MATH 334 - DIFFERENTIAL
EQUATIONS - First-Order Differential
Equations
Question Bank - Set 2
Liberty University
Question 1
Question
Find the particular solution of the differential equation dy
dx =2x+y
xgiven that
y= 1 when x= 1.
Solution
Step 1: Rewrite the differential equation in the form dy
dx +P(x)y=Q(x), if
possible.
dy
dx =2x+y
x=⇒dy
dx −y
x= 2
Step 2: Identify the integrating factor I(x) = eRP(x)dx.
P(x) = −1
x=⇒I(x) = eR−1
xdx =e−ln |x|=1
|x|
Step 3: Multiply the differential equation by the integrating factor.
1
|x|
dy
dx −1
|x|y= 2 1
|x|
Step 4: Rewrite the left side as the derivative of the product of the integrating
factor and y.
d
dx 1
|x|y= 2 1
|x|
Step 5: Integrate both sides with respect to x.
Zd
dx 1
|x|ydx =Z21
|x|dx
1
|x|y= 2 ln |x|+C
Step 6: Solve for yusing the initial condition y= 1 when x= 1.
1
1|1|= 2 ln |1|+C
1 = C
Step 7: Substitute C= 1 back into the equation to find the particular
solution. 1
|x|y= 2 ln |x|+ 1
y= 2xln |x|+|x|
Question 2
Question
Solve the following first-order differential equation:
dy
dx =y
x+x2
y
Solution
Step 1: Recognize the differential equation is separable and rearrange it as
follows: dy
dx =y
x+x2
y
dy
dx =y2+x3
xy
Step 2: Separate the variables by multiplying both sides by dx and dividing
by the expression involving yon the right side:
y
y2+x3dy =1
xdx
Step 3: Integrate both sides with respect to their independent variables:
Integrating the left side:
Zy
y2+x3dy =Z1
xdx
Z1
y2+x3dy =Z1
xdx
Step 4: Solve the integrals on both sides:
2
Let’s first solve the integral on the left side by using the substitution method.
Let u=y2+x3and du = 2ydy, then the integral becomes:
1
2Z1
udu =Z1
xdx
1
2ln |u|+C1= ln |x|+C2
where C1and C2are constants of integration.
Step 5: Substitute back u=y2+x3into the equation:
1
2ln |y2+x3|= ln |x|+C
ln |y2+x3|= 2 ln |x|+ 2C
ln |y2+x3|= ln |x2|+ lne2C
ln |y2+x3|= ln |x2| · k
where k=e2Cis a new constant.
Step 6: Exponentiate both sides to get rid of logarithms:
|y2+x3|=k|x2|
Step 7: Finally, solve for yby removing the absolute value:
y2+x3=kx2
y2=kx2−x3
y=±pkx2−x3
Question 3
Question
Solve the following first-order differential equation: dy
dx =3x2+2y
2x+3y.
Solution
Step 1: Let’s rewrite the given differential equation in the standard form dy
dx +
P(x)y=Q(x).
We have dy
dx =3x2+2y
2x+3y.
Separating variables, we get:
dy
3x2+ 2y=dx
2x+ 3y
Step 2: Integrate both sides with respect to x.
3
Integrating the left-hand side with respect to yand the right-hand side with
respect to x, we get:
Z1
3x2+ 2ydy =Z1
2x+ 3ydx
Step 3:
Let u= 3x2+ 2yand v= 2x+ 3y. This implies du = 6xdx + 2dy and
dv = 2dx + 3dy.
Therefore, the equation becomes:
1
2Z1
udu =1
3Z1
vdv
Step 4:
Integrating both sides, we get:
1
2ln |u|=1
3ln |v|+C
Simplifying, we have:
ln |u|1/2= ln |v|1/3+C
Step 5:
Taking the exponential of both sides, we get:
u1/2=C·v1/3
Substitute u= 3x2+ 2yand v= 2x+ 3yback in:
(3x2+ 2y)1/2=C·(2x+ 3y)1/3
Thus, the general solution to the differential equation is p3x2+ 2y=C(2x+
3y)1/3where Cis a constant.
Question 4
Question
Find the general solution to the first-order differential equation: dy
dx =3y
x.
Solution
Step 1: This is a first-order separable differential equation, so we can separate
the variables:
dy
y=3dx
x
4
Step 2: Integrate both sides with respect to their variables:
Zdy
y=Z3dx
x
Step 3: Solving the integrals gives:
ln |y|= 3 ln |x|+C1
Step 4: Simplify the logarithmic expression by using the properties of loga-
rithms:
ln |y|= ln |x3|+C1
Step 5: Rewrite the expression using the properties of logarithms:
ln |y|= lneC1x3
Step 6: Eliminate the natural logarithm by exponentiation:
y=eC1x3
Step 7: Let k=eC1where kis a non-zero constant:
y=kx3
Therefore, the general solution to the differential equation is given by y=
kx3, where kis an arbitrary non-zero constant.
Question 5
Question
Solve the following first-order differential equation:
dy
dx =x2+y
Solution
Step 1: Rewrite the equation with all terms involving yon one side:
dy
dx −y=x2
Step 2: This is a linear first-order differential equation in the form of dy
dx +
P(x)y=Q(x). We can find an integrating factor I(x) where I(x) = eRP(x)dx.
In this case, P(x) = −1, so I(x) = eR−1dx =e−x.
Step 3: Multiply the entire equation by the integrating factor e−x:
e−xdy
dx −e−xy=x2e−x
5
Step 4: Recognize the left side as the result of applying the product rule in
reverse: d
dx (e−xy) = x2e−x
Step 5: Integrate both sides with respect to x:
Zd
dx (e−xy)dx =Zx2e−xdx
Step 6: Solve the integrals on both sides:
e−xy=−x2e−x−2xe−x−2e−x+C
where Cis the constant of integration.
Step 7: Multiply by exto isolate y:
y=−x2−2x−2 + Cex
Therefore, the general solution to the differential equation is y=−x2−2x−
2 + Cex, where Cis an arbitrary constant.
Question 6
Question
Solve the first-order differential equation:
y′+ 2xy =x2
Solution
We will solve this first-order differential equation using the integrating factor
method.
Step 1: Write the differential equation in standard form y′+P(x)y=Q(x).
In this case, P(x)=2xand Q(x) = x2, so the equation becomes:
y′+ 2xy =x2
Step 2: Find the integrating factor I(x) = eRP(x)dx. In this case, P(x) =
2x, so:
Z2x dx =x2+C
So the integrating factor is I(x) = ex2.
Step 3: Multiply the integrating factor with the original equation and
rewrite it in the form of a total derivative. We multiply the integrating fac-
tor ex2to the original equation to get:
ex2y′+ 2xex2y=x2ex2
6
This can be written as the derivative of the product ex2ywith respect to x:
d
dx (ex2y) = x2ex2
Step 4: Integrate both sides of the equation. Integrating both sides gives
us: Zd
dx (ex2y)dx =Zx2ex2dx
ex2y=Zx2ex2dx
Step 5: Solve for y. To solve for y, we evaluate the integral on the right
side:
ex2y=Zx2ex2dx
Integrating by parts, with u=x2and dv =ex2dx:
u′= 2x, v =ex2
The integral becomes:
ex2y=x2ex2−Z2xex2dx
ex2y=x2ex2−ex2+C
Solving for ygives:
y=x2−1 + Ce−x2
where Cis an arbitrary constant.
