MATH 334 - DIFFERENTIAL EQUATIONS First-Order Differential Equations Question Bank Set 3

MATH 334 - DIFFERENTIAL

EQUATIONS - First-Order Differential

Equations

Question Bank - Set 3

Liberty University

Question 1

Question

Solve the first-order differential equation: dy

dx =e2xy−y.

Solution

Step 1: We can rewrite the given differential equation in the standard form:

dx = (e2x−1)y

Step 2: This is a linear first-order differential equation. To solve it, we can

use an integrating factor. Let’s find the integrating factor I(x):

I(x) = eR(e2x−1)dx

I(x) = eRe2x−dx

I(x) = ee2x−x=e(e2xx)

Step 3: Multiply both sides of the original differential equation by the inte-

grating factor e(e2xx):

e(e2xx)dy

dx =e2xe(e2xx)y−e(e2xx)y

dx(e(e2xx)y) = e3xy

Step 4: Integrate both sides with respect to x:

dx(e(e2xx)y)dx =Ze3xydx

e(e2xx)y=Ze3xydx

Step 5: Now, let’s solve the integral on the right-hand side by rearranging

the equation:

e(e2xx)=1

dx =d

dx(e2x)

Integrating both sides:

e2x=Ze3xdx

e2x=1

3e3x+C

Step 6: Finally, solve for yby rearranging the terms:

e2xy=1

3e3xy+Cy

2y=1

3ex+Cy

y=1

6ex+Cy

Question 2

Question

Find the general solution to the first-order differential equation: dy

dx =2x+3y+5

3x+2y+7 .

Solution

Step 1: Rearrange the equation in the form dy

dx =f(x) + g(y).

dx =2x+ 3y+ 5

3x+ 2y+ 7

dx =2x+ 3y+ 5

3x+ 2y+ 7

dx =2(x+5

2) + 3(y+5

3(x+7

3) + 2(y+7

Step 2: Recognize that this equation is exact, and find the auxiliary equation

by comparing it to dy

dx =P(x) + Q(y).

2P(x)+3Q(y)=0

∂P

∂y =∂Q

∂x

2 = 3

Step 3: Integrate P(x) with respect to xto find f(x).

f(x) = 2(x+5

2) = 2x+ 5

Step 4: Integrate Q(y) with respect to yto find g(y).

g(y) = 3(y+5

3)=3y+ 5

Step 5: Write down the general solution to the differential equation.

y=−5−2x+Ce−3x

Therefore, the general solution to the first-order differential equation is y=

−5−2x+Ce−3x.

Question 3

Question

Consider the differential equation:

dx =2x+ 3y+ 7

x−4y−5

Find the solution to the initial value problem y(0) = −3.

Solution

Step 1: Rewrite the differential equation in standard form by separating vari-

ables. dy

dx =2x+ 3y+ 7

x−4y−5=⇒(x−4y−5)dy = (2x+ 3y+ 7)dx

Step 2: Integrate both sides to solve for y.

Z(x−4y−5)dy =Z(2x+ 3y+ 7)dx

Step 3: Simplify the integrals.

Zxdy −4Zydy −5Zdy = 2 Zxdx + 3 Zydx + 7 Zdx

Step 4: Integrate each term separately.

xy −2y2−5y=x2+ 3xy + 7x+C

where Cis the constant of integration.

Step 5: Use the initial condition y(0) = −3 to find the value of the constant

0(−3) −2(−3)2−5(−3) = 02+ 3(0)(−3) + 7(0) + C

9−18 + 15 = C

C= 6

Step 6: Substitute C= 6 back into the equation to get the final solution.

xy −2y2−5y=x2+ 3xy + 7x+ 6

x(−3) −2(−3)2−5(−3) = x2+ 3x(−3) + 7x+ 6

−3x+ 18 −15 = x2−9x+ 7x+ 6

x2−11x+ 9 = 0

(x−1)(x−9) = 0

x= 1,9

Therefore, the solution to the initial value problem is y=2x

x−4+7

x−4.

Question 4

Question

Solve the following first-order differential equation:

dx −2y

x=x3

Solution

Step 1: This is a first-order linear differential equation in the form

dx +P(x)y=Q(x)

where P(x) = −2

xand Q(x) = x3. We begin by finding the integrating factor

µ(x) = eRP(x)dx.

Step 2: Calculate the integrating factor µ(x):

µ(x) = eR−2

xdx =e−2 ln |x|=x−2

Step 3: Multiply both sides of the differential equation by the integrating

factor:

x−2dy

dx −2x−3y=x

Step 4: Rewrite the left-hand side of the equation as the derivative of the

product of x−2and y:d

dx(x−2y) = x

Step 5: Integrate both sides with respect to x:

dx(x−2y)dx =Zxdx

Step 6: Solve the integrals:

x−2y=1

2x2+C

Step 7: Solve for y:

y=1

2x4+Cx2

Thus, the general solution to the differential equation is y=1

2x4+Cx2,

where Cis an arbitrary constant.

Question 5

Question

Solve the following first-order differential equation using an integrating factor:

dx + 2y=e4x

Solution

Step 1: Write the given equation in the standard form dy

dx +P(x)y=Q(x). In

this case, P(x) = 2 and Q(x) = e4x, so the equation becomes:

dx + 2y=e4x

Step 2: Identify the integrating factor by multiplying both sides of the equa-

tion by the integrating factor µ(x) = eRP(x)dx. The integrating factor in this

case is µ(x) = eR2dx =e2x.

Step 3: Multiply both sides of the differential equation by the integrating

factor.

e2xdy

dx + 2e2xy=e2xe4x

Step 4: Rewrite the left side of the equation as the derivative of the product

of the integrating factor and y.

dx(e2xy) = e6x

Step 5: Integrate both sides with respect to x.

dx(e2xy)dx =Ze6xdx

Step 6: Solve the integrals.

e2xy=e6x

6+C

Step 7: Solve for y.

y=1

6e4x+Ce−2x

where Cis an arbitrary constant.

Question 6

Question

Consider the first-order differential equation:

dx =y

x+ex

Find the general solution to this differential equation.

Solution

Step 1: Rewrite the equation in a standard form:

dx −y

x=ex

Step 2: This is a linear first-order differential equation. We can solve it using

an integrating factor. The integrating factor, denoted by I(x), is calculated as:

I(x) = eR−1

xdx =e−ln |x|=1

|x|

Step 3: Multiply both sides of the equation by the integrating factor:

|x|

dx −1

|x|

x=1

|x|ex

Step 4: Write the left side as the derivative of the product of the integrating

factor and y:

dx y

|x|=1

|x|ex

Step 5: Integrate both sides with respect to x:

dx y

|x|dx =Z1

|x|exdx

Step 6: Simplify the integrals:

|x|=Z1

|x|exdx

Step 7: The integral on the right side can be solved by using integration by

parts:

u=ex, dv =1

|x|dx

du =exdx, v =Z1

|x|dx = ln |x|

|x|exdx =exln |x| − Zex·ln |x|dx

Step 8: Express the solution in terms of y:

y=|x|exln |x|−|x|ex+C

Therefore, the general solution to the first-order differential equation is:

y=|x|exln |x|−|x|ex+C

where Cis an arbitrary constant.

Question 7

Question

Find the general solution to the first-order differential equation: dy

dx =1

x2+y2.

Solution

Step 1: We can rewrite the given differential equation as dy

dx =1

x2(1+(y

x)2).

Step 2: Let u=y

x, then y=ux and dy

dx =u+xdu

dx .

Step 3: Substitute y=ux and dy

dx =u+xdu

dx into the differential equation

to get du

dx x2(1 + u2) = 1.

Step 4: Rearrange the equation to separate variables: du

1+u2=1

x2dx.

Step 5: Integrate both sides with respect to their respective variables: Rdu

1+u2=

x2dx.

arctan(u) = −1

x+C

Step 6: Substitute back u=y

xinto the equation to get arctan y

x=−1

x+C.

Step 7: The general solution is y=xtan −1

x+C).

Question 8

Question

Find the general solution to the first-order differential equation: dy

dx =2y2−5

3x.

