MATH 334 - DIFFERENTIAL
EQUATIONS - First-Order Differential
Equations
Question Bank - Set 1
Liberty University
Question 1
Question
Solve the following first-order ordinary differential equation:
y′= 2xy2
Solution
To solve the differential equation y′= 2xy2, we can separate variables and then
integrate both sides.
Step 1: Separate variables
dy
dx = 2xy2
dy
y2= 2xdx
Step 2: Integrate both sides Integrating the left side with respect to y:
Z1
y2dy =Z2xdx
Zy−2dy =Z2xdx
−1/y =x2+C
where Cis the constant of integration.
Step 3: Solve for yMultiplying by yon both sides:
y=1
−x2−C
Therefore, the solution to the differential equation y′= 2xy2is y=1
−x2−C,
where Cis an arbitrary constant.
Question 2
Question
Find the general solution to the first-order differential equation:
y′+ 2xy =x
Solution
Step 1: This is a first-order linear differential equation in the form y′+P(x)y=
Q(x). To solve this, we will use an integrating factor µ(x) = eRP(x)dx.
Step 2: First, identify P(x)=2xand Q(x) = x.
Step 3: Calculate the integrating factor µ(x) = eR2xdx.
=ex2+C
Step 4: Multiply both sides of the differential equation by the integrating
factor µ(x):
ex2+Cy′+ 2xex2+Cy=xex2+C
Step 5: Recognize that the left side is the result of the product rule on the
left side of the differential equation, i.e., d
dx [ex2+Cy]. Thus, we can rewrite the
equation as: d
dx[ex2+Cy] = xex2+C
Step 6: Integrate both sides with respect to x:
Zd
dx[ex2+Cy]dx =Zxex2+Cdx
ex2+Cy=Zxex2+Cdx +K
Step 7: Solve the integral on the right side with a substitution u=x2+
C, du = 2xdx:
ex2+Cy=1
2Zeudu +K
=1
2eu+K
2
=1
2ex2+C+K
Step 8: Divide by ex2+Cto solve for y:
y=1
2+Ke−x2−C
Therefore, the general solution to the differential equation y′+ 2xy =xis
y=1
2+Ke−x2−C.
Question 3
Question
Find the particular solution to the differential equation dy
dx = 3x2−3 given that
y(0) = 5.
Solution
Step 1: First, integrate the differential equation to find y.
Zdy
dx dx =Z(3x2−3) dx
Step 2: Integrate both sides with respect to x.
y=x3−3x+C
Step 3: Next, use the initial condition y(0) = 5 to find the value of the
constant C.
5=03−3(0) + C
C= 5
Step 4: Substitute C= 5 back into the general solution.
y=x3−3x+ 5
Therefore, the particular solution to the differential equation dy
dx = 3x2−3
with the initial condition y(0) = 5 is y=x3−3x+ 5.
Question 4
Question
Find the general solution to the first-order differential equation:
y′+2x
x2+y= 0
3
Solution
Step 1: We first rewrite the equation in the form of separable variables:
y′=−2x
x2+y
Step 2: Next, we separate the variables by multiplying both sides by dx and
dividing by 2x:dy
dx =−1
2·2x
x2+ydx
dy
dx =−1
2·2x
x2+ydx
dy
dx =−x
x2+ydx
Step 3: Now, we rearrange terms to get all yterms on one side and xterms
on the other side: dy
x2+y=−dx
Step 4: We integrate both sides with respect to their respective variables:
Z1
x2+ydy =Z−dx
Step 5: Integrating the left side using the substitution u=x2+y: Let
u=x2+y; then du = 2xdx or du
2x=dx
Z1
udu =−Zdx
ln |u|=−x+C1
ln |x2+y|=−x+C1
Step 6: Solve for yby taking the exponential of both sides:
x2+y=e−x+C1
x2+y=e−x·eC1
y=e−x·eC1−x2
y=Ae−x−x2
Therefore, the general solution to the differential equation y′+2x
x2+y= 0 is
y=Ae−x−x2, where Ais an arbitrary constant.
4
Question 5
Question
Consider the first-order differential equation:
y′+ycos(x) = 1 + sin(x)
Find the general solution to this differential equation.
Solution
To solve the given first-order differential equation, we can use an integrating
factor. The integrating factor is µ=eRcos(x)dx =esin(x).
Step 1: Find the integrating factor We calculate the integrating factor
µ=eRcos(x)dx =esin(x).
Step 2: Multiply both sides by the integrating factor We multiply
both sides of the original differential equation by the integrating factor µ=
esin(x):
esin(x)(y′+ycos(x)) = esin(x)(1 + sin(x))
Step 3: Rewrite the left side using the product rule Using the product
rule, we rewrite the left side as:
d
dx(esin(x)y) = esin(x)(1 + sin(x))
Step 4: Integrate both sides Integrating both sides of the equation gives:
Zd
dx(esin(x)y)dx =Zesin(x)(1 + sin(x)) dx
By integrating both sides, we have:
esin(x)y=Zesin(x)(1 + sin(x)) dx +C
Step 5: Solve for y Finally, we solve for yby dividing by esin(x):
y=Resin(x)(1 + sin(x)) dx +C
esin(x)
This gives the general solution to the first-order differential equation y′+
ycos(x) = 1 + sin(x).
Question 6
Question
Consider the first-order differential equation:
dy
dx =1 + y
1 + x.
Find the particular solution to the initial condition y(0) = 1.
5
Solution
Step 1: Rewrite the differential equation in separated variables form:
dy
1 + y=dx
1 + x.
Step 2: Integrate both sides of the equation:
Z1
1 + ydy =Z1
1 + xdx.
Step 3: Integrate the left side with respect to y:
ln |1 + y|=Z1
1 + xdx = ln |1 + x|+C1,
where C1is the constant of integration.
Step 4: Solve for yby exponentiating both sides:
|1 + y|=|1 + x|eC1=C(1 + x),
where C=±eC1.
Step 5: Apply the initial condition y(0) = 1:
|1+1|=C(1 + 0) ⇒2 = C⇒C= 2.
Step 6: Determine the particular solution:
1 + y= 2(1 + x)⇒y= 2x+ 1.
Therefore, the particular solution to the initial condition y(0) = 1 is y=
2x+ 1.
Question 7
Question
Solve the first-order differential equation:
dy
dx =2x+y
x−y
Solution
Step 1: Rewrite the given differential equation in a more standard form:
(x−y)dy
dx = 2x+y
Step 2: Let’s make the substitution v=x−y, then dv
dx = 1 −dy
dx .
6
Step 3: Substitute into the differential equation:
vdv
dx = 2x+ (x−v)
Step 4: Simplify the equation:
vdv
dx = 3x−v
Step 5: Separate variables and integrate both sides:
v dv = (3x−v)dx
Step 6: Integrate both sides:
Zv dv =Z(3x−v)dx
v2
2=3x2
2−v2
2+C
Step 7: Rearrange the equation to solve for v:
2v2= 3x2+C
Step 8: Substitute back v=x−y, and rewrite the equation:
2(x−y)2= 3x2+C
Step 9: Expand and simplify the equation:
2x2−4xy + 2y2= 3x2+C
Step 10: Rearrange the equation in standard form:
y=−x2+ 4xy +C
2
Therefore, the solution to the given first-order differential equation is:
y=−x2+ 4xy +C
2
Question 8
Question
Solve the following first-order linear differential equation:
y′+y=e−x
.