Question 7
Question
Solve the first-order differential equation: dy
dx =x+y
x−y.
Solution
Step 1: Rewrite the differential equation in a different form by separating vari-
ables. dy
dx =x+y
x−y
⇒dy
dx =1 + y
x
1−y
x
7
Step 2: Let u=y
x, then y=ux.
dy
dx =u+xdu
dx
Step 3: Substitute y=ux and dy
dx =u+xdu
dx into the differential equation.
u+xdu
dx =1 + u
1−u
Step 4: Solve for du
dx .
u+xdu
dx =1 + u
1−u
⇒u(1 −u) + xdu
dx (1 −u) = 1 + u
⇒u−u2+xdu
dx −ux du
dx = 1 + u
⇒u−u2+xdu
dx (1 −u) = 1 + u
⇒xdu
dx = 1 + u
⇒du
1 + u=dx
x
Step 5: Integrate both sides.
Zdu
1 + u=Zdx
x
⇒ln |1 + u|= ln |x|+C
⇒1 + u=Cx
⇒1 + y
x=Cx
⇒y= (Cx −1)x
Question 8
Question
Consider the first-order differential equation:
dy
dx + 2y= 4x
Find the general solution for the differential equation.
8
Solution
Step 1: This is a first-order linear differential equation in the form dy
dx +P(x)y=
Q(x). Here, P(x) = 2 and Q(x) = 4x.
Step 2: To solve this type of equation, we can use an integrating factor
µ(x) = eRP(x)dx.
Step 3: Calculate the integrating factor:
µ(x) = eR2dx =e2x
Step 4: Multiply the integrating factor by the original differential equation:
e2xdy
dx + 2e2xy= 4xe2x
Step 5: Notice that the left side is now the result of the product rule:
d
dx (e2xy) = 4xe2x
Step 6: Integrate both sides with respect to x:
Zd
dx (e2xy)dx =Z4xe2xdx
Step 7: Simplify and solve the integrals:
e2xy=Z4xe2xdx = 2xe2x−Z2e2xdx = 2xe2x−e2x+C
Step 8: Solve for y:
y=2xe2x−e2x+C
e2x= 2x−1 + Ce−2x
Step 9: Therefore, the general solution to the differential equation is y=
2x−1 + Ce−2x, where Cis an arbitrary constant.
Question 9
Question
Find the general solution to the first-order differential equation (2y2−3x)dx +
(4xy)dy = 0.
Solution
Step 1: First, we rewrite the given differential equation in a more recognizable
form. We recognize that the equation is not exact, so we will check if it is exact
by computing ∂M
∂y and ∂N
∂x where M= 2y2−3xand N= 4xy. If these partial
9
derivatives are equal, the equation is exact; otherwise, it requires an integrating
factor. ∂M
∂y = 4yand ∂N
∂x = 4y
Since ∂M
∂y =∂N
∂x , the given equation is exact.
Step 2: To find the general solution, we integrate the expression M dx +
Ndy = 0.
Z(2y2−3x)dx +Z(4xy)dy =C
y2x−3x2+C= 0
So, the general solution to the given first-order differential equation is y2x−
3x2+C= 0, where Cis the constant of integration.
Question 10
Question
Solve the following first-order differential equation:
dy
dx =2x+y
x
Solution
Step 1: Rewrite the equation in the standard form of a first-order linear homo-
geneous differential equation: dy
dx −y
x= 2.
Step 2: The integrating factor µ(x) is given by µ(x) = eR−1
xdx =e−ln |x|=
1
|x|.
Step 3: Multiply both sides of the differential equation by the integrating
factor: 1
|x|
dy
dx −1
|x|
y
x=2
|x|
Step 4: Recognize the left-hand side as the derivative of y
|x|with respect to
x:d
dx y
|x|=2
|x|
Step 5: Integrate both sides with respect to x:
Zd
dx y
|x|dx =Z2
|x|dx
Step 6: Simplify the integrals:
y
|x|= 2 ln |x|+C
10
Step 7: Multiply both sides by |x|to solve for y:
y= 2xln |x|+C|x|
Therefore, the general solution to the differential equation is y= 2xln |x|+
C|x|, where Cis an arbitrary constant.
Question 11
Question
Consider the first-order differential equation:
dy
dx =2x2−x
y
Find the general solution to this differential equation.
Solution
Step 1: Rewrite the given differential equation in a separable form. We can
achieve this by multiplying both sides by yand dividing by 2x2−x:
y dy =2x2−x
ydx
Step 2: Integrate both sides of the equation:
Zy dy =Z2x2−x
ydx
Step 3: Solve the integrals on both sides:
1
2y2=Z2x2−x
ydx
Step 4: To solve the integral on the right side, we can rewrite it as:
Z2x2−x
ydx =Z2x2
ydx −Zx
ydx
Step 5: Integrate each term separately:
1
2y2=Z2x2
ydx −Zx
ydx
Step 6: Let’s solve the integrals, starting with R2x2
ydx:
Z2x2
ydx = 2 Zx2
ydx
11
Step 7: Rewrite the integral as:
2Zx2
ydx =Z2x2
ydx =Z2x2
ydy
Step 8: Similarly, solve Rx
ydx:
Zx
ydx =Zx
ydy
Step 9: Substituting these integrals back into the equation and simplifying,
we get:
1
2y2= 2 Zx2
ydy −Zx
ydy
Step 10: At this stage, we have reduced the original differential equation
to an integral form. Proceed with integrating the terms to find the general
solution.
Question 12
Question
Consider the first-order differential equation (4x+ 2y)dx + (1 −2x)dy = 0. Find
the general solution to this differential equation.
Solution
To solve the given first-order differential equation, we can first check if it is
exact by verifying the equality of the mixed partial derivatives of the function
with respect to xand y.
Step 1: Check for Exactness Let M= 4x+ 2yand N= 1 −2x. We
need to find ∂M
∂y and ∂N
∂x .
∂M
∂y = 2 and ∂N
∂x =−2
Since ∂M
∂y =∂N
∂x , the differential equation is not exact and we need to find
an integrating factor.
Step 2: Find the Integrating Factor To find the integrating factor, we
use the formula: µ=eRMy−Nx
Ndx.
My−Nx
N=2−(−2)
1−2x=4
1−2x=−4
2x−1
So, µ=eR−4
2x−1dx =e−4 ln |2x−1|=1
(2x−1)4.
Step 3: Multiply by the Integrating Factor Multiply the given differ-
ential equation by the integrating factor µ.