Solution

Step 1: Rearrange the equation to separate variables:

2y2−5=dx

Step 2: Integrate both sides with respect to yand x:

2y2−5dy =Z1

3xdx

Step 3: Factor the denominator of the left integral and use partial fractions

to integrate:

2y2−5dy =Z1

3xdx

(y+p5/2)(y−p5/2)dy =Z1

3xdx

2p5/2Z 1

y+p5/2−1

y−p5/2!dy =1

3Z1

xdx

Step 4: Integrate both sides:

2p5/2(ln |y+p5/2| − ln |y−p5/2|) = 1

3ln |x|+C

Step 5: Simplify the natural logarithms:

2p5/2ln 

y+p5/2

y−p5/2

3ln |x|+C

Step 6: Exponentiate both sides to eliminate the natural logarithms:



y+p5/2

y−p5/2

1/2√5/2

=|x|1/3eC

Step 7: Simplify by taking the absolute value on both sides and reusing the

constant term:

y+p5/2

y−p5/2=±x1/3eC

Step 8: Solve for y:

y+p5/2 = ±x1/3eC(y−p5/2)

y+p5/2 = ±x1/3eCy∓x1/3p5/2eC

y(1 ∓x1/3eC) = ±x1/3p5/2eC−p5/2

Step 9: Finally, solve for yto get the general solution:

y=±x1/3p5/2eC−p5/2

1∓x1/3eC

Question 9

Question

Find the general solution to the first-order differential equation: dy

dx =2x+y−1

x+2y.

Solution

To solve this first-order differential equation, we will rearrange the terms, sep-

arate variables, and integrate.

Step 1: Write the given differential equation in the form dy

dx =f(x, y).

dx =2x+y−1

x+ 2y

Step 2: Multiply through by dx and separate variables.

(x+ 2y)dy = (2x+y−1)dx

Step 3: Integrate both sides.

Z(x+ 2y)dy =Z(2x+y−1)dx

Step 4: Perform the integrations.

Z(x+ 2y)dy =Z(2x+y−1)dx

2y2+xy =x2+1

2y2−x+C

Step 5: Simplify the equation.

xy =x2−x+C

Step 6: Rearrange the equation to find the general solution.

y=x−1 + C

Therefore, the general solution to the given first-order differential equation

is y=x−1 + C

x, where Cis an arbitrary constant.

Question 10

Question

Consider the first-order differential equation: dy

dx =x3−e2x. Solve the differen-

tial equation using the method of separation of variables.

Solution

Step 1: Separate the variables by writing the equation in the form dy

dx =

f(x)g(y).

We have: dy

dx =x3−e2x

Step 2: Rewrite the equation as:

dx =x3−e2x

dy = (x3−e2x)dx

Step 3: Integrate both sides with respect to their respective variables.

Zdy =Z(x3−e2x)dx

Step 4: Integrate both sides:

Zdy =Zx3dx −Ze2xdx

y=x4

4−1

2e2x+C

Therefore, the general solution to the differential equation dy

dx =x3−e2xis

y=x4

4−1

2e2x+C, where C is the constant of integration.

Question 11

Question

Solve the following first-order differential equation:

y′+ 4y= 8e−2x

Solution

Step 1: We first identify the integrating factor which is given by:

µ(x) = eRP(x)dx

where P(x) is the coefficient of yin the differential equation. In this case,

P(x) = 4, so

µ(x) = eR4dx

µ(x) = e4x

Step 2: We multiply both sides of the differential equation by the integrating

factor:

e4x·(y′+ 4y) = e4x·8e−2x

e4xy′+ 4e4xy= 8

Step 3: Notice the left side can be written as the derivative of a product:

(e4xy)′= 8

Step 4: Integrating both sides with respect to x:

Z(e4xy)′dx =Z8dx

e4xy= 8x+C

where Cis the constant of integration.

Step 5: Solving for y:

y=8x+C

e4x

y= 8xe−4x+Ce−4x

y=x(8 −4x) + Ce−4x

where C= 0 because it was absorbed into the constant of integration.

Therefore, the solution to the differential equation is:

y=x(8 −4x)

Question 12

Question

Solve the initial value problem:

y′−2xy =x, y(0) = 1

Solution

Step 1: First, we identify the integrating factor µ(x) = eR−2x dx =e−x2.

Step 2: Multiply both sides of the differential equation by the integrating

factor:

e−x2y′−2xe−x2y=xe−x2

Step 3: Recognize that the left side is now the derivative of the product of

yand the integrating factor:

(e−x2y)′=xe−x2

Step 4: Integrate both sides with respect to x:

Z(e−x2y)dx =Zxe−x2dx

Step 5: On the left side, we simplify the integral to get e−x2y. And on the

right side, we have to do a substitution u=−x2:

−1

2Zeudu =−1

2e−x2+C1

Step 6: Simplifying the integration on both sides gives us:

e−x2y=−1

2e−x2+C1

Step 7: To find y, multiply through by ex2:

y=−1

2+C1ex2

Step 8: Apply the initial condition y(0) = 1 to find the value of the constant

C1:

1 = −1

2+C1

C1=3

Step 9: Substitute C1=3

2back into the solution to get the particular

solution:

y=−1

2+3

2ex2

Question 13

Question

Consider the first-order differential equation given by dy

dx =2x+y

x. Find the

general solution to this differential equation.

Solution

Step 1: Rewrite the differential equation in standard form by multiplying through

by dx:dy

dx =2x+y

x=⇒xdy

dx −y= 2x

Step 2: This is now a linear first-order differential equation in the form

y′+P(x)y=Q(x). To solve this, we need to find an integrating factor I(x)

defined by I(x) = eRP(x)dx. Here, P(x) = −1

x, so I(x) = eR−1

xdx =e−ln |x|=

|x|.

Step 3: Multiply our differential equation by the integrating factor:

|x|xdy

dx −y=1

|x|·2x

This simplifies to: 1

|x|·xdy

dx −1

|x|·y= 2

Step 4: Recognize that the left side of the equation is the derivative of y

|x|

with respect to x:

dx y

|x|= 2

Step 5: Integrate both sides with respect to xto find the general solution:

dx y

|x|dx =Z2dx

|x|= 2x+C

where Cis the constant of integration.

Step 6: Multiply through by |x|to get the general solution in a more standard

form:

y= 2x|x|+C|x|

y= 2x|x|+Cx (for x= 0)

Thus, the general solution to the differential equation is y= 2x|x|+Cx for

x= 0, where Cis an arbitrary constant.

Question 14

Question

Solve the first-order differential equation:

dx =x+y

Solution

Step 1: Rewrite the differential equation in standard form:

dx −y

x=1

Step 2: This is a linear first-order differential equation. To solve it, we first

find the integrating factor, denoted by µ(x), which is given by:

µ(x) = eR−1

xdx

Step 3: Calculate the integrating factor:

µ(x) = e−ln |x|=1

|x|=1

Step 4: Multiply the integrating factor with the differential equation:

dx −y

x2= 1

Step 5: Rewrite the equation in terms of the total derivative:

dx y

x= 1

Step 6: Integrate both sides with respect to x:

dx y

xdx =Z1dx

Step 7: Integrate the right side with respect to x:

x=x+C

Step 8: Multiply through by xto obtain the general solution:

y=x2+Cx

Question 15

Question

Find the general solution to the following first-order differential equation:

dx + 2y=e−2x

Solution

Step 1: First, we identify the standard form of a first-order linear differential

equation:

dx +P(x)y=Q(x)

where P(x) = 2 and Q(x) = e−2x.

Step 2: To solve the differential equation, we need to find an integrating

factor I(x) defined by:

I(x) = eRP(x)dx

In this case, P(x) = 2, so:

ZP(x)dx =Z2dx = 2x

Thus, the integrating factor is I(x) = e2x.

Step 3: Multiply both sides of the differential equation by the integrating

factor:

e2xdy

dx + 2e2xy=e2xe−2x

Simplify to get:

dx(e2xy) = 1

Step 4: Integrate both sides with respect to x:

dx(e2xy)dx =Z1dx

This gives us:

e2xy=x+C

where Cis the constant of integration.

Step 5: Solve for yby dividing through by e2x:

y=x

e2x+C

e2x

which simplifies to:

y=xe−2x+Ce−2x

So, the general solution to the differential equation is y=xe−2x+Ce−2x,

where Cis an arbitrary constant.

Question 16

Question

Find the particular solution to the differential equation dy

dx −2y= 6xgiven that

y= 5 when x= 2.

Solution

Step 1: First, we rewrite the given differential equation in the standard form of

a first-order linear ODE: dy

dx −2y= 6x.

Step 2: We identify the integrating factor, which is eR−2dx =e−2x.

Step 3: Multiply both sides of the differential equation by the integrating

factor e−2x:

e−2xdy

dx −2e−2xy= 6xe−2x

Step 4: The left side can be rewritten as the derivative of the integrating

factor multiplied by y:d

dx(e−2xy) = 6xe−2x

Step 5: Integrate both sides with respect to x:

dx(e−2xy)dx =Z6xe−2xdx

Step 6: Integrate the right side using integration by parts: Let u=xand

dv = 6e−2xdx. Then, du =dx and v=−3e−2x.