7
Solution
To solve the given first-order linear differential equation, we will use an inte-
grating factor method. The standard form of a first-order linear differential
equation is:
y′+P(x)y=Q(x).
Comparing the given equation with the standard form, we have P(x)=1
and Q(x) = e−x.
Step 1: Find the integrating factor I(x) = eRP(x)dx.
I(x) = eR1dx =ex.
Step 2: Multiply the integrating factor with the given differential equation.
exy′+exy=exe−x.
Step 3: Rewrite the left-hand side of the equation as a derivative of a
product. d
dx(exy) = exe−x.
Step 4: Integrate both sides with respect to x.
Zd
dx(exy)dx =Zexe−xdx.
Step 5: Solve the integrals on both sides.
exy=Zedx =e+C,
where Cis the constant of integration.
Step 6: Solve for y.
y=e−x(e+C) = e+Ce−x.
Therefore, the general solution to the given differential equation is y=e+
Ce−x.
Question 9
Question
Find the particular solution to the differential equation: dy
dx = 2y2with the
initial condition y(0) = 1.
8
Solution
Step 1: Write the differential equation in the form dy
dx =f(x, y).
The given differential equation is dy
dx = 2y2.
Step 2: Separate the variables.
Separating variables, we have dy
y2= 2dx.
Step 3: Integrate both sides.
Integrating both sides, we get
Z1
y2dy =Z2dx
=−1
y= 2x+C
where Cis the constant of integration.
Step 4: Use the initial condition to find the value of the constant.
Applying the initial condition y(0) = 1, we have
−1
1= 2(0) + C
C=−1
Step 5: Write the particular solution by substituting the value of the constant
back into the equation.
Substitute C=−1 back into the equation to obtain
−1
y= 2x−1
y=1
1−2x
Therefore, the particular solution to the differential equation dy
dx = 2y2with
the initial condition y(0) = 1 is y=1
1−2x.
Question 10
Question
Find the general solution to the first-order differential equation:
dy
dx =2x+y+ 1
x+ 2y−2
9
Solution
Step 1: Rewrite the equation in the standard form dy/dx +P(x)y=Q(x).
dy
dx =2x+y+ 1
x+ 2y−2
dy
dx =2x+y+ 1
x+ 2y−2−1
dy
dx =2x+y+ 1
x+ 2y−2−x+ 2y−2
x+ 2y−2
dy
dx =x+y+ 3
x+ 2y−2
Step 2: Multiply through by the integrating factor µ(x) = eRP(x)dx.
µ(x) = eR1dx =ex
Step 3: Multiply the differential equation by the integrating factor.
exdy
dx −exy=ex(x+y+ 3)
Step 4: Rewrite the left side as a perfect derivative.
d
dx(exy)=(x+y+ 3)ex
Step 5: Integrate both sides with respect to x.
exy=Z(x+y+ 3)exdx
exy=Zxexdx +Zyexdx +Z3exdx
Step 6: Solve the integrals on the right side.
exy=xex−Zexdx +yex−Zexdx + 3 Zexdx
exy=xex−ex+yex−ex+ 3ex+C
Step 7: Combine like terms and solve for y.
y=x−1 + y+ 3 + 2ex+Ce−x
y−y=x+ 2 + 2ex+Ce−x
y=x+ 2 + 2ex+Ce−x
10
Question 11
Question
Solve the following first-order differential equation:
dy
dx =2x+y
x+ 2y
Solution
Step 1: Rewrite the differential equation in standard form by separating vari-
ables.
(x+ 2y)dy = (2x+y)dx
Step 2: Integrate both sides with respect to their respective variables.
Z(x+ 2y)dy =Z(2x+y)dx
Step 3: Simplify and integrate.
Zx dy + 2 Zy dy = 2 Zx dx +Zy dx
Step 4: Solve the integrals.
xy +y2=x2+C
Step 4: This is the general solution to the differential equation:
xy +y2=x2+C
Question 12
Question
Solve the following first-order differential equation:
dy
dx =y2cos(x)
Solution
Step 1: Rewrite the differential equation in a separable form.
dy
y2= cos(x)dx
Step 2: Integrate both sides of the equation.
Z1
y2dy =Zcos(x)dx
11
Step 3: Solve the integrals.
−1
y= sin(x) + C
where Cis the constant of integration.
Step 4: Solve for y.
y=−1
sin(x) + C
Therefore, the solution to the differential equation is y=−1
sin(x)+C.
Question 13
Question
Solve the following first-order differential equation:
y′+ 2xy =ex2
Solution
Step 1: Identify the integrating factor. The integrating factor for the equation
y′+p(x)y=g(x) is given by I(x) = eRp(x)dx. In this case, p(x) = 2x, so the
integrating factor is I(x) = eR2x dx.
=ex2
Step 2: Multiply both sides of the differential equation by the integrating
factor I(x).
ex2y′+ 2xex2y=e2x2
Step 3: Rewrite the left-hand side as the derivative of a product.
d
dx ex2y=e2x2
Step 4: Integrate both sides with respect to x.
Zd
dx ex2ydx =Ze2x2dx
ex2y=1
2e2x2+C
Step 5: Solve for y.
y=1
2ex2+Ce−x2
12
Question 14
Question
Solve the first-order differential equation:
y′+ 2xy =x2
subject to the initial condition y(0) = 1.
Solution
Step 1: First, we identify the form of the differential equation as linear, where
y′+P(x)y=Q(x). Here, P(x) = 2xand Q(x) = x2.
Step 2: To solve the linear differential equation, we use the integrating factor
µ(x) = eRP(x)dx. In this case, µ(x) = eR2xdx =ex2.
Step 3: We multiply both sides of the original differential equation by the
integrating factor µ(x):
ex2y′+ 2xex2y=x2ex2
Step 4: Notice that the left-hand side is the derivative of the product ex2y:
d
dx(ex2y) = x2ex2
Step 5: Integrating both sides with respect to xgives:
ex2y=Zx2ex2dx +C
Step 6: We calculate the integral on the right-hand side by substituting
u=x2,du = 2xdx:
ex2y=1
2Zex2du +C=1
2ex2+C
Step 7: Solving for y, we have:
y=1
2+Ce−x2
Step 8: To find the constant C, we use the initial condition y(0) = 1:
1 = 1
2+C
C=1
2
Step 9: Therefore, the solution to the differential equation with the initial
condition is:
y=1
2+1
2e−x2
13
Question 15
Question
Find the general solution to the first-order differential equation: dy
dx +y=
exsin(x).
Solution
Step 1: First, we rewrite the given differential equation in the standard form of
a linear first-order ODE, which is dy
dx +P(x)y=Q(x). In this case, P(x) = −1
and Q(x) = exsin(x).
Step 2: We then find the integrating factor, which is given by the formula
eRP(x)dx. In this case, the integrating factor is eR−1dx =e−x.
Step 3: Next, we multiply both sides of the differential equation by the
integrating factor e−x:
e−xdy
dx +e−xy=e−xexsin(x)
Step 4: Simplify the equation to get:
d
dx(e−xy) = sin(x)
Step 5: Integrate both sides with respect to x:
Zd
dx(e−xy)dx =Zsin(x)dx
Step 6: Solve the integrals on both sides:
e−xy=−cos(x) + C
Step 7: Multiply through by exto isolate y:
y=−excos(x) + Cex
Therefore, the general solution to the differential equation is y=−excos(x)+
Cex, where Cis an arbitrary constant.