12
1
(2x−1)4(4x+ 2y)dx +1
(2x−1)4(1 −2x)dy = 0
Simplify to get:
2x+y
(2x−1)3dx −1
(2x−1)4dy = 0
Step 4: Identify Exactness Let P=2x+y
(2x−1)3and Q=−1
(2x−1)4. Check
if ∂P
∂y =∂Q
∂x .
Calculate: ∂P
∂y =1
(2x−1)3=−41
(2x−1)4=∂Q
∂x .
Since ∂P
∂y =∂Q
∂x , the differential equation is exact.
Step 5: Solve the Differential Equation Integrating Pwith respect to
xand setting it equal to a constant C, we get:
Z2x+y
(2x−1)3dx =C
Solving the integral yields:
−1
(2x−1)2+y=C
Finally, we rearrange the equation to obtain the general solution:
y=C+1
(2x−1)2
Question 13
Question
Consider the first-order differential equation:
dy
dx + 2xy =x2
Find the general solution to this differential equation.
Solution
Step 1: This is a first-order linear differential equation in the form dy
dx +P(x)y=
Q(x). We can identify P(x)=2xand Q(x) = x2.
Step 2: To find the integrating factor I(x), we use the formula I(x) =
eRP(x)dx. In this case, I(x) = eR2xdx =ex2.
Step 3: Multiply both sides of the differential equation by the integrating
factor I(x):
ex2dy
dx + 2xex2y=x2ex2
13
Step 4: Rewrite the left-hand side as the derivative of a product:
d
dx (ex2y) = x2ex2
Step 5: Integrate both sides with respect to x:
Zd
dx (ex2y)dx =Zx2ex2dx
Step 6: Simplifying the integrals gives:
ex2y=Zx2ex2dx
Step 7: To solve the integral on the right-hand side, we use a substitution
u=x2,du = 2xdx:
Zx2ex2dx =1
2Zeudu =1
2eu+C=1
2ex2+C
Step 8: Substituting back, we get:
ex2y=1
2ex2+C
Step 9: Solve for yby dividing by ex2:
y=1
2+Ce−x2
Therefore, the general solution to the differential equation is y=1
2+Ce−x2,
where Cis an arbitrary constant.
Question 14
Question
Consider the first-order differential equation:
dy
dx =ycos(x) + sin(x)
Solve the differential equation using an integrating factor.
Solution
Step 1: Rewrite the differential equation in standard form:
dy
dx −ycos(x) = sin(x)
14
Step 2: Identify the integrating factor, denoted by I(x), as eR−cos(x)dx =
esin(x).
Step 3: Multiply both sides of the differential equation by the integrating
factor:
esin(x)dy
dx −esin(x)ycos(x) = esin(x)sin(x)
Step 4: Recognize the left-hand side as the result of the product rule for
derivatives (specifically, the derivative of yesin(x)):
d
dx (yesin(x)) = esin(x)sin(x)
Step 5: Integrate both sides with respect to x:
Zd
dx (yesin(x))dx =Zesin(x)sin(x)dx
yesin(x)=−esin(x)cos(x) + C
Step 6: Solve for yby dividing through by esin(x):
y=−cos(x) + Ce−sin(x)
Therefore, the general solution to the differential equation is y=−cos(x) +
Ce−sin(x), where Cis an arbitrary constant.
Question 15
Question
Find the general solution to the first-order differential equation:
dy
dx =3x2+ 4y
2x
Solution
Step 1: Write the equation in a standard form. Separate the variables by moving
all terms involving yto one side and all terms involving xto the other side.
dy
dx =3x2
2x+4y
2x
dy
dx =3
2x+ 2y
Step 2: Rewrite the equation using Leibniz notation.
dy
2y=3
2x dx
15
Step 3: Integrate both sides of the equation.
Z1
2ydy =Z3
2x dx
1
2ln |y|=3
4x2+C
ln |y|=3
2x2+C1
Step 4: Solve for yby taking the exponential of both sides.
|y|=e3
2x2+C1
|y|=eC1·e3
2x2
|y|=C·e3
2x2
y=±C·e3
2x2
Therefore, the general solution to the differential equation is:
y=±Ce 3
2x2
Question 16
Question
Solve the first-order differential equation:
y′=1
x+1
y
Solution
Step 1: Recognize that this is a separable differential equation, so we can sepa-
rate variables and integrate both sides.
Step 2: Separate variables to get:
dy
dx =1
x+1
y
Step 3: Rearrange the equation to isolate all terms involving ywith dy, and
all terms involving xwith dx:
dy
y=dx
x
Step 4: Integrate both sides:
Zdy
y=Zdx
x
16
Step 5: Evaluate the integrals to get:
ln |y|= ln |x|+C
Step 6: Rewrite the equation with an exponential in order to solve for y:
|y|=eln |x|+C
Step 7: Simplify and remove the absolute value:
y=±eCeln |x|
Step 8: Simplify further to obtain the general solution:
y=±eC·x
Therefore, the general solution to the differential equation y′=1
x+1
yis
y=±eC·x, where Cis an arbitrary constant.
Question 17
Question
Consider the first-order differential equation:
dy
dx =2y−x
x+y
Find the general solution for the differential equation.
Solution
To solve the given first-order differential equation, we will use the method of
separation of variables.
Step 1: Rewrite the equation We rewrite the given differential equation
as:
(x+y)dy = (2y−x)dx
Step 2: Separate the variables We separate the variables by dividing
both sides by (2y−x)(x+y):
1
2y−xdy =1
x+ydx
Step 3: Integrate both sides Integrating both sides, we get:
Z1
2y−xdy =Z1
x+ydx
17
Step 4: Solve the integrals We solve the integrals:
ln |2y−x|= ln |x+y|+C
Where Cis the constant of integration.
Step 5: Eliminate the logarithms Taking the exponential of both sides,
we have:
2y−x=k(x+y)
Where k=eC.
Step 6: Simplify the equation Expanding the right side, we get:
2y−x=kx +ky
2y−ky =x+kx
y(2 −k) = x(1 + k)
y=x(1 + k)
2−k
Therefore, the general solution to the given differential equation is:
y=x(1 + k)
2−k
Question 18
Question
Solve the following first-order differential equation:
y′+y=exsin x
Solution
Step 1: First, we identify the integrating factor, which is given by µ(x) =
eR1dx =ex.
Step 2: We multiply both sides of the differential equation by the integrating
factor:
ex(y′+y) = exexsin x
Step 3: Rearranging the equation, we get:
(yex)′=e2xsin x
Step 4: Integrating both sides with respect to x, we find:
yex=−1
2e2xcos x+C
Step 5: Finally, dividing by exto solve for y, we obtain:
y=−1
2excos x+Ce−x
where Cis the constant of integration.
18
Question 19
Question
Solve the following first-order differential equation:
dy
dx =2xy
x2+ 1
Solution
Step 1: Rewrite the equation in a separable form.
dy
dx =2xy
x2+ 1
dy
y=2x
x2+ 1dx
Step 2: Integrate both sides.
Z1
ydy =Z2x
x2+ 1dx
ln |y|=Z2x
x2+ 1dx
Step 3: Perform the integration on the right side.
ln |y|=Z2x
x2+ 1dx
ln |y|= lnx2+ 1+C
where Cis the constant of integration.