Z6xe−2xdx =−3xe−2x−Z(−3e−2xdx)

Step 7: Simplify the integration on the right side:

Z6xe−2xdx =−3xe−2x+Z6e−2xdx

Step 8: Integrate the remaining integral on the right side:

Z6e−2xdx =−3xe−2x−3

2e−2x+C, where Cis the constant of integration

Step 9: Substitute back into the differential equation and solve for y:

e−2xy=−3xe−2x−3

2e−2x+C

Step 10: Finally, rearrange the equation to solve for y:

y=−3x−3

2+Ce2x

Step 11: Use the initial condition y(2) = 5 to find the value of the constant

5 = −6−3

2+Ce4

Step 12: Solve for C:

C=17

2e−4

Therefore, the particular solution to the differential equation is y=−3x−

2+17

2e2x.

Question 17

Question

Solve the following first-order differential equation:

dx +ysin(x) = ecos(x)

Solution

Step 1: This is a first-order linear differential equation in the standard form

y′+P(x)y=Q(x), where P(x) = sin(x) and Q(x) = ecos(x). To solve this type

of equation, we can use an integrating factor u(x) = eRP(x)dx.

Step 2: First, we find the integrating factor u(x):

u(x) = eRsin(x)dx =e−cos(x)

Step 3: Multiply both sides of the differential equation by the integrating

factor u(x):

e−cos(x)dy

dx +e−cos(x)ysin(x) = e−cos(x)ecos(x)

Step 4: Simplify the equation:

dx(e−cos(x)y)=1

Step 5: Integrate both sides with respect to x:

dx(e−cos(x)y)dx =Z1dx

e−cos(x)y=x+C

Step 6: Solve for y:

y=ecos(x)(x+C)

y=xecos(x)+Cecos(x)

Therefore, the general solution to the differential equation is y=xecos(x)+

Cecos(x), where Cis an arbitrary constant.

Question 18

Question

Consider the first-order differential equation:

dt =1

t−2y

Solve the differential equation using an integrating factor method.

Solution

Step 1: Write the differential equation in standard form dy

dt +P(t)y=Q(t)

where P(t) and Q(t) are functions of t.

dt +2

ty=1

Step 2: Identify P(t) and Q(t).

P(t) = 2

Q(t) = 1

Step 3: Find the integrating factor µ(t) = eRP(t)dt.

µ(t) = eR2

tdt

=e2 ln |t|

=eln |t|2

=t2

Step 4: Multiply the differential equation by the integrating factor µ(t).

t2dy

dt + 2ty =t

Step 5: Rewrite the equation as the derivative of a product.

dt(t2y) = t

Step 6: Integrate both sides of the equation with respect to t.

dt(t2y)dt =Ztdt

t2y=1

2t2+C

Step 7: Solve for y.

y=1

2+C

Therefore, the solution to the differential equation is y=1

2+C

t2.

Question 19

Question

Solve the first-order differential equation:

dx + 2y= 4x2, y(0) = 3

Solution

Step 1: This is a first-order linear differential equation in the form dy

dx +P(x)y=

Q(x), where P(x) = 2 and Q(x)=4x2.

Step 2: First, we find the integrating factor IF :

IF =eRP(x)dx =eR2dx =e2x

Step 3: Next, we multiply both sides of the equation by the integrating

factor:

e2xdy

dx + 2e2xy= 4x2e2x

Step 4: Notice that the left side can be written as the derivative of the

product e2xyusing the product rule:

dx(e2xy)=4x2e2x

Step 5: Integrating both sides with respect to xgives:

e2xy=Z4x2e2xdx

Step 6: Simplify the right side by integrating:

e2xy= 2x2e2x−Z2xe2xdx

Step 7: Integrate by parts on the remaining integral: Let u= 2xand dv =

e2xdx. Then, du = 2 dx and v=1

2e2x.

e2xy= 2x2e2x−2xe2x−Z2e2xdx

Step 8: Evaluate the remaining integral:

e2xy= 2x2e2x−2xe2x+1

2e2x+C

Step 9: Finally, solve for yby dividing by e2x:

y= 2x2−2x+1

2+Ce−2x

Step 10: Using the initial condition y(0) = 3, we find the value of the

constant C:

3=0−0 + 1

2+C⇒C=5

Step 11: Therefore, the solution to the initial value problem is:

y= 2x2−2x+1

2+5

2e−2x

Question 20

Question

Solve the first-order differential equation: y′+ytan(x) = cos(x) with the initial

condition y(0) = 1.

Solution

Step 1: Rewrite the differential equation in standard form.

y′+ytan(x) = cos(x)

Step 2: Multiply both sides by sec(x) to simplify the equation.

sec(x)y′+ysec(x) tan(x) = sec(x) cos(x)

Step 3: Notice that the left-hand side can be written as the derivative of ysec(x)

using the product rule.

dx(ysec(x)) = sec(x) cos(x)

Step 4: Integrate both sides with respect to x.

dx(ysec(x)) dx =Zsec(x) cos(x)dx

ysec(x) = Zsec(x) cos(x)dx +C

Step 5: Evaluate the integral on the right-hand side using substitution (u=

sin(x)).

ysec(x) = Zsec(x) cos(x)dx +C=Zdu = sin(x) + C

Step 6: Solve for y.

y=sin(x) + C

sec(x)= sin(x) cos(x) + Ccos(x)

Step 7: Use the initial condition y(0) = 1 to find the value of C.

1 = sin(0) cos(0) + Ccos(0)

1 = 0 + C(1)

C= 1

Step 8: Substitute C= 1 back into the equation to get the final solution.

y= sin(x) cos(x) + cos(x)

Therefore, the solution to the differential equation y′+ytan(x) = cos(x)

with the initial condition y(0) = 1 is y= sin(x) cos(x) + cos(x).

Question 21

Question

Consider the first-order differential equation

y′=y2+x2

xy .

Find the general solution to this differential equation.

Solution

Step 1: Rewrite the equation in a standard form.

y′=y2+x2

dx =y2+x2

dx =y

x+x

ydy =xdx +ydx

Step 2: Integrate both sides.

Zydy =Zxdx +Zy

xdx

2y2=1

2x2+yln |x|+C

where Cis the constant of integration.

Step 3: Simplify the equation.

y2=x2+ 2yln |x|+C1

where C1= 2C.

Step 4: Solve for y.

y2−2yln |x|=x2+C1

y2−2yln |x|+ ln2|x|=x2+C1+ ln2|x|

(y−ln |x|)2=x2+C2

where C2=C1+ ln2|x|.

Step 5: Take the square root to find y.

y−ln |x|=±px2+C2

y= ln |x| ± px2+C2

Therefore, the general solution to the given differential equation is y=

ln |x| ± √x2+C2, where C2is an arbitrary constant.

Question 22

Question

Given the first-order differential equation xy′=y+xcos(y), find the general

solution.

Solution

Step 1: Rearrange the equation in the standard form dy

dx +P(x)y=Q(x) where

P(x) and Q(x) are functions of x.

Step 2: Divide both sides by xto write the equation in the standard form.

dx =y

x+ cos(y)

Step 3: Separate the variables by moving all the y’s to one side and all the

x’s to the other side. dy

dx =y

x+ cos(y)

y=dx

x+ cos(y)dy

Step 4: Integrate both sides.

ydy =Z1

xdx +Zcos(y)dy

ln |y|= ln |x|+ sin(y) + C1

Step 5: Combine the logs and apply the exponential function to both sides

to remove the absolute value.

y=Aesin(y)x

where A=±eC1is a constant.

Therefore, the general solution to the differential equation is y=Aesin(y)x,

where Ais an arbitrary constant.

Question 23

Question

Consider the first-order differential equation:

dx =2x2+y

Find the general solution to the differential equation.

Solution

Step 1: Rewrite the given differential equation in a form suitable for separation

of variables. dy

dx =2x2+y

x=⇒dy

dx = 2x+y

Step 2: Separate the variables by rewriting the equation as follows:

dx = 2x+y

x=⇒dy

dx = 2xdx +y

xdx

Step 3: Integrate both sides with respect to their respective variables.

Zdy

dxdx =Z(2xdx +y

xdx)

Step 4: Solve the integrals on the right-hand side:

y=x2+Cx

Thus, the general solution of the given differential equation is y=x2+Cx,

where Cis an arbitrary constant.

Question 24

Question

Solve the first-order differential equation:

dx =2x+y+ 1

x−y−2

Solution

Step 1: Rewrite the equation in a more standard form by separating variables.

(x−y−2)dy = (2x+y+ 1)dx

Step 2: Integrate both sides of the equation.

Z(x−y−2)dy =Z(2x+y+ 1)dx

Step 3: Perform the integrations.