Question 16
Question
Solve the following first-order differential equation:
dy
dx =2x2+y
x(2y−1)
14
Solution
To solve this first-order differential equation, we will rewrite it in a more suitable
form for separation of variables.
Step 1: Rewrite the equation in terms of dy and dx.
(2y−1)dy = (2x2+y)dx
x
Step 2: Separate the variables by moving all yterms to the left side and
all xterms to the right side.
dy
y+dx
x=2dx
x+dx
x= 2dx
Step 3: Integrate both sides of the equation.
Zdy
y+Zdx
x=Z2dx
Step 4: Solve the integrals to get:
ln |y|+ ln |x|= 2x+C
where Cis the constant of integration.
Step 5: Combine the logarithms and simplify.
ln |xy|= 2x+C
Step 6: Exponentiate both sides to eliminate the natural logarithm.
|xy|=e2x+C
Step 7: Combine the constant term with e2x.
|xy|=Ke2x
where K=eCis the new constant of integration.
Step 8: Solve for yin terms of x.
y=Ke2x
x
This is the general solution to the given first-order differential equation.
Question 17
Question
Solve the following first-order differential equation:
y′+2x
y= 0
15
Solution
Step 1: Rewrite the differential equation in the form y′=f(x, y).
dy
dx +2x
y= 0
Step 2: Rearrange the equation to separate variables.
dy
y=−2xdx
Step 3: Integrate both sides with respect to x.
Z1
ydy =Z−2xdx
Step 4: Solve the integrals.
ln |y|=−x2+C
where Cis the constant of integration.
Step 5: Exponentiate both sides to eliminate the natural logarithm.
y=e−x2+C
Step 6: Simplify the expression using properties of exponents.
y=Ce−x2
where C=eCis the constant of integration.
Therefore, the general solution to the given differential equation is y=
Ce−x2.
Question 18
Question
Solve the first-order differential equation: dy
dx =2x+y−1
x−y.
Solution
Step 1: Rewrite the differential equation in a standard form by multiplying both
sides by (x−y):
(x−y)dy
dx = 2x+y−1
Step 2: Expand the left side using the product rule:
xdy
dx −ydy
dx = 2x+y−1
16
Step 3: Rearrange terms to isolate dy
dx :
xdy
dx −ydy
dx = 2x+y−1
(x−y)dy
dx = 2x+y−1
dy
dx =2x+y−1
x−y
Step 4: Now the differential equation is separable. Let’s rewrite it as follows:
dy
2x+y−1=dx
x−y
Step 5: Integrate both sides with respect to x:
Zdy
2x+y−1=Zdx
x−y
Step 6: Perform the integrals. The left integral can be solved by making the
substitution u= 2x+y−1 and the right integral can be solved by making the
substitution v=x−y.
Step 7: After performing the integrals, we obtain:
ln |2x+y−1|= ln |x−y|+C
Step 8: Exponentiate both sides to eliminate the natural logarithms:
|2x+y−1|=|x−y|eC
Step 9: Further simplify the equation using properties of exponents:
|2x+y−1|=k|x−y|
where kis a positive constant (k=eC).
Step 10: Consider two cases for the absolute value: Case 1: 2x+y−1 =
k(x−y) Case 2: 2x+y−1 = −k(x−y)
Step 11: Solve each case separately by expanding and rearranging to obtain
the general solution.
Question 19
Question
Solve the differential equation (2x+y)dx + (x+ 2y)dy = 0.
17
Solution
Step 1: First, rewrite the given differential equation in the form M(x, y)dx +
N(x, y)dy = 0. In this case, M(x, y)=2x+yand N(x, y) = x+ 2y.
Step 2: Check if the partial derivatives of Mand Nwith respect to yand x
are equal: ∂M
∂y = 1 and ∂N
∂x = 1 which are equal. Therefore, the given differential
equation is exact.
Step 3: To find the solution, we need to find a function ψsuch that ∂ψ
∂x =M
and ∂ψ
∂y =N. Integrating the first equation with respect to xgives ψ=R(2x+
y)dx =x2+xy +f(y), where f(y) is an arbitrary function of y.
Step 4: Now, differentiate ψwith respect to yand set it equal to Nto find
f(y). We have ∂ψ
∂y =x+f′(y) = x+f′(y) = x+ 2y, which implies f′(y)=2y.
Integrating f′(y) gives f(y) = y2+C, where Cis an arbitrary constant.
Step 5: Substitute f(y) back into the expression for ψto get ψ=x2+xy +
y2+C.
Step 6: The solution to the differential equation is given by ψ=C, where
in this case ψ=x2+xy +y2+C.
Therefore, the solution to the differential equation (2x+y)dx+(x+2y)dy = 0
is x2+xy +y2=C.
Question 20
Question
Solve the following first-order differential equation:
y′= (2x+ 3)y2
Solution
Step 1: We start by separating the variables. To do this, we can write the
equation as: dy
dx = (2x+ 3)y2
Step 2: Next, we divide both sides by y2and multiply by dx to separate the
variables: 1
y2dy = (2x+ 3)dx
Step 3: Now, we integrate both sides with respect to their respective vari-
ables. Integrating the left side gives:
Z1
y2dy =Z(2x+ 3)dx
Step 4: Integrating the left side yields:
−1
y=x2+ 3x+C
18
where Cis the constant of integration.
Step 5: To solve for y, we multiply through by -1:
y=−1
x2+ 3x+C
or equivalently:
y=1
x2+ 3x+C
Question 21
Question
Find the general solution to the first-order differential equation:
(x+y2)dx + (2xy + 3y)dy = 0
Solution
We can rewrite the given differential equation in the form:
M(x, y)dx +N(x, y)dy = 0
where M(x, y) = x+y2and N(x, y)=2xy + 3y.
Step 1: Check for exactness by computing the partial derivatives of Mand
N.
∂M
∂y = 2yand ∂N
∂x = 2y
Since ∂M
∂y =∂N
∂x , the equation is exact.
Step 2: Find the general solution by solving the equation ∂u
∂x =Min terms
of xand u.
Integrating M(x, y) with respect to x, we get:
u(x, y) = Z(x+y2)dx =1
2x2+xy2+f(y)
where f(y) is an arbitrary function of y.
Step 3: Differentiate u(x, y) with respect to yand set it equal to N(x, y).
∂u
∂y =x(2y)+2xy +f′(y)=2xy + 3y
Comparing coefficients, we find f′(y) = 3y. Integrating both sides gives
f(y) = 3
2y2+C, where Cis an arbitrary constant.
Step 4: Substitute the expression for u(x, y) back into the equation u(x, y) =
Cto find the general solution.
Therefore, the general solution to the differential equation is:
19
1
2x2+xy2+3
2y2=C
Question 22
Question
Solve the first-order differential equation: dy
dx =y2+ 2x, given that y(0) = 1.
Solution
Step 1: Write the differential equation in the form dy
dx −y2= 2x.
Step 2: This is a first-order linear differential equation which can be solved
using an integrating factor. The integrating factor (IF ) is given by eR−1dx =
e−x.
Step 3: Multiply the integrating factor by both sides of the differential equa-
tion: e−xdy
dx −e−xy2= 2xe−x.
Step 4: Recognize the left side as the derivative of (ye−x): d
dx (ye−x) =
2xe−x.
Step 5: Integrate both sides with respect to x:ye−x=R2xe−xdx.
Step 6: Solve the integral on the right side: ye−x=−2xe−x−R−2e−xdx.