Step 4: Exponentiate both sides to solve for y.
|y|=eln(x2+1)+C
|y|=eln(x2+1)·eC
y=±eC(x2+ 1)
y=C(x2+ 1)
So, the general solution to the differential equation is y=C(x2+ 1), where
Cis an arbitrary constant.
Question 20
Question
Find the particular solution to the differential equation: dy
dx =x2−y2
2xy given that
y= 1 when x= 1.
19
Solution
Step 1: Rewrite the given differential equation in a more convenient form: Let’s
rewrite the given differential equation dy
dx =x2−y2
2xy by separating variables to
simplify the solving process.
2y dy = (x2−y2)dx
Step 2: Integrate both sides of the equation: Integrate both sides of the
equation to solve for y in terms of x.
Z2y dy =Z(x2−y2)dx
Step 3: Integrate the left side:
Z2y dy =y2+C1
Step 4: Integrate the right side: To integrate R(x2−y2)dx, we can simplify
it by expressing yin terms of xusing the initial condition given.
y= 1 when x= 1
Plugging in the initial condition gives us y= 1 when x= 1.
1=12−(1)2
Solving this gives C1= 0. Therefore, the integral of (x2−y2)dx simplifies to
R(x2−1) dx.
Z(x2−1) dx =x3
3−x+C2
Step 5: Set the integrals equal and solve for y: Equating the left and right
side integrals gives:
y2=x3
3−x+C2
Using the initial condition y= 1 when x= 1 to solve for C2gives:
1 = 1
3−1 + C2
Thus, C2=5
3. Therefore, the particular solution to the differential equation is:
y2=x3
3−x+5
3
Question 21
Question
Consider the first-order differential equation: dy
dx =1
x2+y2.
Find the particular solution to the differential equation that passes through
the point (1,0).
20
Solution
Step 1: Rewrite the differential equation in separated variables form:
dy
dx =1
x2+y2
Step 2: Rearrange the terms to separate the variables:
y2dy =dx
Step 3: Integrate both sides to solve for y:
Zy2dy =Zdx
y3
3=x+C
where Cis the constant of integration.
Step 4: To find the particular solution, substitute the initial condition (1,0)
into the equation:
03
3= 1 + C
C=−1
Step 5: Substitute C=−1 back into the equation to find the particular
solution: y3
3=x−1
y3= 3x−3
y=3
√3x−3
Therefore, the particular solution passing through the point (1,0) is y=
3
√3x−3.
Question 22
Question
Find the general solution to the first-order differential equation: dy
dx =x+y+1
x−y−1.
21
Solution
Step 1: Rewrite the given differential equation in a standard form.
dy
dx =x+y+ 1
x−y−1
dy =x+y+ 1
x−y−1dx
Step 2: Separate the variables xand y.
dy =x+y+ 1
x−y−1dx
dy =x+ 1
x−y−1dx +y
x−y−1dx
Step 3: Integrate both sides.
Zdy =Zx+ 1
x−y−1dx +Zy
x−y−1dx
Step 4: For the first integral, substitute u=x−y−1.
Zx+ 1
x−y−1dx =Zu+ 3
udu =Z(1 + 3
u)du
Step 5: Integrate the above expression.
Z(1 + 3
u)du =u+ 3 ln |u|+C
=x−y−1 + 3 ln |x−y−1|+C1
Step 6: For the second integral, substitute v=x−y−1.
Zy
x−y−1dx =Zv+ 1 −y
vdv =Z(1 + 1−y
v)dv
Step 7: Integrate the above expression.
Z(1 + 1−y
v)dv =v+ ln |v| − y+C
=x−y−1 + ln |x−y−1| − y+C2
Step 8: Putting it all together,
y=x−y−1 + 3 ln |x−y−1|+C1+x−y−1 + ln |x−y−1| − y+C2
Step 9: Simplifying the expression gives the general solution:
2y= 2x−2 + 4 ln |x−y−1|+C
22
Question 23
Question
Find the general solution to the following first-order differential equation:
dy
dx =2y+x−1
x2
Solution
Step 1: Rewrite the equation in standard form. To do this, we can multiply
both sides by x2to get:
x2dy
dx = 2y+x−1
Step 2: Rewrite in terms of dy and dx separately by multiplying both sides
by dx:
x2dy = (2y+x−1)dx
Step 3: Rearrange terms:
x2dy −2ydx −xdx =−dx
Step 4: Divide through by x2:
dy
dx −2y
x−1 = −1
x2
Step 5: Rewrite as a homogeneous first-order linear differential equation:
dy
dx −2y
x= 1 −1
x2
Step 6: Find the integrating factor µ(x):
µ(x) = eR−2
xdx =e−2 ln |x|=eln 1
x2=1
x2
Step 7: Multiply both sides of the differential equation by the integrating
factor: 1
x2
dy
dx −2
x3y=1
x2−1
Step 8: Rewrite the equation incorporating the integrating factor:
d
dx y
x2=1
x2−1
Step 9: Integrate both sides:
Zd
dx y
x2dx =Z1
x2−1dx
23
Step 10: Solve the integrals:
y
x2=−1
x−x+C
Step 11: Multiply through by x2to find the general solution:
y=−1−x3+Cx2
where Cis the constant of integration.
Question 24
Question
Find the particular solution to the differential equation y′=ex−ythat satisfies
the initial condition y(0) = 2.
Solution
Step 1: First, let’s rewrite the differential equation in the form dy
dx +y=ex.
Step 2: This is a first-order linear differential equation. To solve it, we can
use an integrating factor µ(x) = eR1dx =ex.
Step 3: Multiply both sides of the differential equation by the integrating
factor:
exdy
dx +exy=e2x
Step 4: Recognize that the left-hand side is the derivative of the product
exy, so we have:
d
dx (exy) = e2x
Step 5: Integrate both sides with respect to x:
Zd
dx (exy)dx =Ze2xdx
Step 6: Solving the integrals gives us:
exy=1
2e2x+C
Step 7: Now, apply the initial condition y(0) = 2 to find the constant:
e0·2 = 1
2e0+C
2 = 1
2+C
C=3
2
24
Step 8: Substitute the value of C back into our equation to find the particular
solution:
exy=1
2e2x+3
2
y=1
2ex+3
2e−x
Therefore, the particular solution to the differential equation with the initial
condition is y=1
2ex+3
2e−x.
Question 25
Question
Solve the following first-order differential equation:
dy
dx =x2−1
y
Solution
Step 1: Rewrite the differential equation in a more convenient form by separating
the variables.
y dy = (x2−1) dx
Step 2: Integrate both sides with respect to their respective variables.
Zy dy =Z(x2−1) dx
Step 3: Simplify the integrals and evaluate them.
1
2y2=1
3x3−x+C
Step 4: Solve for yto obtain the general solution.
y2=2
3x3−2x+ 2C
y=±r2
3x3−2x+ 2C
Therefore, the general solution for the given first-order differential equation
is y=±q2
3x3−2x+ 2C.