Z(x−y−2)dy =Z(2x+y+ 1)dx

⇒xy −1

2y2−2y=x2+1

2y2+x+C

⇒2xy −y2−4y= 2x2+y2+ 2x+C

Step 4: Rearrange the terms to simplify the equation.

y2+ 2xy +y2−y2+ 4y= 2x2−2x−C

⇒2y2+ 2xy + 4y= 2x2−2x−C

Step 5: Complete the square on the left side of the equation.

y2+ 2y+ 1 + 2xy + 4y+ 4 = 2x2−2x−C+ 1 + 4

⇒(y+ 1)2+ 2(y+ 2)x+ 4 = 2x2−2x−C+ 5

Step 6: Rewrite the equation in a more compact form.

(y+ 1)2+ 2(y+ 2)x+ 4 = 2x2−2x+K(where K=−C+ 5)

Step 7: This is the general solution to the differential equation.

Question 25

Question

Consider the first-order differential equation:

y′=y2−y−2

Find the particular solution to the differential equation that satisfies the initial

condition y(0) = 1.

Solution

Step 1: First, let’s rewrite the given differential equation in the form:

dx =y2−y−2

Step 2: This equation can be factored as:

dx = (y−2)(y+ 1)

Step 3: Next, we can separate the variables:

(y−2)(y+ 1) =dx

Step 4: We can decompose the left side into partial fractions:

y−2+B

y+ 1 =dx

Step 5: Solving for Aand B, we get A=3

5and B=2

Step 6: Therefore, our integral becomes:

Z3

5(y−2) +2

5(y+ 1)dy =Zdx

Step 7: Integrating both sides, we get:

5ln |y−2|+2

5ln |y+ 1|=x+C

Step 8: Simplifying, we have:

ln |y−2|3+ ln |y+ 1|2= 5x+ 5C

Step 9: Combine the logs and rewrite in exponential form:

|(y−2)3(y+ 1)2|=e5x+5C

Step 10: Apply the initial condition y(0) = 1 to solve for C:

|(1 −2)3(1 + 1)2|=e5(0)+5C

| − 1|3·22=e5C

8 = e5C

Step 11: Find Cby taking the natural logarithm of both sides:

ln 8 = 5C

C=ln 8

Step 12: Finally, the particular solution is:

(y−2)3(y+ 1)2= 8e5x+5 ln 8

(y−2)3(y+ 1)2= 8ex+ln 8

(y−2)3(y+ 1)2= 8ex·8

(y−2)3(y+ 1)2= 64ex

Question 26

Question

Consider the first-order differential equation: dy

dx =x2+y. Find the particular

solution that satisfies the initial condition y(0) = 1.

Solution

Step 1: Rewrite the given first-order differential equation in the form dy

dx −y=x2.

Step 2: This is a linear first-order differential equation of the form dy

dx +

P(x)y=Q(x), where P(x) = −1 and Q(x) = x2.

Step 3: Find the integrating factor I(x) = eRP(x)dx.

I(x) = eR−1dx =e−x

Step 4: Multiply the integrating factor by the original equation.

e−xdy

dx −e−xy=x2e−x

Step 5: Rewrite the left-hand side as a derivative of a product.

dx(e−xy) = x2e−x

Step 6: Integrate both sides with respect to x.

dx(e−xy)dx =Zx2e−xdx

Step 7: Simplify the integrals.

e−xy=Zx2e−xdx

Step 8: Solve the integral on the right-hand side.

e−xy=−x2e−x−2xe−x−2e−x+C

Step 9: Multiply through by exto solve for y.

y=−x2−2x−2 + Cex

Step 10: Apply the initial condition y(0) = 1 to find the value of the constant

1 = −0−2(0) −2 + C

C= 3

Step 11: Thus, the particular solution to the differential equation that sat-

isfies the initial condition y(0) = 1 is y=−x2−2x−2+3ex.

Question 27

Question

Solve the following first-order differential equation:

dx =y

x+x

Solution

To solve the given first-order differential equation, we can rewrite it in the form

dx =P(x,y)

Q(x,y), where P(x, y) and Q(x, y) are functions of xand y.

Step 1: Identify P(x, y)and Q(x, y) In this case, P(x, y) = yand

Q(x, y) = x+x2

Step 2: Check for exactness Let M(x, y) = yand N(x, y) = x+x2

y. We

need to check if ∂M

∂y =∂N

∂x .

∂M

∂y = 1 and ∂N

∂x = 1 + 2x

Since ∂M

∂y =∂N

∂x , the equation is exact.

Step 3: Find the general solution To find the general solution, we need

to integrate the function M(x, y) = ywith respect to xand N(x, y) = x+x2

with respect to y.

Integrating M(x, y) with respect to x, we get:

Zy dx =xy +C(y)

Where C(y) is a constant of integration that may also be a function of y.

Integrating N(x, y) with respect to y, we get:

Zx+x2

ydy =xy +xln |y|+D(x)

Where D(x) is a constant of integration that may also be a function of x.

Step 4: Equate the two expressions Setting the integrals equal to each

other gives:

xy +C(y) = xy +xln |y|+D(x)

Step 5: Simplify the equation From the equation above, we find that

C(y) = xln |y|+D(x).

Step 6: Find the general solution Therefore, the general solution to the

differential equation is:

xy =xln |y|+D(x)

Where D(x) is an arbitrary function of x.

Question 28

Question

Consider the first-order differential equation:

dx +y=excos(x)

Find the general solution to this differential equation.

Solution

Step 1: First, we need to find the integrating factor, denoted by µ(x), which

is given by µ(x) = eRP(x)dx where P(x) is the coefficient of y. In this case,

P(x) = 1, so

µ(x) = eR1dx =ex

Step 2: Multiply both sides of the differential equation by the integrating

factor µ(x):

exdy

dx +exy=e2xcos(x)

Step 3: Recognize that the left side is the result of the product rule for

derivatives. Rewrite the left side as the derivative of the product exy:

dx(exy) = e2xcos(x)

Step 4: Integrate both sides with respect to x:

dx(exy)dx =Ze2xcos(x)dx

exy=Ze2xcos(x)dx

Step 5: Use integration by parts to solve the integral on the right side. Let

u= cos(x) and dv =e2xdx. Then, du =−sin(x)dx and v=1

2e2x.

exy=1

2e2xcos(x)−1

2Ze2xsin(x)dx

Step 6: Integrate the remaining integral on the right side using integration

by parts again. Let u= sin(x) and dv =e2xdx. Then, du = cos(x)dx and

v=1

2e2x.

exy=1

2e2xcos(x)−1

21

2e2xsin(x) + 1

2Ze2xcos(x)dx

Step 7: Solve the final integral using substitution or another method to find

the general solution

exy=1

2e2xcos(x)−1

4e2xsin(x)−1

4e2xcos(x) + C

Step 8: Divide by exto solve for y:

y=1

2cos(x)−1

4sin(x)−1

4cos(x) + C

Therefore, the general solution to the differential equation is:

y=−1

4cos(x)−1

4sin(x) + C

Question 29

Question

Find the general solution to the first-order differential equation:

dx + 2xy = 1.

Solution

Step 1: This is a first-order linear differential equation of the form dy

dx +P(x)y=

Q(x). We can use an integrating factor to solve this equation.

Step 2: Identify P(x) and Q(x). Here, P(x)=2xand Q(x) = 1.

Step 3: Find the integrating factor µ(x) = eRP(x)dx =eR2x dx =ex2.

Step 4: Multiply both sides of the differential equation by the integrating

factor µ(x):

ex2dy

dx + 2xex2y=ex2.

Step 5: Rewrite the left side of the equation as the derivative of the product

y·ex2using the product rule:

dx(y·ex2) = ex2.

Step 6: Integrate both sides with respect to xto get:

y·ex2=Zex2dx =1

2ex2+C,

where Cis the constant of integration.

Step 7: Solve for yby dividing both sides by ex2:

y=1

2+Ce−x2.

Therefore, the general solution to the differential equation dy

dx + 2xy = 1 is

y=1

2+Ce−x2, where Cis an arbitrary constant.

Question 30

Question

Consider the first-order differential equation:

dx =2x−y

x+ 2y

Find the general solution to this differential equation.

Solution

Step 1: Rewrite the differential equation in a more familiar form by multiplying

both sides by (x+ 2y) to get rid of the denominator.

(x+ 2y)dy

dx = 2x−y

Step 2: Rearrange the equation to isolate dy

dx .

(x+ 2y)dy = (2x−y)dx

Step 3: Integrate both sides with respect to yand xrespectively.

Z(x+ 2y)dy =Z(2x−y)dx

Step 4: Perform the integrations.

xy +y2=x2−xy +C

Step 5: Rearrange the equation into a standard form.

x2+ 2xy +y2=C

Step 6: Recognize that the left side is a perfect square.

