x), we get:
ex2y′+ 2xex2y=x2ex2.
Step 4: We can rewrite the left side of the equation as the derivative of the
product ex2yusing the product rule:
(ex2y)′=x2ex2.
Step 5: Integrating both sides, we have:
ex2y=Zx2ex2dx +C,
where Cis the constant of integration.
Step 6: Solving the integral on the right-hand side by substitution (u=
x2, du = 2xdx), we get:
Zx2ex2dx =1
2ex2+D,
32
where Dis another constant.
Step 7: Substituting this back into the equation from step 5, we have:
ex2y=1
2ex2+D+C.
Step 8: Simplifying, we find:
y=1
2+D
ex2+Ce−x2.
Step 9: Applying the initial condition y(0) = −1, we get:
−1 = 1
2+D+C.
Step 10: Solving for C, we have C=−3
2−D.
Step 11: Substituting back into the general solution, we get the particular
solution:
y=1
2+D
ex2−3
2e−x2−De−x2.
33