25
Question 26
Question
Solve the following first-order differential equation:
dy
dx =2x+y−1
x−2y+ 4
Solution
Step 1: First, let’s rewrite the given differential equation in a more standard
form by multiplying both sides by (x−2y+ 4):
(x−2y+ 4)dy = (2x+y−1)dx
Step 2: Next, we integrate both sides of the equation:
Z(x−2y+ 4)dy =Z(2x+y−1)dx
Step 3: Integrating the left side gives us:
Z(x−2y+ 4)dy =xy −y2+ 4y+C1
Step 4: Integrating the right side gives us:
Z(2x+y−1)dx =x2+1
2y−x+C2
Step 5: So the equation now becomes:
xy −y2+ 4y+C1=x2+1
2y−x+C2
Step 6: Rearranging terms gives us:
y2−xy +1
2y−x+ 4y−x=C2−C1
Step 7: Simplifying further:
y2−3
2y−2x=C
Therefore, the solution to the given first-order differential equation is y2−
3
2y−2x=C, where Cis the constant of integration.
Question 27
Question
Find the general solution to the differential equation: dy
dx =y
x−xln x.
26
Solution
Step 1: Write the differential equation in standard form. Let’s multiply through
by dx to separate variables:
dy
dx =y
x−xln x
dy =y
x−xln xdx
Step 2: Rewrite the equation in a more convenient form. Let’s separate the
variables on the right-hand side and rewrite the equation as:
dy
y=dx
x−ln xdx
Step 3: Integrate both sides of the equation. Integrating the left-hand side
with respect to yand the right-hand side with respect to x, we have:
Z1
ydy =Z1
xdx −Zln xdx
ln |y|= ln |x| − xln x−x+C
Step 4: Clear the natural logarithm on the left-hand side. Exponentiate
both sides to get rid of the natural logarithms:
|y|=eln |x|−xln x−x+C
|y|=eln |x|e−xln xe−xeC
|y|=|x|e−xln xKe−x, where K=eC
Step 5: Consider the absolute value on the right-hand side. We remove
the absolute value by splitting the equation into two cases: y=xKe−xln xor
y=−xKe−xln x. Therefore, the general solution to the differential equation is:
y=±xKe−xln x
Question 28
Question
Consider the first-order differential equation:
y′+ (2x+ 3)y= 4x2+ 6x
Find the general solution of the given differential equation.
27
Solution
Step 1: First, we identify the integrating factor which is given by:
IF (x) = eR(2x+3) dx
Step 2: Compute the integrating factor:
IF (x) = eR(2x+3) dx =ex2+3x
Step 3: Multiply the integrating factor with the given differential equation:
ex2+3x(y′+ (2x+ 3)y) = ex2+3x(4x2+ 6x)
Step 4: Rewrite the left-hand side as the derivative of a product:
(ex2+3xy)′= (4x2+ 6x)ex2+3x
Step 5: Integrate both sides of the equation to find the general solution:
ex2+3xy=Z(4x2+ 6x)ex2+3xdx
Step 6: Solve the integral on the right-hand side:
Z(4x2+ 6x)ex2+3xdx = 2ex2+3x(2x2+ 3x+ 1) + C
Step 7: Multiply by the integrating factor to find the general solution for y:
y=e−(x2+3x)(2x2+ 3x+ 1) + Ce−(x2+3x)
Question 29
Question
Consider the first-order differential equation given by dy
dx =y2−x2
xy . Find the
general solution to this differential equation.
Solution
Step 1: Rewrite the differential equation in a more convenient form. We have
dy
dx =y2−x2
xy . This can be rewritten as 1
y
dy
dx =y
x−x
y.
Step 2: Integrate both sides with respect to x. Integrating the left side
with respect to x, we get:
Z1
y
dy
dx dx =Zy
x−x
ydx.
28
This simplifies to:
ln |y|=y2
2−x2
2+C,
where Cis the constant of integration.
Step 3: Solve for y. Exponentiating both sides, we get:
|y|=ey2
2−x2
2+C.
This further simplifies to:
y=±ey2
2−x2
2+C.
Therefore, the general solution to the differential equation is:
y=±ey2
2−x2
2+C.
Question 30
Question
Find the general solution to the first-order differential equation:
dy
dx =y
x+3x2
y
Solution
Step 1: Recognize that the given differential equation is a Bernoulli equation,
which can be transformed into a linear first-order differential equation by making
a suitable substitution. Let z=y1−n, where nis different from 0 and 1. In this
case, n=−1 will be used:
y=z−1and dy
dx =dz
dx (−z−2)
Step 2: Substitute y=z−1and dy
dx =dz
dx (−z−2) into the given differential
equation:
dz
dx (−z−2) = 1
xz−1+3x2
z−1
Step 3: Simplify the equation by multiplying through by −z2to get rid of
negative exponents:
dz
dx =−z
x−3x2z3
Step 4: This is now a linear first-order ordinary differential equation. We
can solve it using an integrating factor. First, find the integrating factor I(x)
by integrating the function in front of z:
I(x) = eR−1
xdx =e−ln |x|=1
|x|
29
1
|x|y= 2 ln |x|+C
Step 6: Solve for yusing the initial condition y= 1 when x= 1.
1
1|1|= 2 ln |1|+C
1 = C
Step 7: Substitute C= 1 back into the equation to find the particular
solution. 1
|x|y= 2 ln |x|+ 1
y= 2xln |x|+|x|
Question 2
Question
Solve the following first-order differential equation:
dy
dx =y
x+x2
y
Solution
Step 1: Recognize the differential equation is separable and rearrange it as
follows: dy
dx =y
x+x2
y
dy
dx =y2+x3
xy
Step 2: Separate the variables by multiplying both sides by dx and dividing
by the expression involving yon the right side:
y
y2+x3dy =1
xdx
Step 3: Integrate both sides with respect to their independent variables:
Integrating the left side:
Zy
y2+x3dy =Z1
xdx
Z1
y2+x3dy =Z1
xdx
Step 4: Solve the integrals on both sides:
2
Let’s first solve the integral on the left side by using the substitution method.
Let u=y2+x3and du = 2ydy, then the integral becomes:
1
2Z1
udu =Z1
xdx
1
2ln |u|+C1= ln |x|+C2
where C1and C2are constants of integration.
Step 5: Substitute back u=y2+x3into the equation:
1
2ln |y2+x3|= ln |x|+C
ln |y2+x3|= 2 ln |x|+ 2C
ln |y2+x3|= ln |x2|+ lne2C
ln |y2+x3|= ln |x2| · k
where k=e2Cis a new constant.
Step 6: Exponentiate both sides to get rid of logarithms:
|y2+x3|=k|x2|
Step 7: Finally, solve for yby removing the absolute value:
y2+x3=kx2
y2=kx2−x3
y=±pkx2−x3
Question 3
Question
Solve the following first-order differential equation: dy
dx =3x2+2y
2x+3y.