(x+y)2=C

Step 7: Take the square root of both sides.

x+y=±√C

Step 8: Let C=k2where kis a constant.

x+y=±k

Step 9: Solve for y.

y=−x±k

Therefore, the general solution to the differential equation is y=−x±k,

where kis a constant.

Step 4: Integrate both sides with respect to x:

dx(e(e2xx)y)dx =Ze3xydx

e(e2xx)y=Ze3xydx

Step 5: Now, let’s solve the integral on the right-hand side by rearranging

the equation:

e(e2xx)=1

dx =d

dx(e2x)

Integrating both sides:

e2x=Ze3xdx

e2x=1

3e3x+C

Step 6: Finally, solve for yby rearranging the terms:

e2xy=1

3e3xy+Cy

2y=1

3ex+Cy

y=1

6ex+Cy

Question 2

Question

Find the general solution to the first-order differential equation: dy

dx =2x+3y+5

3x+2y+7 .

Solution

Step 1: Rearrange the equation in the form dy

dx =f(x) + g(y).

dx =2x+ 3y+ 5

3x+ 2y+ 7

dx =2x+ 3y+ 5

3x+ 2y+ 7

dx =2(x+5

2) + 3(y+5

3(x+7

3) + 2(y+7

Step 2: Recognize that this equation is exact, and find the auxiliary equation

by comparing it to dy

dx =P(x) + Q(y).

2P(x)+3Q(y)=0

∂P

∂y =∂Q

∂x

2 = 3

Step 3: Integrate P(x) with respect to xto find f(x).

f(x) = 2(x+5

2) = 2x+ 5

Step 4: Integrate Q(y) with respect to yto find g(y).

g(y) = 3(y+5

3)=3y+ 5

Step 5: Write down the general solution to the differential equation.

y=−5−2x+Ce−3x

Therefore, the general solution to the first-order differential equation is y=

−5−2x+Ce−3x.

Question 3

Question

Consider the differential equation:

dx =2x+ 3y+ 7

x−4y−5

Find the solution to the initial value problem y(0) = −3.

Solution

Step 1: Rewrite the differential equation in standard form by separating vari-

ables. dy

dx =2x+ 3y+ 7

x−4y−5=⇒(x−4y−5)dy = (2x+ 3y+ 7)dx

Step 2: Integrate both sides to solve for y.

Z(x−4y−5)dy =Z(2x+ 3y+ 7)dx

Step 3: Simplify the integrals.

Zxdy −4Zydy −5Zdy = 2 Zxdx + 3 Zydx + 7 Zdx

Step 4: Integrate each term separately.

xy −2y2−5y=x2+ 3xy + 7x+C

where Cis the constant of integration.

Step 5: Use the initial condition y(0) = −3 to find the value of the constant

0(−3) −2(−3)2−5(−3) = 02+ 3(0)(−3) + 7(0) + C

9−18 + 15 = C

C= 6

Step 6: Substitute C= 6 back into the equation to get the final solution.

xy −2y2−5y=x2+ 3xy + 7x+ 6

x(−3) −2(−3)2−5(−3) = x2+ 3x(−3) + 7x+ 6

−3x+ 18 −15 = x2−9x+ 7x+ 6

x2−11x+ 9 = 0

(x−1)(x−9) = 0

x= 1,9

Therefore, the solution to the initial value problem is y=2x

x−4+7

x−4.

Question 4

Question

Solve the following first-order differential equation:

dx −2y

x=x3

Solution

Step 1: This is a first-order linear differential equation in the form

dx +P(x)y=Q(x)

where P(x) = −2

xand Q(x) = x3. We begin by finding the integrating factor

µ(x) = eRP(x)dx.

Step 2: Calculate the integrating factor µ(x):

µ(x) = eR−2

xdx =e−2 ln |x|=x−2

Step 3: Multiply both sides of the differential equation by the integrating

factor:

x−2dy

dx −2x−3y=x

Step 4: Rewrite the left-hand side of the equation as the derivative of the

product of x−2and y:d

dx(x−2y) = x

Step 5: Integrate both sides with respect to x:

dx(x−2y)dx =Zxdx

Step 6: Solve the integrals:

x−2y=1

2x2+C

Step 7: Solve for y:

y=1

2x4+Cx2

Thus, the general solution to the differential equation is y=1

2x4+Cx2,

where Cis an arbitrary constant.

Question 5

Question

Solve the following first-order differential equation using an integrating factor:

dx + 2y=e4x

Solution

Step 1: Write the given equation in the standard form dy

dx +P(x)y=Q(x). In

this case, P(x) = 2 and Q(x) = e4x, so the equation becomes:

dx + 2y=e4x

Step 2: Identify the integrating factor by multiplying both sides of the equa-

tion by the integrating factor µ(x) = eRP(x)dx. The integrating factor in this

case is µ(x) = eR2dx =e2x.

Step 3: Multiply both sides of the differential equation by the integrating

factor.

e2xdy

dx + 2e2xy=e2xe4x

Step 4: Rewrite the left side of the equation as the derivative of the product

of the integrating factor and y.

dx(e2xy) = e6x

Step 5: Integrate both sides with respect to x.

dx(e2xy)dx =Ze6xdx

Step 6: Solve the integrals.

e2xy=e6x

6+C

Step 7: Solve for y.

y=1

6e4x+Ce−2x

where Cis an arbitrary constant.

Question 6

Question

Consider the first-order differential equation:

dx =y

x+ex

Find the general solution to this differential equation.

Solution

Step 1: Rewrite the equation in a standard form:

dx −y

x=ex

Step 2: This is a linear first-order differential equation. We can solve it using

an integrating factor. The integrating factor, denoted by I(x), is calculated as:

I(x) = eR−1

xdx =e−ln |x|=1

|x|

Step 3: Multiply both sides of the equation by the integrating factor:

|x|

dx −1

|x|

x=1

|x|ex

Step 4: Write the left side as the derivative of the product of the integrating

factor and y:

dx y

|x|=1

|x|ex

Step 5: Integrate both sides with respect to x:

dx y

|x|dx =Z1

|x|exdx

Step 6: Simplify the integrals:

|x|=Z1

|x|exdx

Step 7: The integral on the right side can be solved by using integration by

parts:

u=ex, dv =1

|x|dx

du =exdx, v =Z1

|x|dx = ln |x|

|x|exdx =exln |x| − Zex·ln |x|dx

Step 8: Express the solution in terms of y:

y=|x|exln |x|−|x|ex+C

Therefore, the general solution to the first-order differential equation is:

y=|x|exln |x|−|x|ex+C

where Cis an arbitrary constant.

Question 7

Question

Find the general solution to the first-order differential equation: dy

dx =1

x2+y2.

Solution

Step 1: We can rewrite the given differential equation as dy

dx =1

x2(1+(y

x)2).

Step 2: Let u=y

x, then y=ux and dy

dx =u+xdu

dx .

Step 3: Substitute y=ux and dy

dx =u+xdu

dx into the differential equation

to get du

dx x2(1 + u2) = 1.

Step 4: Rearrange the equation to separate variables: du

1+u2=1

x2dx.

Step 5: Integrate both sides with respect to their respective variables: Rdu

1+u2=

x2dx.

arctan(u) = −1

x+C

Step 6: Substitute back u=y

xinto the equation to get arctan y

x=−1

x+C.

Step 7: The general solution is y=xtan −1

x+C).

Question 8

Question

Find the general solution to the first-order differential equation: dy

dx =2y2−5

3x.

Solution

Step 1: Rearrange the equation to separate variables:

2y2−5=dx

Step 2: Integrate both sides with respect to yand x:

2y2−5dy =Z1

3xdx

Step 3: Factor the denominator of the left integral and use partial fractions

to integrate:

2y2−5dy =Z1

3xdx

(y+p5/2)(y−p5/2)dy =Z1

3xdx

2p5/2Z 1

y+p5/2−1

y−p5/2!dy =1

3Z1

xdx

Step 4: Integrate both sides:

2p5/2(ln |y+p5/2| − ln |y−p5/2|) = 1

3ln |x|+C

Step 5: Simplify the natural logarithms:

2p5/2ln 

y+p5/2

y−p5/2

3ln |x|+C

Step 6: Exponentiate both sides to eliminate the natural logarithms:



y+p5/2

y−p5/2

1/2√5/2

=|x|1/3eC

Step 7: Simplify by taking the absolute value on both sides and reusing the

constant term:

y+p5/2

y−p5/2=±x1/3eC

Step 8: Solve for y:

y+p5/2 = ±x1/3eC(y−p5/2)

y+p5/2 = ±x1/3eCy∓x1/3p5/2eC

y(1 ∓x1/3eC) = ±x1/3p5/2eC−p5/2

Step 9: Finally, solve for yto get the general solution:

y=±x1/3p5/2eC−p5/2

1∓x1/3eC

Question 9

Question

Find the general solution to the first-order differential equation: dy

dx =2x+y−1

x+2y.