Step 7: Integrate the remaining integral on the right side: ye−x=−2xe−x+
2e−x+C, where Cis the constant of integration.
Step 8: Divide by e−xto solve for y:y=−2x+2+Cex.
Step 9: Apply the initial condition y(0) = 1 to find the value of the constant
C: 1 = −2(0) + 2 + Ce01 = 2 + C C =−1.
Step 10: Substitute the value of Cback into the equation to get the final
solution: y=−2x+ 2 −ex.
Question 23
Question
Find the general solution to the first-order differential equation: dy
dx =3x2
y.
Solution
Step 1: Rewrite the differential equation in separated variables form by multi-
plying both sides by yand dividing by 3x2:
dy
y=3x2
ydx.
Step 2: Integrate both sides:
Z1
ydy =Z3x2dx.
20
Step 3: Solve the integrals:
ln |y|=x3+C1,
where C1is the constant of integration.
Step 4: Eliminate the absolute value by taking eof both sides:
|y|=ex3+C1.
Step 5: Rewrite C1as eC1=C, where Cis a new constant:
|y|=Cex3.
Step 6: Considering the two possible cases of ybeing positive or negative,
we have the general solution:
y=Cex3or y=−Cex3,
where Cis an arbitrary constant.
Question 24
Question
Consider the first-order differential equation:
y′=2x
y−y
x
Find the general solution to this differential equation.
Solution
Step 1: Rewrite the differential equation in a more standard form.
The given differential equation can be rewritten as:
y′=2x
y−y
x
Step 2: Multiply through by yto separate variables.
y dy = 2x dx −y dx
Step 3: Integrate both sides.
Integrating both sides gives:
Zy dy =Z(2x−y)dx
21
Step 4: Evaluate the integrals.
Integrating the left side gives: 1
2y2+C1
Integrating the right side gives:
x2−1
2y2+C2
Step 5: Combine the constants of integration.
Combining the constants C1and C2, we get a new constant C:
1
2y2=x2+C
Step 6: Solve for y.
y2= 2(x2+C)
y=±p2(x2+C)
Therefore, the general solution to the differential equation is:
y=±p2(x2+C)
Question 25
Question
Consider the differential equation dy
dx =y(x2−1). Find the general solution to
this differential equation.
Solution
Step 1: This is a first-order differential equation that is separable. We’ll first
separate variables and then integrate both sides.
dy
y= (x2−1)dx
Step 2: Integrate both sides with respect to their respective variables.
Zdy
y=Z(x2−1)dx
ln |y|=x3
3−x+C
Step 3: Simplify the natural logarithm on the left side using the property of
logarithms.
ln |y|= lnex3
3−x+C
22
ln |y|=x3
3−x+C
Step 4: Exponentiate both sides to eliminate the natural logarithm.
|y|=ex3
3−x+C
Step 5: Remove the absolute value, introducing a constant of integration K.
y=±ex3
3−x+K
Step 6: Finally, we can write the general solution by combining the ±into
a constant A.
y=Aex3
3−x
Thus, the general solution to the differential equation is y=Aex3
3−x.
Question 26
Question
Solve the first-order differential equation: y′=2x+y
x.
Solution
To solve the given first-order differential equation y′=2x+y
x, we will use the
method of integrating factors.
Step 1: Find the integrating factor The integrating factor, denoted by
I(x), is given by I(x) = eRP(x)dx, where P(x) is the coefficient of yin the given
differential equation.
In this case, P(x) = 1
x, so the integrating factor is
I(x) = eR1
xdx =eln |x|=|x|.
Step 2: Multiply the differential equation by the integrating factor
Multiply both sides of the differential equation by the integrating factor |x|:
|x|y′= 2x+y.
Step 3: Rewrite the left-hand side as a derivative Notice that the
left-hand side can be rewritten as the derivative of the product |x|yusing the
product rule: d
dx(|x|y)=2x+y.
Step 4: Integrate both sides Integrate both sides of the equation with
respect to x:
Zd
dx(|x|y)dx =Z(2x+y)dx.
23
This simplifies to:
|x|y=x2+xy +C,
where Cis the constant of integration.
Step 5: Solve for y Solving for y, we get:
y=x+C
|x|.
Therefore, the general solution to the differential equation y′=2x+y
xis
y=x+C
|x|, where Cis an arbitrary constant.
Question 27
Question
Find the general solution to the first-order differential equation: dy
dx =3x+2y+1
2x+y−4.
Solution
To solve this differential equation, we will first rewrite it in a more standard
form and then find the integrating factor.
Step 1: Rewrite the equation
Given differential equation is: dy
dx =3x+2y+1
2x+y−4. We can rewrite it as:
(2x+y−4)dy −(3x+ 2y+ 1)dx = 0
Step 2: Find the integrating factor
Let M= 2x+y−4 and N=−3x−2y−1. The integrating factor µ(x) is given
by:
µ(x) = eRMy−Nx
Ndx =eR−1−(−3)
−3x−2y−1dx
=eR2
3x+2y+1 dx =e2 ln |3x+2y+1|= (3x+ 2y+ 1)2
Step 3: Multiply by integrating factor and simplify
Multiplying the given differential equation by the integrating factor, we get:
(2x+y−4)(3x+ 2y+ 1)2dy −(3x+ 2y+ 1)(3x+ 2y+ 1)2dx = 0
(2x+y−4)(3x+ 2y+ 1)2dy −(3x+ 2y+ 1)3dx = 0
Step 4: Integrate to find general solution
Integrating the equation, we get:
Z(2x+y−4)(3x+ 2y+ 1)2dy −Z(3x+ 2y+ 1)3dx =C
This is the general solution to the given first-order differential equation.
24
Question 28
Question
Consider the first-order differential equation:
dy
dx =3y+ 2x
x−2y
Find the general solution to the differential equation.
Solution
Step 1: First, rewrite the given differential equation in a standard form by
separating variables:
(x−2y)dy = (3y+ 2x)dx
Step 2: Next, integrate both sides of the equation:
Z(x−2y)dy =Z(3y+ 2x)dx
Step 3: Integrate the left side with respect to y and the right side with
respect to x:
xy −2Zydy = 3 Zydx + 2 Zxdx
Step 4: Simplify the integrals:
xy −2y2
2= 3 y2
2+ 2 x2
2
Step 5: Further simplify the equation:
xy −y2=3y2
2+x2
2
Step 6: Rearrange the equation to a standard form:
2xy −2y2= 3y2+x2
Step 7: Rearrange the terms to obtain the general solution:
x2−2xy −5y2= 0
So, the general solution to the differential equation is given by:
x2−2xy −5y2= 0
25
Question 29
Question
Find the general solution to the first-order differential equation:
dy
dx =e2xy2
Solution
Step 1: Rearrange the equation by dividing both sides by y2:
1
y2
dy
dx =e2x
Step 2: Recognize that the left-hand side can be expressed as the derivative
of −1
ywith respect to x:
−d
dx 1
y=e2x
Step 3: Integrate both sides with respect to x:
−Zd
dx 1
ydx =Ze2xdx
Step 4: Integrate the right-hand side:
−1
y=1
2e2x+C
Step 5: Solve for yby taking the reciprocal of both sides:
y=−1
1
2e2x+C
Step 6: Simplify the expression to obtain the general solution:
y=1
1
2e2x+C
Question 30
Question
Find the general solution to the differential equation: dy
dx =y2given that y(0) =
−2.