Solution
Step 1: Let’s rewrite the given differential equation in the standard form dy
dx +
P(x)y=Q(x).
We have dy
dx =3x2+2y
2x+3y.
Separating variables, we get:
dy
3x2+ 2y=dx
2x+ 3y
Step 2: Integrate both sides with respect to x.
3
Integrating the left-hand side with respect to yand the right-hand side with
respect to x, we get:
Z1
3x2+ 2ydy =Z1
2x+ 3ydx
Step 3:
Let u= 3x2+ 2yand v= 2x+ 3y. This implies du = 6xdx + 2dy and
dv = 2dx + 3dy.
Therefore, the equation becomes:
1
2Z1
udu =1
3Z1
vdv
Step 4:
Integrating both sides, we get:
1
2ln |u|=1
3ln |v|+C
Simplifying, we have:
ln |u|1/2= ln |v|1/3+C
Step 5:
Taking the exponential of both sides, we get:
u1/2=C·v1/3
Substitute u= 3x2+ 2yand v= 2x+ 3yback in:
(3x2+ 2y)1/2=C·(2x+ 3y)1/3
Thus, the general solution to the differential equation is p3x2+ 2y=C(2x+
3y)1/3where Cis a constant.
Question 4
Question
Find the general solution to the first-order differential equation: dy
dx =3y
x.
Solution
Step 1: This is a first-order separable differential equation, so we can separate
the variables:
dy
y=3dx
x
4
Step 2: Integrate both sides with respect to their variables:
Zdy
y=Z3dx
x
Step 3: Solving the integrals gives:
ln |y|= 3 ln |x|+C1
Step 4: Simplify the logarithmic expression by using the properties of loga-
rithms:
ln |y|= ln |x3|+C1
Step 5: Rewrite the expression using the properties of logarithms:
ln |y|= lneC1x3
Step 6: Eliminate the natural logarithm by exponentiation:
y=eC1x3
Step 7: Let k=eC1where kis a non-zero constant:
y=kx3
Therefore, the general solution to the differential equation is given by y=
kx3, where kis an arbitrary non-zero constant.
Question 5
Question
Solve the following first-order differential equation:
dy
dx =x2+y
Solution
Step 1: Rewrite the equation with all terms involving yon one side:
dy
dx −y=x2
Step 2: This is a linear first-order differential equation in the form of dy
dx +
P(x)y=Q(x). We can find an integrating factor I(x) where I(x) = eRP(x)dx.
In this case, P(x) = −1, so I(x) = eR−1dx =e−x.
Step 3: Multiply the entire equation by the integrating factor e−x:
e−xdy
dx −e−xy=x2e−x
5
Step 4: Recognize the left side as the result of applying the product rule in
reverse: d
dx (e−xy) = x2e−x
Step 5: Integrate both sides with respect to x:
Zd
dx (e−xy)dx =Zx2e−xdx
Step 6: Solve the integrals on both sides:
e−xy=−x2e−x−2xe−x−2e−x+C
where Cis the constant of integration.
Step 7: Multiply by exto isolate y:
y=−x2−2x−2 + Cex
Therefore, the general solution to the differential equation is y=−x2−2x−
2 + Cex, where Cis an arbitrary constant.
Question 6
Question
Solve the first-order differential equation:
y′+ 2xy =x2
Solution
We will solve this first-order differential equation using the integrating factor
method.
Step 1: Write the differential equation in standard form y′+P(x)y=Q(x).
In this case, P(x)=2xand Q(x) = x2, so the equation becomes:
y′+ 2xy =x2
Step 2: Find the integrating factor I(x) = eRP(x)dx. In this case, P(x) =
2x, so:
Z2x dx =x2+C
So the integrating factor is I(x) = ex2.
Step 3: Multiply the integrating factor with the original equation and
rewrite it in the form of a total derivative. We multiply the integrating fac-
tor ex2to the original equation to get:
ex2y′+ 2xex2y=x2ex2
6
This can be written as the derivative of the product ex2ywith respect to x:
d
dx (ex2y) = x2ex2
Step 4: Integrate both sides of the equation. Integrating both sides gives
us: Zd
dx (ex2y)dx =Zx2ex2dx
ex2y=Zx2ex2dx
Step 5: Solve for y. To solve for y, we evaluate the integral on the right
side:
ex2y=Zx2ex2dx
Integrating by parts, with u=x2and dv =ex2dx:
u′= 2x, v =ex2
The integral becomes:
ex2y=x2ex2−Z2xex2dx
ex2y=x2ex2−ex2+C
Solving for ygives:
y=x2−1 + Ce−x2
where Cis an arbitrary constant.
Question 7
Question
Solve the first-order differential equation: dy
dx =x+y
x−y.
Solution
Step 1: Rewrite the differential equation in a different form by separating vari-
ables. dy
dx =x+y
x−y
⇒dy
dx =1 + y
x
1−y
x
7
Step 2: Let u=y
x, then y=ux.
dy
dx =u+xdu
dx
Step 3: Substitute y=ux and dy
dx =u+xdu
dx into the differential equation.
u+xdu
dx =1 + u
1−u
Step 4: Solve for du
dx .
u+xdu
dx =1 + u
1−u
⇒u(1 −u) + xdu
dx (1 −u) = 1 + u
⇒u−u2+xdu
dx −ux du
dx = 1 + u
⇒u−u2+xdu
dx (1 −u) = 1 + u
⇒xdu
dx = 1 + u
⇒du
1 + u=dx
x
Step 5: Integrate both sides.
Zdu
1 + u=Zdx
x
⇒ln |1 + u|= ln |x|+C
⇒1 + u=Cx
⇒1 + y
x=Cx
⇒y= (Cx −1)x
Question 8
Question
Consider the first-order differential equation:
dy
dx + 2y= 4x
Find the general solution for the differential equation.
8
Solution
Step 1: This is a first-order linear differential equation in the form dy
dx +P(x)y=
Q(x). Here, P(x) = 2 and Q(x) = 4x.
Step 2: To solve this type of equation, we can use an integrating factor
µ(x) = eRP(x)dx.
Step 3: Calculate the integrating factor:
µ(x) = eR2dx =e2x
Step 4: Multiply the integrating factor by the original differential equation:
e2xdy
dx + 2e2xy= 4xe2x
Step 5: Notice that the left side is now the result of the product rule:
d
dx (e2xy) = 4xe2x
Step 6: Integrate both sides with respect to x:
Zd
dx (e2xy)dx =Z4xe2xdx
Step 7: Simplify and solve the integrals:
e2xy=Z4xe2xdx = 2xe2x−Z2e2xdx = 2xe2x−e2x+C
Step 8: Solve for y:
y=2xe2x−e2x+C
e2x= 2x−1 + Ce−2x
Step 9: Therefore, the general solution to the differential equation is y=
2x−1 + Ce−2x, where Cis an arbitrary constant.
Question 9
Question
Find the general solution to the first-order differential equation (2y2−3x)dx +
(4xy)dy = 0.