Solution

To solve this first-order differential equation, we will rearrange the terms, sep-

arate variables, and integrate.

Step 1: Write the given differential equation in the form dy

dx =f(x, y).

dx =2x+y−1

x+ 2y

Step 2: Multiply through by dx and separate variables.

(x+ 2y)dy = (2x+y−1)dx

Step 3: Integrate both sides.

Z(x+ 2y)dy =Z(2x+y−1)dx

Step 4: Perform the integrations.

Z(x+ 2y)dy =Z(2x+y−1)dx

2y2+xy =x2+1

2y2−x+C

Step 5: Simplify the equation.

xy =x2−x+C

Step 6: Rearrange the equation to find the general solution.

y=x−1 + C

Therefore, the general solution to the given first-order differential equation

is y=x−1 + C

x, where Cis an arbitrary constant.

Question 10

Question

Consider the first-order differential equation: dy

dx =x3−e2x. Solve the differen-

tial equation using the method of separation of variables.

Solution

Step 1: Separate the variables by writing the equation in the form dy

dx =

f(x)g(y).

We have: dy

dx =x3−e2x

Step 2: Rewrite the equation as:

dx =x3−e2x

dy = (x3−e2x)dx

Step 3: Integrate both sides with respect to their respective variables.

Zdy =Z(x3−e2x)dx

Step 4: Integrate both sides:

Zdy =Zx3dx −Ze2xdx

y=x4

4−1

2e2x+C

Therefore, the general solution to the differential equation dy

dx =x3−e2xis

y=x4

4−1

2e2x+C, where C is the constant of integration.

Question 11

Question

Solve the following first-order differential equation:

y′+ 4y= 8e−2x

Solution

Step 1: We first identify the integrating factor which is given by:

µ(x) = eRP(x)dx

where P(x) is the coefficient of yin the differential equation. In this case,

P(x) = 4, so

µ(x) = eR4dx

µ(x) = e4x

Step 2: We multiply both sides of the differential equation by the integrating

factor:

e4x·(y′+ 4y) = e4x·8e−2x

e4xy′+ 4e4xy= 8

Step 3: Notice the left side can be written as the derivative of a product:

(e4xy)′= 8

Step 4: Integrating both sides with respect to x:

Z(e4xy)′dx =Z8dx

e4xy= 8x+C

where Cis the constant of integration.

Step 5: Solving for y:

y=8x+C

e4x

y= 8xe−4x+Ce−4x

y=x(8 −4x) + Ce−4x

where C= 0 because it was absorbed into the constant of integration.

Therefore, the solution to the differential equation is:

y=x(8 −4x)

Question 12

Question

Solve the initial value problem:

y′−2xy =x, y(0) = 1

Solution

Step 1: First, we identify the integrating factor µ(x) = eR−2x dx =e−x2.

Step 2: Multiply both sides of the differential equation by the integrating

factor:

e−x2y′−2xe−x2y=xe−x2

Step 3: Recognize that the left side is now the derivative of the product of

yand the integrating factor:

(e−x2y)′=xe−x2

Step 4: Integrate both sides with respect to x:

Z(e−x2y)dx =Zxe−x2dx

Step 5: On the left side, we simplify the integral to get e−x2y. And on the

right side, we have to do a substitution u=−x2:

−1

2Zeudu =−1

2e−x2+C1

Step 6: Simplifying the integration on both sides gives us:

e−x2y=−1

2e−x2+C1

Step 7: To find y, multiply through by ex2:

y=−1

2+C1ex2

Step 8: Apply the initial condition y(0) = 1 to find the value of the constant

C1:

1 = −1

2+C1

C1=3

Step 9: Substitute C1=3

2back into the solution to get the particular

solution:

y=−1

2+3

2ex2

Question 13

Question

Consider the first-order differential equation given by dy

dx =2x+y

x. Find the

general solution to this differential equation.

Solution

Step 1: Rewrite the differential equation in standard form by multiplying through

by dx:dy

dx =2x+y

x=⇒xdy

dx −y= 2x

Step 2: This is now a linear first-order differential equation in the form

y′+P(x)y=Q(x). To solve this, we need to find an integrating factor I(x)

defined by I(x) = eRP(x)dx. Here, P(x) = −1

x, so I(x) = eR−1

xdx =e−ln |x|=

|x|.

Step 3: Multiply our differential equation by the integrating factor:

|x|xdy

dx −y=1

|x|·2x

This simplifies to: 1

|x|·xdy

dx −1

|x|·y= 2

Step 4: Recognize that the left side of the equation is the derivative of y

|x|

with respect to x:

dx y

|x|= 2

Step 5: Integrate both sides with respect to xto find the general solution:

dx y

|x|dx =Z2dx

|x|= 2x+C

where Cis the constant of integration.

Step 6: Multiply through by |x|to get the general solution in a more standard

form:

y= 2x|x|+C|x|

y= 2x|x|+Cx (for x= 0)

Thus, the general solution to the differential equation is y= 2x|x|+Cx for

x= 0, where Cis an arbitrary constant.

Question 14

Question

Solve the first-order differential equation:

dx =x+y

Solution

Step 1: Rewrite the differential equation in standard form:

dx −y

x=1

Step 2: This is a linear first-order differential equation. To solve it, we first

find the integrating factor, denoted by µ(x), which is given by:

µ(x) = eR−1

xdx

Step 3: Calculate the integrating factor:

µ(x) = e−ln |x|=1

|x|=1

Step 4: Multiply the integrating factor with the differential equation:

dx −y

x2= 1

Step 5: Rewrite the equation in terms of the total derivative:

dx y

x= 1

Step 6: Integrate both sides with respect to x:

dx y

xdx =Z1dx

Step 7: Integrate the right side with respect to x:

x=x+C

Step 8: Multiply through by xto obtain the general solution:

y=x2+Cx

Question 15

Question

Find the general solution to the following first-order differential equation:

dx + 2y=e−2x

Solution

Step 1: First, we identify the standard form of a first-order linear differential

equation:

dx +P(x)y=Q(x)

where P(x) = 2 and Q(x) = e−2x.

Step 2: To solve the differential equation, we need to find an integrating

factor I(x) defined by:

I(x) = eRP(x)dx

In this case, P(x) = 2, so:

ZP(x)dx =Z2dx = 2x

Thus, the integrating factor is I(x) = e2x.

Step 3: Multiply both sides of the differential equation by the integrating

factor:

e2xdy

dx + 2e2xy=e2xe−2x

Simplify to get:

dx(e2xy) = 1

Step 4: Integrate both sides with respect to x:

dx(e2xy)dx =Z1dx

This gives us:

e2xy=x+C

where Cis the constant of integration.

Step 5: Solve for yby dividing through by e2x:

y=x

e2x+C

e2x

which simplifies to:

y=xe−2x+Ce−2x

So, the general solution to the differential equation is y=xe−2x+Ce−2x,

where Cis an arbitrary constant.

Question 16

Question

Find the particular solution to the differential equation dy

dx −2y= 6xgiven that

y= 5 when x= 2.

Solution

Step 1: First, we rewrite the given differential equation in the standard form of

a first-order linear ODE: dy

dx −2y= 6x.

Step 2: We identify the integrating factor, which is eR−2dx =e−2x.

Step 3: Multiply both sides of the differential equation by the integrating

factor e−2x:

e−2xdy

dx −2e−2xy= 6xe−2x

Step 4: The left side can be rewritten as the derivative of the integrating

factor multiplied by y:d

dx(e−2xy) = 6xe−2x

Step 5: Integrate both sides with respect to x:

dx(e−2xy)dx =Z6xe−2xdx

Step 6: Integrate the right side using integration by parts: Let u=xand

dv = 6e−2xdx. Then, du =dx and v=−3e−2x.

Z6xe−2xdx =−3xe−2x−Z(−3e−2xdx)

Step 7: Simplify the integration on the right side:

Z6xe−2xdx =−3xe−2x+Z6e−2xdx

Step 8: Integrate the remaining integral on the right side:

Z6e−2xdx =−3xe−2x−3

2e−2x+C, where Cis the constant of integration

Step 9: Substitute back into the differential equation and solve for y:

e−2xy=−3xe−2x−3

2e−2x+C

Step 10: Finally, rearrange the equation to solve for y:

y=−3x−3

2+Ce2x

Step 11: Use the initial condition y(2) = 5 to find the value of the constant

5 = −6−3

2+Ce4

Step 12: Solve for C:

C=17

2e−4

Therefore, the particular solution to the differential equation is y=−3x−

2+17

2e2x.