26
Step 3: Solve for yMultiplying by yon both sides:
y=1
−x2−C
Therefore, the solution to the differential equation y′= 2xy2is y=1
−x2−C,
where Cis an arbitrary constant.
Question 2
Question
Find the general solution to the first-order differential equation:
y′+ 2xy =x
Solution
Step 1: This is a first-order linear differential equation in the form y′+P(x)y=
Q(x). To solve this, we will use an integrating factor µ(x) = eRP(x)dx.
Step 2: First, identify P(x)=2xand Q(x) = x.
Step 3: Calculate the integrating factor µ(x) = eR2xdx.
=ex2+C
Step 4: Multiply both sides of the differential equation by the integrating
factor µ(x):
ex2+Cy′+ 2xex2+Cy=xex2+C
Step 5: Recognize that the left side is the result of the product rule on the
left side of the differential equation, i.e., d
dx [ex2+Cy]. Thus, we can rewrite the
equation as: d
dx[ex2+Cy] = xex2+C
Step 6: Integrate both sides with respect to x:
Zd
dx[ex2+Cy]dx =Zxex2+Cdx
ex2+Cy=Zxex2+Cdx +K
Step 7: Solve the integral on the right side with a substitution u=x2+
C, du = 2xdx:
ex2+Cy=1
2Zeudu +K
=1
2eu+K
2
=1
2ex2+C+K
Step 8: Divide by ex2+Cto solve for y:
y=1
2+Ke−x2−C
Therefore, the general solution to the differential equation y′+ 2xy =xis
y=1
2+Ke−x2−C.
Question 3
Question
Find the particular solution to the differential equation dy
dx = 3x2−3 given that
y(0) = 5.
Solution
Step 1: First, integrate the differential equation to find y.
Zdy
dx dx =Z(3x2−3) dx
Step 2: Integrate both sides with respect to x.
y=x3−3x+C
Step 3: Next, use the initial condition y(0) = 5 to find the value of the
constant C.
5=03−3(0) + C
C= 5
Step 4: Substitute C= 5 back into the general solution.
y=x3−3x+ 5
Therefore, the particular solution to the differential equation dy
dx = 3x2−3
with the initial condition y(0) = 5 is y=x3−3x+ 5.
Question 4
Question
Find the general solution to the first-order differential equation:
y′+2x
x2+y= 0
3
Solution
Step 1: We first rewrite the equation in the form of separable variables:
y′=−2x
x2+y
Step 2: Next, we separate the variables by multiplying both sides by dx and
dividing by 2x:dy
dx =−1
2·2x
x2+ydx
dy
dx =−1
2·2x
x2+ydx
dy
dx =−x
x2+ydx
Step 3: Now, we rearrange terms to get all yterms on one side and xterms
on the other side: dy
x2+y=−dx
Step 4: We integrate both sides with respect to their respective variables:
Z1
x2+ydy =Z−dx
Step 5: Integrating the left side using the substitution u=x2+y: Let
u=x2+y; then du = 2xdx or du
2x=dx
Z1
udu =−Zdx
ln |u|=−x+C1
ln |x2+y|=−x+C1
Step 6: Solve for yby taking the exponential of both sides:
x2+y=e−x+C1
x2+y=e−x·eC1
y=e−x·eC1−x2
y=Ae−x−x2
Therefore, the general solution to the differential equation y′+2x
x2+y= 0 is
y=Ae−x−x2, where Ais an arbitrary constant.
4
Question 5
Question
Consider the first-order differential equation:
y′+ycos(x) = 1 + sin(x)
Find the general solution to this differential equation.
Solution
To solve the given first-order differential equation, we can use an integrating
factor. The integrating factor is µ=eRcos(x)dx =esin(x).
Step 1: Find the integrating factor We calculate the integrating factor
µ=eRcos(x)dx =esin(x).
Step 2: Multiply both sides by the integrating factor We multiply
both sides of the original differential equation by the integrating factor µ=
esin(x):
esin(x)(y′+ycos(x)) = esin(x)(1 + sin(x))
Step 3: Rewrite the left side using the product rule Using the product
rule, we rewrite the left side as:
d
dx(esin(x)y) = esin(x)(1 + sin(x))
Step 4: Integrate both sides Integrating both sides of the equation gives:
Zd
dx(esin(x)y)dx =Zesin(x)(1 + sin(x)) dx
By integrating both sides, we have:
esin(x)y=Zesin(x)(1 + sin(x)) dx +C
Step 5: Solve for y Finally, we solve for yby dividing by esin(x):
y=Resin(x)(1 + sin(x)) dx +C
esin(x)
This gives the general solution to the first-order differential equation y′+
ycos(x) = 1 + sin(x).
Question 6
Question
Consider the first-order differential equation:
dy
dx =1 + y
1 + x.
Find the particular solution to the initial condition y(0) = 1.
5
Solution
Step 1: Rewrite the differential equation in separated variables form:
dy
1 + y=dx
1 + x.
Step 2: Integrate both sides of the equation:
Z1
1 + ydy =Z1
1 + xdx.
Step 3: Integrate the left side with respect to y:
ln |1 + y|=Z1
1 + xdx = ln |1 + x|+C1,
where C1is the constant of integration.
Step 4: Solve for yby exponentiating both sides:
|1 + y|=|1 + x|eC1=C(1 + x),
where C=±eC1.
Step 5: Apply the initial condition y(0) = 1:
|1+1|=C(1 + 0) ⇒2 = C⇒C= 2.
Step 6: Determine the particular solution:
1 + y= 2(1 + x)⇒y= 2x+ 1.
Therefore, the particular solution to the initial condition y(0) = 1 is y=
2x+ 1.
Question 7
Question
Solve the first-order differential equation:
dy
dx =2x+y
x−y
Solution
Step 1: Rewrite the given differential equation in a more standard form:
(x−y)dy
dx = 2x+y
Step 2: Let’s make the substitution v=x−y, then dv
dx = 1 −dy
dx .
6
Step 3: Substitute into the differential equation:
vdv
dx = 2x+ (x−v)
Step 4: Simplify the equation:
vdv
dx = 3x−v
Step 5: Separate variables and integrate both sides:
v dv = (3x−v)dx
Step 6: Integrate both sides:
Zv dv =Z(3x−v)dx
v2
2=3x2
2−v2
2+C
Step 7: Rearrange the equation to solve for v:
2v2= 3x2+C
Step 8: Substitute back v=x−y, and rewrite the equation:
2(x−y)2= 3x2+C
Step 9: Expand and simplify the equation:
2x2−4xy + 2y2= 3x2+C
Step 10: Rearrange the equation in standard form:
y=−x2+ 4xy +C
2
Therefore, the solution to the given first-order differential equation is:
y=−x2+ 4xy +C
2
Question 8
Question
Solve the following first-order linear differential equation:
y′+y=e−x
.
7
Solution
To solve the given first-order linear differential equation, we will use an inte-
grating factor method. The standard form of a first-order linear differential
equation is:
y′+P(x)y=Q(x).
Comparing the given equation with the standard form, we have P(x)=1
and Q(x) = e−x.
Step 1: Find the integrating factor I(x) = eRP(x)dx.
I(x) = eR1dx =ex.
Step 2: Multiply the integrating factor with the given differential equation.
exy′+exy=exe−x.
Step 3: Rewrite the left-hand side of the equation as a derivative of a
product. d
dx(exy) = exe−x.
Step 4: Integrate both sides with respect to x.