Solution
Step 1: First, we rewrite the given differential equation in a more recognizable
form. We recognize that the equation is not exact, so we will check if it is exact
by computing ∂M
∂y and ∂N
∂x where M= 2y2−3xand N= 4xy. If these partial
9
derivatives are equal, the equation is exact; otherwise, it requires an integrating
factor. ∂M
∂y = 4yand ∂N
∂x = 4y
Since ∂M
∂y =∂N
∂x , the given equation is exact.
Step 2: To find the general solution, we integrate the expression M dx +
Ndy = 0.
Z(2y2−3x)dx +Z(4xy)dy =C
y2x−3x2+C= 0
So, the general solution to the given first-order differential equation is y2x−
3x2+C= 0, where Cis the constant of integration.
Question 10
Question
Solve the following first-order differential equation:
dy
dx =2x+y
x
Solution
Step 1: Rewrite the equation in the standard form of a first-order linear homo-
geneous differential equation: dy
dx −y
x= 2.
Step 2: The integrating factor µ(x) is given by µ(x) = eR−1
xdx =e−ln |x|=
1
|x|.
Step 3: Multiply both sides of the differential equation by the integrating
factor: 1
|x|
dy
dx −1
|x|
y
x=2
|x|
Step 4: Recognize the left-hand side as the derivative of y
|x|with respect to
x:d
dx y
|x|=2
|x|
Step 5: Integrate both sides with respect to x:
Zd
dx y
|x|dx =Z2
|x|dx
Step 6: Simplify the integrals:
y
|x|= 2 ln |x|+C
10
Step 7: Multiply both sides by |x|to solve for y:
y= 2xln |x|+C|x|
Therefore, the general solution to the differential equation is y= 2xln |x|+
C|x|, where Cis an arbitrary constant.
Question 11
Question
Consider the first-order differential equation:
dy
dx =2x2−x
y
Find the general solution to this differential equation.
Solution
Step 1: Rewrite the given differential equation in a separable form. We can
achieve this by multiplying both sides by yand dividing by 2x2−x:
y dy =2x2−x
ydx
Step 2: Integrate both sides of the equation:
Zy dy =Z2x2−x
ydx
Step 3: Solve the integrals on both sides:
1
2y2=Z2x2−x
ydx
Step 4: To solve the integral on the right side, we can rewrite it as:
Z2x2−x
ydx =Z2x2
ydx −Zx
ydx
Step 5: Integrate each term separately:
1
2y2=Z2x2
ydx −Zx
ydx
Step 6: Let’s solve the integrals, starting with R2x2
ydx:
Z2x2
ydx = 2 Zx2
ydx
11
Step 7: Rewrite the integral as:
2Zx2
ydx =Z2x2
ydx =Z2x2
ydy
Step 8: Similarly, solve Rx
ydx:
Zx
ydx =Zx
ydy
Step 9: Substituting these integrals back into the equation and simplifying,
we get:
1
2y2= 2 Zx2
ydy −Zx
ydy
Step 10: At this stage, we have reduced the original differential equation
to an integral form. Proceed with integrating the terms to find the general
solution.
Question 12
Question
Consider the first-order differential equation (4x+ 2y)dx + (1 −2x)dy = 0. Find
the general solution to this differential equation.
Solution
To solve the given first-order differential equation, we can first check if it is
exact by verifying the equality of the mixed partial derivatives of the function
with respect to xand y.
Step 1: Check for Exactness Let M= 4x+ 2yand N= 1 −2x. We
need to find ∂M
∂y and ∂N
∂x .
∂M
∂y = 2 and ∂N
∂x =−2
Since ∂M
∂y =∂N
∂x , the differential equation is not exact and we need to find
an integrating factor.
Step 2: Find the Integrating Factor To find the integrating factor, we
use the formula: µ=eRMy−Nx
Ndx.
My−Nx
N=2−(−2)
1−2x=4
1−2x=−4
2x−1
So, µ=eR−4
2x−1dx =e−4 ln |2x−1|=1
(2x−1)4.
Step 3: Multiply by the Integrating Factor Multiply the given differ-
ential equation by the integrating factor µ.
12
1
(2x−1)4(4x+ 2y)dx +1
(2x−1)4(1 −2x)dy = 0
Simplify to get:
2x+y
(2x−1)3dx −1
(2x−1)4dy = 0
Step 4: Identify Exactness Let P=2x+y
(2x−1)3and Q=−1
(2x−1)4. Check
if ∂P
∂y =∂Q
∂x .
Calculate: ∂P
∂y =1
(2x−1)3=−41
(2x−1)4=∂Q
∂x .
Since ∂P
∂y =∂Q
∂x , the differential equation is exact.
Step 5: Solve the Differential Equation Integrating Pwith respect to
xand setting it equal to a constant C, we get:
Z2x+y
(2x−1)3dx =C
Solving the integral yields:
−1
(2x−1)2+y=C
Finally, we rearrange the equation to obtain the general solution:
y=C+1
(2x−1)2
Question 13
Question
Consider the first-order differential equation:
dy
dx + 2xy =x2
Find the general solution to this differential equation.
Solution
Step 1: This is a first-order linear differential equation in the form dy
dx +P(x)y=
Q(x). We can identify P(x)=2xand Q(x) = x2.
Step 2: To find the integrating factor I(x), we use the formula I(x) =
eRP(x)dx. In this case, I(x) = eR2xdx =ex2.
Step 3: Multiply both sides of the differential equation by the integrating
factor I(x):
ex2dy
dx + 2xex2y=x2ex2
13
Step 4: Rewrite the left-hand side as the derivative of a product:
d
dx (ex2y) = x2ex2
Step 5: Integrate both sides with respect to x:
Zd
dx (ex2y)dx =Zx2ex2dx
Step 6: Simplifying the integrals gives:
ex2y=Zx2ex2dx
Step 7: To solve the integral on the right-hand side, we use a substitution
u=x2,du = 2xdx:
Zx2ex2dx =1
2Zeudu =1
2eu+C=1
2ex2+C
Step 8: Substituting back, we get:
ex2y=1
2ex2+C
Step 9: Solve for yby dividing by ex2:
y=1
2+Ce−x2
Therefore, the general solution to the differential equation is y=1
2+Ce−x2,
where Cis an arbitrary constant.
Question 14
Question
Consider the first-order differential equation:
dy
dx =ycos(x) + sin(x)
Solve the differential equation using an integrating factor.
Solution
Step 1: Rewrite the differential equation in standard form:
dy
dx −ycos(x) = sin(x)
14
Step 2: Identify the integrating factor, denoted by I(x), as eR−cos(x)dx =
esin(x).
Step 3: Multiply both sides of the differential equation by the integrating
factor:
esin(x)dy
dx −esin(x)ycos(x) = esin(x)sin(x)
Step 4: Recognize the left-hand side as the result of the product rule for
derivatives (specifically, the derivative of yesin(x)):
d
dx (yesin(x)) = esin(x)sin(x)
Step 5: Integrate both sides with respect to x:
Zd
dx (yesin(x))dx =Zesin(x)sin(x)dx
yesin(x)=−esin(x)cos(x) + C
Step 6: Solve for yby dividing through by esin(x):
y=−cos(x) + Ce−sin(x)
Therefore, the general solution to the differential equation is y=−cos(x) +
Ce−sin(x), where Cis an arbitrary constant.