Question 17

Question

Solve the following first-order differential equation:

dx +ysin(x) = ecos(x)

Solution

Step 1: This is a first-order linear differential equation in the standard form

y′+P(x)y=Q(x), where P(x) = sin(x) and Q(x) = ecos(x). To solve this type

of equation, we can use an integrating factor u(x) = eRP(x)dx.

Step 2: First, we find the integrating factor u(x):

u(x) = eRsin(x)dx =e−cos(x)

Step 3: Multiply both sides of the differential equation by the integrating

factor u(x):

e−cos(x)dy

dx +e−cos(x)ysin(x) = e−cos(x)ecos(x)

Step 4: Simplify the equation:

dx(e−cos(x)y)=1

Step 5: Integrate both sides with respect to x:

dx(e−cos(x)y)dx =Z1dx

e−cos(x)y=x+C

Step 6: Solve for y:

y=ecos(x)(x+C)

y=xecos(x)+Cecos(x)

Therefore, the general solution to the differential equation is y=xecos(x)+

Cecos(x), where Cis an arbitrary constant.

Question 18

Question

Consider the first-order differential equation:

dt =1

t−2y

Solve the differential equation using an integrating factor method.

Solution

Step 1: Write the differential equation in standard form dy

dt +P(t)y=Q(t)

where P(t) and Q(t) are functions of t.

dt +2

ty=1

Step 2: Identify P(t) and Q(t).

P(t) = 2

Q(t) = 1

Step 3: Find the integrating factor µ(t) = eRP(t)dt.

µ(t) = eR2

tdt

=e2 ln |t|

=eln |t|2

=t2

Step 4: Multiply the differential equation by the integrating factor µ(t).

t2dy

dt + 2ty =t

Step 5: Rewrite the equation as the derivative of a product.

dt(t2y) = t

Step 6: Integrate both sides of the equation with respect to t.

dt(t2y)dt =Ztdt

t2y=1

2t2+C

Step 7: Solve for y.

y=1

2+C

Therefore, the solution to the differential equation is y=1

2+C

t2.

Question 19

Question

Solve the first-order differential equation:

dx + 2y= 4x2, y(0) = 3

Solution

Step 1: This is a first-order linear differential equation in the form dy

dx +P(x)y=

Q(x), where P(x) = 2 and Q(x)=4x2.

Step 2: First, we find the integrating factor IF :

IF =eRP(x)dx =eR2dx =e2x

Step 3: Next, we multiply both sides of the equation by the integrating

factor:

e2xdy

dx + 2e2xy= 4x2e2x

Step 4: Notice that the left side can be written as the derivative of the

product e2xyusing the product rule:

dx(e2xy)=4x2e2x

Step 5: Integrating both sides with respect to xgives:

e2xy=Z4x2e2xdx

Step 6: Simplify the right side by integrating:

e2xy= 2x2e2x−Z2xe2xdx

Step 7: Integrate by parts on the remaining integral: Let u= 2xand dv =

e2xdx. Then, du = 2 dx and v=1

2e2x.

e2xy= 2x2e2x−2xe2x−Z2e2xdx

Step 8: Evaluate the remaining integral:

e2xy= 2x2e2x−2xe2x+1

2e2x+C

Step 9: Finally, solve for yby dividing by e2x:

y= 2x2−2x+1

2+Ce−2x

Step 10: Using the initial condition y(0) = 3, we find the value of the

constant C:

3=0−0 + 1

2+C⇒C=5

Step 11: Therefore, the solution to the initial value problem is:

y= 2x2−2x+1

2+5

2e−2x

Question 20

Question

Solve the first-order differential equation: y′+ytan(x) = cos(x) with the initial

condition y(0) = 1.

Solution

Step 1: Rewrite the differential equation in standard form.

y′+ytan(x) = cos(x)

Step 2: Multiply both sides by sec(x) to simplify the equation.

sec(x)y′+ysec(x) tan(x) = sec(x) cos(x)

Step 3: Notice that the left-hand side can be written as the derivative of ysec(x)

using the product rule.

dx(ysec(x)) = sec(x) cos(x)

Step 4: Integrate both sides with respect to x.

dx(ysec(x)) dx =Zsec(x) cos(x)dx

ysec(x) = Zsec(x) cos(x)dx +C

Step 5: Evaluate the integral on the right-hand side using substitution (u=

sin(x)).

ysec(x) = Zsec(x) cos(x)dx +C=Zdu = sin(x) + C

Step 6: Solve for y.

y=sin(x) + C

sec(x)= sin(x) cos(x) + Ccos(x)

Step 7: Use the initial condition y(0) = 1 to find the value of C.

1 = sin(0) cos(0) + Ccos(0)

1 = 0 + C(1)

C= 1

Step 8: Substitute C= 1 back into the equation to get the final solution.

y= sin(x) cos(x) + cos(x)

Therefore, the solution to the differential equation y′+ytan(x) = cos(x)

with the initial condition y(0) = 1 is y= sin(x) cos(x) + cos(x).

Question 21

Question

Consider the first-order differential equation

y′=y2+x2

xy .

Find the general solution to this differential equation.

Solution

Step 1: Rewrite the equation in a standard form.

y′=y2+x2

dx =y2+x2

dx =y

x+x

ydy =xdx +ydx

Step 2: Integrate both sides.

Zydy =Zxdx +Zy

xdx

2y2=1

2x2+yln |x|+C

where Cis the constant of integration.

Step 3: Simplify the equation.

y2=x2+ 2yln |x|+C1

where C1= 2C.

Step 4: Solve for y.

y2−2yln |x|=x2+C1

y2−2yln |x|+ ln2|x|=x2+C1+ ln2|x|

(y−ln |x|)2=x2+C2

where C2=C1+ ln2|x|.

Step 5: Take the square root to find y.

y−ln |x|=±px2+C2

y= ln |x| ± px2+C2

Therefore, the general solution to the given differential equation is y=

ln |x| ± √x2+C2, where C2is an arbitrary constant.

Question 22

Question

Given the first-order differential equation xy′=y+xcos(y), find the general

solution.

Solution

Step 1: Rearrange the equation in the standard form dy

dx +P(x)y=Q(x) where

P(x) and Q(x) are functions of x.

Step 2: Divide both sides by xto write the equation in the standard form.

dx =y

x+ cos(y)

Step 3: Separate the variables by moving all the y’s to one side and all the

x’s to the other side. dy

dx =y

x+ cos(y)

y=dx

x+ cos(y)dy

Step 4: Integrate both sides.

ydy =Z1

xdx +Zcos(y)dy

ln |y|= ln |x|+ sin(y) + C1

Step 5: Combine the logs and apply the exponential function to both sides

to remove the absolute value.

y=Aesin(y)x

where A=±eC1is a constant.

Therefore, the general solution to the differential equation is y=Aesin(y)x,

where Ais an arbitrary constant.

Question 23

Question

Consider the first-order differential equation:

dx =2x2+y

Find the general solution to the differential equation.

Solution

Step 1: Rewrite the given differential equation in a form suitable for separation

of variables. dy

dx =2x2+y

x=⇒dy

dx = 2x+y

Step 2: Separate the variables by rewriting the equation as follows:

dx = 2x+y

x=⇒dy

dx = 2xdx +y

xdx

Step 3: Integrate both sides with respect to their respective variables.

Zdy

dxdx =Z(2xdx +y

xdx)

Step 4: Solve the integrals on the right-hand side:

y=x2+Cx

Thus, the general solution of the given differential equation is y=x2+Cx,

where Cis an arbitrary constant.

Question 24

Question

Solve the first-order differential equation:

dx =2x+y+ 1

x−y−2

Solution

Step 1: Rewrite the equation in a more standard form by separating variables.

(x−y−2)dy = (2x+y+ 1)dx

Step 2: Integrate both sides of the equation.

Z(x−y−2)dy =Z(2x+y+ 1)dx

Step 3: Perform the integrations.

Z(x−y−2)dy =Z(2x+y+ 1)dx

⇒xy −1

2y2−2y=x2+1

2y2+x+C

⇒2xy −y2−4y= 2x2+y2+ 2x+C

Step 4: Rearrange the terms to simplify the equation.

y2+ 2xy +y2−y2+ 4y= 2x2−2x−C

⇒2y2+ 2xy + 4y= 2x2−2x−C

Step 5: Complete the square on the left side of the equation.

y2+ 2y+ 1 + 2xy + 4y+ 4 = 2x2−2x−C+ 1 + 4

⇒(y+ 1)2+ 2(y+ 2)x+ 4 = 2x2−2x−C+ 5

Step 6: Rewrite the equation in a more compact form.