Zd
dx(exy)dx =Zexe−xdx.
Step 5: Solve the integrals on both sides.
exy=Zedx =e+C,
where Cis the constant of integration.
Step 6: Solve for y.
y=e−x(e+C) = e+Ce−x.
Therefore, the general solution to the given differential equation is y=e+
Ce−x.
Question 9
Question
Find the particular solution to the differential equation: dy
dx = 2y2with the
initial condition y(0) = 1.
8
Solution
Step 1: Write the differential equation in the form dy
dx =f(x, y).
The given differential equation is dy
dx = 2y2.
Step 2: Separate the variables.
Separating variables, we have dy
y2= 2dx.
Step 3: Integrate both sides.
Integrating both sides, we get
Z1
y2dy =Z2dx
=−1
y= 2x+C
where Cis the constant of integration.
Step 4: Use the initial condition to find the value of the constant.
Applying the initial condition y(0) = 1, we have
−1
1= 2(0) + C
C=−1
Step 5: Write the particular solution by substituting the value of the constant
back into the equation.
Substitute C=−1 back into the equation to obtain
−1
y= 2x−1
y=1
1−2x
Therefore, the particular solution to the differential equation dy
dx = 2y2with
the initial condition y(0) = 1 is y=1
1−2x.
Question 10
Question
Find the general solution to the first-order differential equation:
dy
dx =2x+y+ 1
x+ 2y−2
9
Solution
Step 1: Rewrite the equation in the standard form dy/dx +P(x)y=Q(x).
dy
dx =2x+y+ 1
x+ 2y−2
dy
dx =2x+y+ 1
x+ 2y−2−1
dy
dx =2x+y+ 1
x+ 2y−2−x+ 2y−2
x+ 2y−2
dy
dx =x+y+ 3
x+ 2y−2
Step 2: Multiply through by the integrating factor µ(x) = eRP(x)dx.
µ(x) = eR1dx =ex
Step 3: Multiply the differential equation by the integrating factor.
exdy
dx −exy=ex(x+y+ 3)
Step 4: Rewrite the left side as a perfect derivative.
d
dx(exy)=(x+y+ 3)ex
Step 5: Integrate both sides with respect to x.
exy=Z(x+y+ 3)exdx
exy=Zxexdx +Zyexdx +Z3exdx
Step 6: Solve the integrals on the right side.
exy=xex−Zexdx +yex−Zexdx + 3 Zexdx
exy=xex−ex+yex−ex+ 3ex+C
Step 7: Combine like terms and solve for y.
y=x−1 + y+ 3 + 2ex+Ce−x
y−y=x+ 2 + 2ex+Ce−x
y=x+ 2 + 2ex+Ce−x
10
Question 11
Question
Solve the following first-order differential equation:
dy
dx =2x+y
x+ 2y
Solution
Step 1: Rewrite the differential equation in standard form by separating vari-
ables.
(x+ 2y)dy = (2x+y)dx
Step 2: Integrate both sides with respect to their respective variables.
Z(x+ 2y)dy =Z(2x+y)dx
Step 3: Simplify and integrate.
Zx dy + 2 Zy dy = 2 Zx dx +Zy dx
Step 4: Solve the integrals.
xy +y2=x2+C
Step 4: This is the general solution to the differential equation:
xy +y2=x2+C
Question 12
Question
Solve the following first-order differential equation:
dy
dx =y2cos(x)
Solution
Step 1: Rewrite the differential equation in a separable form.
dy
y2= cos(x)dx
Step 2: Integrate both sides of the equation.
Z1
y2dy =Zcos(x)dx
11
Step 3: Solve the integrals.
−1
y= sin(x) + C
where Cis the constant of integration.
Step 4: Solve for y.
y=−1
sin(x) + C
Therefore, the solution to the differential equation is y=−1
sin(x)+C.
Question 13
Question
Solve the following first-order differential equation:
y′+ 2xy =ex2
Solution
Step 1: Identify the integrating factor. The integrating factor for the equation
y′+p(x)y=g(x) is given by I(x) = eRp(x)dx. In this case, p(x) = 2x, so the
integrating factor is I(x) = eR2x dx.
=ex2
Step 2: Multiply both sides of the differential equation by the integrating
factor I(x).
ex2y′+ 2xex2y=e2x2
Step 3: Rewrite the left-hand side as the derivative of a product.
d
dx ex2y=e2x2
Step 4: Integrate both sides with respect to x.
Zd
dx ex2ydx =Ze2x2dx
ex2y=1
2e2x2+C
Step 5: Solve for y.
y=1
2ex2+Ce−x2
12
Question 14
Question
Solve the first-order differential equation:
y′+ 2xy =x2
subject to the initial condition y(0) = 1.
Solution
Step 1: First, we identify the form of the differential equation as linear, where
y′+P(x)y=Q(x). Here, P(x) = 2xand Q(x) = x2.
Step 2: To solve the linear differential equation, we use the integrating factor
µ(x) = eRP(x)dx. In this case, µ(x) = eR2xdx =ex2.
Step 3: We multiply both sides of the original differential equation by the
integrating factor µ(x):
ex2y′+ 2xex2y=x2ex2
Step 4: Notice that the left-hand side is the derivative of the product ex2y:
d
dx(ex2y) = x2ex2
Step 5: Integrating both sides with respect to xgives:
ex2y=Zx2ex2dx +C
Step 6: We calculate the integral on the right-hand side by substituting
u=x2,du = 2xdx:
ex2y=1
2Zex2du +C=1
2ex2+C
Step 7: Solving for y, we have:
y=1
2+Ce−x2
Step 8: To find the constant C, we use the initial condition y(0) = 1:
1 = 1
2+C
C=1
2
Step 9: Therefore, the solution to the differential equation with the initial
condition is:
y=1
2+1
2e−x2
13
Question 15
Question
Find the general solution to the first-order differential equation: dy
dx +y=
exsin(x).
Solution
Step 1: First, we rewrite the given differential equation in the standard form of
a linear first-order ODE, which is dy
dx +P(x)y=Q(x). In this case, P(x) = −1
and Q(x) = exsin(x).
Step 2: We then find the integrating factor, which is given by the formula
eRP(x)dx. In this case, the integrating factor is eR−1dx =e−x.
Step 3: Next, we multiply both sides of the differential equation by the
integrating factor e−x:
e−xdy
dx +e−xy=e−xexsin(x)
Step 4: Simplify the equation to get:
d
dx(e−xy) = sin(x)
Step 5: Integrate both sides with respect to x:
Zd
dx(e−xy)dx =Zsin(x)dx
Step 6: Solve the integrals on both sides:
e−xy=−cos(x) + C
Step 7: Multiply through by exto isolate y:
y=−excos(x) + Cex
Therefore, the general solution to the differential equation is y=−excos(x)+
Cex, where Cis an arbitrary constant.
Question 16
Question
Solve the following first-order differential equation:
dy
dx =2x2+y
x(2y−1)
14
Solution
To solve this first-order differential equation, we will rewrite it in a more suitable
form for separation of variables.
Step 1: Rewrite the equation in terms of dy and dx.
(2y−1)dy = (2x2+y)dx
x
Step 2: Separate the variables by moving all yterms to the left side and
all xterms to the right side.
dy
y+dx
x=2dx
x+dx
x= 2dx
Step 3: Integrate both sides of the equation.
Zdy
y+Zdx
x=Z2dx
Step 4: Solve the integrals to get:
ln |y|+ ln |x|= 2x+C
where Cis the constant of integration.
Step 5: Combine the logarithms and simplify.
ln |xy|= 2x+C
Step 6: Exponentiate both sides to eliminate the natural logarithm.
|xy|=e2x+C
Step 7: Combine the constant term with e2x.
|xy|=Ke2x
where K=eCis the new constant of integration.
Step 8: Solve for yin terms of x.
y=Ke2x
x
This is the general solution to the given first-order differential equation.
Question 17
Question
Solve the following first-order differential equation:
y′+2x
y= 0
15
Solution
Step 1: Rewrite the differential equation in the form y′=f(x, y).
dy
dx +2x
y= 0
Step 2: Rearrange the equation to separate variables.
dy
y=−2xdx
Step 3: Integrate both sides with respect to x.
Z1
ydy =Z−2xdx
Step 4: Solve the integrals.
ln |y|=−x2+C
where Cis the constant of integration.
Step 5: Exponentiate both sides to eliminate the natural logarithm.
y=e−x2+C
Step 6: Simplify the expression using properties of exponents.
y=Ce−x2
where C=eCis the constant of integration.
Therefore, the general solution to the given differential equation is y=
Ce−x2.
Question 18
Question
Solve the first-order differential equation: dy
dx =2x+y−1
x−y.
Solution
Step 1: Rewrite the differential equation in a standard form by multiplying both
sides by (x−y):
(x−y)dy
dx = 2x+y−1
Step 2: Expand the left side using the product rule:
xdy
dx −ydy
dx = 2x+y−1
16
Step 3: Rearrange terms to isolate dy
dx :
xdy
dx −ydy
dx = 2x+y−1
(x−y)dy
dx = 2x+y−1
dy
dx =2x+y−1
x−y
Step 4: Now the differential equation is separable. Let’s rewrite it as follows:
dy
2x+y−1=dx
x−y
Step 5: Integrate both sides with respect to x:
Zdy
2x+y−1=Zdx
x−y
Step 6: Perform the integrals. The left integral can be solved by making the
substitution u= 2x+y−1 and the right integral can be solved by making the
substitution v=x−y.
Step 7: After performing the integrals, we obtain:
ln |2x+y−1|= ln |x−y|+C
Step 8: Exponentiate both sides to eliminate the natural logarithms:
|2x+y−1|=|x−y|eC
Step 9: Further simplify the equation using properties of exponents:
|2x+y−1|=k|x−y|
where kis a positive constant (k=eC).
Step 10: Consider two cases for the absolute value: Case 1: 2x+y−1 =
k(x−y) Case 2: 2x+y−1 = −k(x−y)
Step 11: Solve each case separately by expanding and rearranging to obtain
the general solution.
Question 19
Question
Solve the differential equation (2x+y)dx + (x+ 2y)dy = 0.
17
Solution
Step 1: First, rewrite the given differential equation in the form M(x, y)dx +
N(x, y)dy = 0. In this case, M(x, y)=2x+yand N(x, y) = x+ 2y.
Step 2: Check if the partial derivatives of Mand Nwith respect to yand x
are equal: ∂M
∂y = 1 and ∂N
∂x = 1 which are equal. Therefore, the given differential
equation is exact.
Step 3: To find the solution, we need to find a function ψsuch that ∂ψ
∂x =M
and ∂ψ
∂y =N. Integrating the first equation with respect to xgives ψ=R(2x+
y)dx =x2+xy +f(y), where f(y) is an arbitrary function of y.
Step 4: Now, differentiate ψwith respect to yand set it equal to Nto find
f(y). We have ∂ψ
∂y =x+f′(y) = x+f′(y) = x+ 2y, which implies f′(y)=2y.
Integrating f′(y) gives f(y) = y2+C, where Cis an arbitrary constant.
Step 5: Substitute f(y) back into the expression for ψto get ψ=x2+xy +
y2+C.
Step 6: The solution to the differential equation is given by ψ=C, where
in this case ψ=x2+xy +y2+C.
Therefore, the solution to the differential equation (2x+y)dx+(x+2y)dy = 0
is x2+xy +y2=C.
Question 20
Question
Solve the following first-order differential equation:
y′= (2x+ 3)y2
Solution
Step 1: We start by separating the variables. To do this, we can write the
equation as: dy
dx = (2x+ 3)y2
Step 2: Next, we divide both sides by y2and multiply by dx to separate the
variables: 1
y2dy = (2x+ 3)dx
Step 3: Now, we integrate both sides with respect to their respective vari-
ables. Integrating the left side gives:
Z1
y2dy =Z(2x+ 3)dx
Step 4: Integrating the left side yields:
−1
y=x2+ 3x+C
18
where Cis the constant of integration.
Step 5: To solve for y, we multiply through by -1:
y=−1
x2+ 3x+C
or equivalently:
y=1
x2+ 3x+C
Question 21
Question
Find the general solution to the first-order differential equation:
(x+y2)dx + (2xy + 3y)dy = 0
Solution
We can rewrite the given differential equation in the form:
M(x, y)dx +N(x, y)dy = 0
where M(x, y) = x+y2and N(x, y)=2xy + 3y.
Step 1: Check for exactness by computing the partial derivatives of Mand
N.
∂M
∂y = 2yand ∂N
∂x = 2y
Since ∂M
∂y =∂N
∂x , the equation is exact.
Step 2: Find the general solution by solving the equation ∂u
∂x =Min terms
of xand u.
Integrating M(x, y) with respect to x, we get:
u(x, y) = Z(x+y2)dx =1
2x2+xy2+f(y)
where f(y) is an arbitrary function of y.
Step 3: Differentiate u(x, y) with respect to yand set it equal to N(x, y).
∂u
∂y =x(2y)+2xy +f′(y)=2xy + 3y
Comparing coefficients, we find f′(y) = 3y. Integrating both sides gives
f(y) = 3
2y2+C, where Cis an arbitrary constant.
Step 4: Substitute the expression for u(x, y) back into the equation u(x, y) =
Cto find the general solution.
Therefore, the general solution to the differential equation is:
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1
2x2+xy2+3
2y2=C
Question 22
Question
Solve the first-order differential equation: dy
dx =y2+ 2x, given that y(0) = 1.
Solution
Step 1: Write the differential equation in the form dy
dx −y2= 2x.
Step 2: This is a first-order linear differential equation which can be solved
using an integrating factor. The integrating factor (IF ) is given by eR−1dx =
e−x.
Step 3: Multiply the integrating factor by both sides of the differential equa-
tion: e−xdy
dx −e−xy2= 2xe−x.
Step 4: Recognize the left side as the derivative of (ye−x): d
dx (ye−x) =
2xe−x.
Step 5: Integrate both sides with respect to x:ye−x=R2xe−xdx.
Step 6: Solve the integral on the right side: ye−x=−2xe−x−R−2e−xdx.
Step 7: Integrate the remaining integral on the right side: ye−x=−2xe−x+
2e−x+C, where Cis the constant of integration.
Step 8: Divide by e−xto solve for y:y=−2x+2+Cex.
Step 9: Apply the initial condition y(0) = 1 to find the value of the constant
C: 1 = −2(0) + 2 + Ce01 = 2 + C C =−1.
Step 10: Substitute the value of Cback into the equation to get the final
solution: y=−2x+ 2 −ex.
Question 23
Question
Find the general solution to the first-order differential equation: dy
dx =3x2
y.