(y+ 1)2+ 2(y+ 2)x+ 4 = 2x2−2x+K(where K=−C+ 5)

Step 7: This is the general solution to the differential equation.

Question 25

Question

Consider the first-order differential equation:

y′=y2−y−2

Find the particular solution to the differential equation that satisfies the initial

condition y(0) = 1.

Solution

Step 1: First, let’s rewrite the given differential equation in the form:

dx =y2−y−2

Step 2: This equation can be factored as:

dx = (y−2)(y+ 1)

Step 3: Next, we can separate the variables:

(y−2)(y+ 1) =dx

Step 4: We can decompose the left side into partial fractions:

y−2+B

y+ 1 =dx

Step 5: Solving for Aand B, we get A=3

5and B=2

Step 6: Therefore, our integral becomes:

Z3

5(y−2) +2

5(y+ 1)dy =Zdx

Step 7: Integrating both sides, we get:

5ln |y−2|+2

5ln |y+ 1|=x+C

Step 8: Simplifying, we have:

ln |y−2|3+ ln |y+ 1|2= 5x+ 5C

Step 9: Combine the logs and rewrite in exponential form:

|(y−2)3(y+ 1)2|=e5x+5C

Step 10: Apply the initial condition y(0) = 1 to solve for C:

|(1 −2)3(1 + 1)2|=e5(0)+5C

| − 1|3·22=e5C

8 = e5C

Step 11: Find Cby taking the natural logarithm of both sides:

ln 8 = 5C

C=ln 8

Step 12: Finally, the particular solution is:

(y−2)3(y+ 1)2= 8e5x+5 ln 8

(y−2)3(y+ 1)2= 8ex+ln 8

(y−2)3(y+ 1)2= 8ex·8

(y−2)3(y+ 1)2= 64ex

Question 26

Question

Consider the first-order differential equation: dy

dx =x2+y. Find the particular

solution that satisfies the initial condition y(0) = 1.

Solution

Step 1: Rewrite the given first-order differential equation in the form dy

dx −y=x2.

Step 2: This is a linear first-order differential equation of the form dy

dx +

P(x)y=Q(x), where P(x) = −1 and Q(x) = x2.

Step 3: Find the integrating factor I(x) = eRP(x)dx.

I(x) = eR−1dx =e−x

Step 4: Multiply the integrating factor by the original equation.

e−xdy

dx −e−xy=x2e−x

Step 5: Rewrite the left-hand side as a derivative of a product.

dx(e−xy) = x2e−x

Step 6: Integrate both sides with respect to x.

dx(e−xy)dx =Zx2e−xdx

Step 7: Simplify the integrals.

e−xy=Zx2e−xdx

Step 8: Solve the integral on the right-hand side.

e−xy=−x2e−x−2xe−x−2e−x+C

Step 9: Multiply through by exto solve for y.

y=−x2−2x−2 + Cex

Step 10: Apply the initial condition y(0) = 1 to find the value of the constant

1 = −0−2(0) −2 + C

C= 3

Step 11: Thus, the particular solution to the differential equation that sat-

isfies the initial condition y(0) = 1 is y=−x2−2x−2+3ex.

Question 27

Question

Solve the following first-order differential equation:

dx =y

x+x

Solution

To solve the given first-order differential equation, we can rewrite it in the form

dx =P(x,y)

Q(x,y), where P(x, y) and Q(x, y) are functions of xand y.

Step 1: Identify P(x, y)and Q(x, y) In this case, P(x, y) = yand

Q(x, y) = x+x2

Step 2: Check for exactness Let M(x, y) = yand N(x, y) = x+x2

y. We

need to check if ∂M

∂y =∂N

∂x .

∂M

∂y = 1 and ∂N

∂x = 1 + 2x

Since ∂M

∂y =∂N

∂x , the equation is exact.

Step 3: Find the general solution To find the general solution, we need

to integrate the function M(x, y) = ywith respect to xand N(x, y) = x+x2

with respect to y.

Integrating M(x, y) with respect to x, we get:

Zy dx =xy +C(y)

Where C(y) is a constant of integration that may also be a function of y.

Integrating N(x, y) with respect to y, we get:

Zx+x2

ydy =xy +xln |y|+D(x)

Where D(x) is a constant of integration that may also be a function of x.

Step 4: Equate the two expressions Setting the integrals equal to each

other gives:

xy +C(y) = xy +xln |y|+D(x)

Step 5: Simplify the equation From the equation above, we find that

C(y) = xln |y|+D(x).

Step 6: Find the general solution Therefore, the general solution to the

differential equation is:

xy =xln |y|+D(x)

Where D(x) is an arbitrary function of x.

Question 28

Question

Consider the first-order differential equation:

dx +y=excos(x)

Find the general solution to this differential equation.

Solution

Step 1: First, we need to find the integrating factor, denoted by µ(x), which

is given by µ(x) = eRP(x)dx where P(x) is the coefficient of y. In this case,

P(x) = 1, so

µ(x) = eR1dx =ex

Step 2: Multiply both sides of the differential equation by the integrating

factor µ(x):

exdy

dx +exy=e2xcos(x)

Step 3: Recognize that the left side is the result of the product rule for

derivatives. Rewrite the left side as the derivative of the product exy:

dx(exy) = e2xcos(x)

Step 4: Integrate both sides with respect to x:

dx(exy)dx =Ze2xcos(x)dx

exy=Ze2xcos(x)dx

Step 5: Use integration by parts to solve the integral on the right side. Let

u= cos(x) and dv =e2xdx. Then, du =−sin(x)dx and v=1

2e2x.

exy=1

2e2xcos(x)−1

2Ze2xsin(x)dx

Step 6: Integrate the remaining integral on the right side using integration

by parts again. Let u= sin(x) and dv =e2xdx. Then, du = cos(x)dx and

v=1

2e2x.

exy=1

2e2xcos(x)−1

21

2e2xsin(x) + 1

2Ze2xcos(x)dx

Step 7: Solve the final integral using substitution or another method to find

the general solution

exy=1

2e2xcos(x)−1

4e2xsin(x)−1

4e2xcos(x) + C

Step 8: Divide by exto solve for y:

y=1

2cos(x)−1

4sin(x)−1

4cos(x) + C

Therefore, the general solution to the differential equation is:

y=−1

4cos(x)−1

4sin(x) + C

Question 29

Question

Find the general solution to the first-order differential equation:

dx + 2xy = 1.

Solution

Step 1: This is a first-order linear differential equation of the form dy

dx +P(x)y=

Q(x). We can use an integrating factor to solve this equation.

Step 2: Identify P(x) and Q(x). Here, P(x)=2xand Q(x) = 1.

Step 3: Find the integrating factor µ(x) = eRP(x)dx =eR2x dx =ex2.

Step 4: Multiply both sides of the differential equation by the integrating

factor µ(x):

ex2dy

dx + 2xex2y=ex2.

Step 5: Rewrite the left side of the equation as the derivative of the product

y·ex2using the product rule:

dx(y·ex2) = ex2.

Step 6: Integrate both sides with respect to xto get:

y·ex2=Zex2dx =1

2ex2+C,

where Cis the constant of integration.

Step 7: Solve for yby dividing both sides by ex2:

y=1

2+Ce−x2.

Therefore, the general solution to the differential equation dy

dx + 2xy = 1 is

y=1

2+Ce−x2, where Cis an arbitrary constant.

Question 30

Question

Consider the first-order differential equation:

dx =2x−y

x+ 2y

Find the general solution to this differential equation.

Solution

Step 1: Rewrite the differential equation in a more familiar form by multiplying

both sides by (x+ 2y) to get rid of the denominator.

(x+ 2y)dy

dx = 2x−y

Step 2: Rearrange the equation to isolate dy

dx .

(x+ 2y)dy = (2x−y)dx

Step 3: Integrate both sides with respect to yand xrespectively.

Z(x+ 2y)dy =Z(2x−y)dx

Step 4: Perform the integrations.

xy +y2=x2−xy +C

Step 5: Rearrange the equation into a standard form.

x2+ 2xy +y2=C

Step 6: Recognize that the left side is a perfect square.

(x+y)2=C

Step 7: Take the square root of both sides.

x+y=±√C

Step 8: Let C=k2where kis a constant.

x+y=±k

Step 9: Solve for y.

y=−x±k

Therefore, the general solution to the differential equation is y=−x±k,

where kis a constant.

MATH 334 - DIFFERENTIAL EQUATIONS First-Order Differential Equations Question Bank Set 3

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