Solution
Step 1: Rewrite the differential equation in separated variables form by multi-
plying both sides by yand dividing by 3x2:
dy
y=3x2
ydx.
Step 2: Integrate both sides:
Z1
ydy =Z3x2dx.
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Step 3: Solve the integrals:
ln |y|=x3+C1,
where C1is the constant of integration.
Step 4: Eliminate the absolute value by taking eof both sides:
|y|=ex3+C1.
Step 5: Rewrite C1as eC1=C, where Cis a new constant:
|y|=Cex3.
Step 6: Considering the two possible cases of ybeing positive or negative,
we have the general solution:
y=Cex3or y=−Cex3,
where Cis an arbitrary constant.
Question 24
Question
Consider the first-order differential equation:
y′=2x
y−y
x
Find the general solution to this differential equation.
Solution
Step 1: Rewrite the differential equation in a more standard form.
The given differential equation can be rewritten as:
y′=2x
y−y
x
Step 2: Multiply through by yto separate variables.
y dy = 2x dx −y dx
Step 3: Integrate both sides.
Integrating both sides gives:
Zy dy =Z(2x−y)dx
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Step 4: Evaluate the integrals.
Integrating the left side gives: 1
2y2+C1
Integrating the right side gives:
x2−1
2y2+C2
Step 5: Combine the constants of integration.
Combining the constants C1and C2, we get a new constant C:
1
2y2=x2+C
Step 6: Solve for y.
y2= 2(x2+C)
y=±p2(x2+C)
Therefore, the general solution to the differential equation is:
y=±p2(x2+C)
Question 25
Question
Consider the differential equation dy
dx =y(x2−1). Find the general solution to
this differential equation.
Solution
Step 1: This is a first-order differential equation that is separable. We’ll first
separate variables and then integrate both sides.
dy
y= (x2−1)dx
Step 2: Integrate both sides with respect to their respective variables.
Zdy
y=Z(x2−1)dx
ln |y|=x3
3−x+C
Step 3: Simplify the natural logarithm on the left side using the property of
logarithms.
ln |y|= lnex3
3−x+C
22
ln |y|=x3
3−x+C
Step 4: Exponentiate both sides to eliminate the natural logarithm.
|y|=ex3
3−x+C
Step 5: Remove the absolute value, introducing a constant of integration K.
y=±ex3
3−x+K
Step 6: Finally, we can write the general solution by combining the ±into
a constant A.
y=Aex3
3−x
Thus, the general solution to the differential equation is y=Aex3
3−x.
Question 26
Question
Solve the first-order differential equation: y′=2x+y
x.
Solution
To solve the given first-order differential equation y′=2x+y
x, we will use the
method of integrating factors.
Step 1: Find the integrating factor The integrating factor, denoted by
I(x), is given by I(x) = eRP(x)dx, where P(x) is the coefficient of yin the given
differential equation.
In this case, P(x) = 1
x, so the integrating factor is
I(x) = eR1
xdx =eln |x|=|x|.
Step 2: Multiply the differential equation by the integrating factor
Multiply both sides of the differential equation by the integrating factor |x|:
|x|y′= 2x+y.
Step 3: Rewrite the left-hand side as a derivative Notice that the
left-hand side can be rewritten as the derivative of the product |x|yusing the
product rule: d
dx(|x|y)=2x+y.
Step 4: Integrate both sides Integrate both sides of the equation with
respect to x:
Zd
dx(|x|y)dx =Z(2x+y)dx.
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This simplifies to:
|x|y=x2+xy +C,
where Cis the constant of integration.
Step 5: Solve for y Solving for y, we get:
y=x+C
|x|.
Therefore, the general solution to the differential equation y′=2x+y
xis
y=x+C
|x|, where Cis an arbitrary constant.
Question 27
Question
Find the general solution to the first-order differential equation: dy
dx =3x+2y+1
2x+y−4.
Solution
To solve this differential equation, we will first rewrite it in a more standard
form and then find the integrating factor.
Step 1: Rewrite the equation
Given differential equation is: dy
dx =3x+2y+1
2x+y−4. We can rewrite it as:
(2x+y−4)dy −(3x+ 2y+ 1)dx = 0
Step 2: Find the integrating factor
Let M= 2x+y−4 and N=−3x−2y−1. The integrating factor µ(x) is given
by:
µ(x) = eRMy−Nx
Ndx =eR−1−(−3)
−3x−2y−1dx
=eR2
3x+2y+1 dx =e2 ln |3x+2y+1|= (3x+ 2y+ 1)2
Step 3: Multiply by integrating factor and simplify
Multiplying the given differential equation by the integrating factor, we get:
(2x+y−4)(3x+ 2y+ 1)2dy −(3x+ 2y+ 1)(3x+ 2y+ 1)2dx = 0
(2x+y−4)(3x+ 2y+ 1)2dy −(3x+ 2y+ 1)3dx = 0
Step 4: Integrate to find general solution
Integrating the equation, we get:
Z(2x+y−4)(3x+ 2y+ 1)2dy −Z(3x+ 2y+ 1)3dx =C
This is the general solution to the given first-order differential equation.
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Question 28
Question
Consider the first-order differential equation:
dy
dx =3y+ 2x
x−2y
Find the general solution to the differential equation.
Solution
Step 1: First, rewrite the given differential equation in a standard form by
separating variables:
(x−2y)dy = (3y+ 2x)dx
Step 2: Next, integrate both sides of the equation:
Z(x−2y)dy =Z(3y+ 2x)dx
Step 3: Integrate the left side with respect to y and the right side with
respect to x:
xy −2Zydy = 3 Zydx + 2 Zxdx
Step 4: Simplify the integrals:
xy −2y2
2= 3 y2
2+ 2 x2
2
Step 5: Further simplify the equation:
xy −y2=3y2
2+x2
2
Step 6: Rearrange the equation to a standard form:
2xy −2y2= 3y2+x2
Step 7: Rearrange the terms to obtain the general solution:
x2−2xy −5y2= 0
So, the general solution to the differential equation is given by:
x2−2xy −5y2= 0
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Question 29
Question
Find the general solution to the first-order differential equation:
dy
dx =e2xy2
Solution
Step 1: Rearrange the equation by dividing both sides by y2:
1
y2
dy
dx =e2x
Step 2: Recognize that the left-hand side can be expressed as the derivative
of −1
ywith respect to x:
−d
dx 1
y=e2x
Step 3: Integrate both sides with respect to x:
−Zd
dx 1
ydx =Ze2xdx
Step 4: Integrate the right-hand side:
−1
y=1
2e2x+C
Step 5: Solve for yby taking the reciprocal of both sides:
y=−1
1
2e2x+C
Step 6: Simplify the expression to obtain the general solution:
y=1
1
2e2x+C
Question 30
Question
Find the general solution to the differential equation: dy
dx =y2given that y(0) =
−2.
26
Solution
Step 1: We will separate variables and integrate both sides. Let’s write the
differential equation in the form dy
y2=dx.
Z1
y2dy =Zdx
Step 2: Integrate both sides.
−1
y=x+C
where Cis the constant of integration.
Step 3: Solve for yby multiplying through by −1.
y=−1
x+C
Step 4: Apply the initial condition y(0) = −2 to find C.
−2 = −1
0 + C
−2 = −1
C
C=1
2
Step 5: Substitute C=1
2back into the general solution.
y=−1
x+1
2
Therefore, the general solution to the differential equation dy
dx =y2with the
initial condition y(0) = −2 is y=−1
x+1
2
.
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