Question 15
Question
Find the general solution to the first-order differential equation:
dy
dx =3x2+ 4y
2x
Solution
Step 1: Write the equation in a standard form. Separate the variables by moving
all terms involving yto one side and all terms involving xto the other side.
dy
dx =3x2
2x+4y
2x
dy
dx =3
2x+ 2y
Step 2: Rewrite the equation using Leibniz notation.
dy
2y=3
2x dx
15
Step 3: Integrate both sides of the equation.
Z1
2ydy =Z3
2x dx
1
2ln |y|=3
4x2+C
ln |y|=3
2x2+C1
Step 4: Solve for yby taking the exponential of both sides.
|y|=e3
2x2+C1
|y|=eC1·e3
2x2
|y|=C·e3
2x2
y=±C·e3
2x2
Therefore, the general solution to the differential equation is:
y=±Ce 3
2x2
Question 16
Question
Solve the first-order differential equation:
y′=1
x+1
y
Solution
Step 1: Recognize that this is a separable differential equation, so we can sepa-
rate variables and integrate both sides.
Step 2: Separate variables to get:
dy
dx =1
x+1
y
Step 3: Rearrange the equation to isolate all terms involving ywith dy, and
all terms involving xwith dx:
dy
y=dx
x
Step 4: Integrate both sides:
Zdy
y=Zdx
x
16
Step 5: Evaluate the integrals to get:
ln |y|= ln |x|+C
Step 6: Rewrite the equation with an exponential in order to solve for y:
|y|=eln |x|+C
Step 7: Simplify and remove the absolute value:
y=±eCeln |x|
Step 8: Simplify further to obtain the general solution:
y=±eC·x
Therefore, the general solution to the differential equation y′=1
x+1
yis
y=±eC·x, where Cis an arbitrary constant.
Question 17
Question
Consider the first-order differential equation:
dy
dx =2y−x
x+y
Find the general solution for the differential equation.
Solution
To solve the given first-order differential equation, we will use the method of
separation of variables.
Step 1: Rewrite the equation We rewrite the given differential equation
as:
(x+y)dy = (2y−x)dx
Step 2: Separate the variables We separate the variables by dividing
both sides by (2y−x)(x+y):
1
2y−xdy =1
x+ydx
Step 3: Integrate both sides Integrating both sides, we get:
Z1
2y−xdy =Z1
x+ydx
17
Step 4: Solve the integrals We solve the integrals:
ln |2y−x|= ln |x+y|+C
Where Cis the constant of integration.
Step 5: Eliminate the logarithms Taking the exponential of both sides,
we have:
2y−x=k(x+y)
Where k=eC.
Step 6: Simplify the equation Expanding the right side, we get:
2y−x=kx +ky
2y−ky =x+kx
y(2 −k) = x(1 + k)
y=x(1 + k)
2−k
Therefore, the general solution to the given differential equation is:
y=x(1 + k)
2−k
Question 18
Question
Solve the following first-order differential equation:
y′+y=exsin x
Solution
Step 1: First, we identify the integrating factor, which is given by µ(x) =
eR1dx =ex.
Step 2: We multiply both sides of the differential equation by the integrating
factor:
ex(y′+y) = exexsin x
Step 3: Rearranging the equation, we get:
(yex)′=e2xsin x
Step 4: Integrating both sides with respect to x, we find:
yex=−1
2e2xcos x+C
Step 5: Finally, dividing by exto solve for y, we obtain:
y=−1
2excos x+Ce−x
where Cis the constant of integration.
18
Question 19
Question
Solve the following first-order differential equation:
dy
dx =2xy
x2+ 1
Solution
Step 1: Rewrite the equation in a separable form.
dy
dx =2xy
x2+ 1
dy
y=2x
x2+ 1dx
Step 2: Integrate both sides.
Z1
ydy =Z2x
x2+ 1dx
ln |y|=Z2x
x2+ 1dx
Step 3: Perform the integration on the right side.
ln |y|=Z2x
x2+ 1dx
ln |y|= lnx2+ 1+C
where Cis the constant of integration.
Step 4: Exponentiate both sides to solve for y.
|y|=eln(x2+1)+C
|y|=eln(x2+1)·eC
y=±eC(x2+ 1)
y=C(x2+ 1)
So, the general solution to the differential equation is y=C(x2+ 1), where
Cis an arbitrary constant.
Question 20
Question
Find the particular solution to the differential equation: dy
dx =x2−y2
2xy given that
y= 1 when x= 1.
19
Solution
Step 1: Rewrite the given differential equation in a more convenient form: Let’s
rewrite the given differential equation dy
dx =x2−y2
2xy by separating variables to
simplify the solving process.
2y dy = (x2−y2)dx
Step 2: Integrate both sides of the equation: Integrate both sides of the
equation to solve for y in terms of x.
Z2y dy =Z(x2−y2)dx
Step 3: Integrate the left side:
Z2y dy =y2+C1
Step 4: Integrate the right side: To integrate R(x2−y2)dx, we can simplify
it by expressing yin terms of xusing the initial condition given.
y= 1 when x= 1
Plugging in the initial condition gives us y= 1 when x= 1.
1=12−(1)2
Solving this gives C1= 0. Therefore, the integral of (x2−y2)dx simplifies to
R(x2−1) dx.
Z(x2−1) dx =x3
3−x+C2
Step 5: Set the integrals equal and solve for y: Equating the left and right
side integrals gives:
y2=x3
3−x+C2
Using the initial condition y= 1 when x= 1 to solve for C2gives:
1 = 1
3−1 + C2
Thus, C2=5
3. Therefore, the particular solution to the differential equation is:
y2=x3
3−x+5
3
Question 21
Question
Consider the first-order differential equation: dy
dx =1
x2+y2.
Find the particular solution to the differential equation that passes through
the point (1,0).
20
Solution
Step 1: Rewrite the differential equation in separated variables form:
dy
dx =1
x2+y2
Step 2: Rearrange the terms to separate the variables:
y2dy =dx
Step 3: Integrate both sides to solve for y:
Zy2dy =Zdx
y3
3=x+C
where Cis the constant of integration.
Step 4: To find the particular solution, substitute the initial condition (1,0)
into the equation:
03
3= 1 + C
C=−1
Step 5: Substitute C=−1 back into the equation to find the particular
solution: y3
3=x−1
y3= 3x−3
y=3
√3x−3
Therefore, the particular solution passing through the point (1,0) is y=
3
√3x−3.
Question 22
Question
Find the general solution to the first-order differential equation: dy
dx =x+y+1
x−y−1.
21
Solution
Step 1: Rewrite the given differential equation in a standard form.
dy
dx =x+y+ 1
x−y−1
dy =x+y+ 1
x−y−1dx
Step 2: Separate the variables xand y.
dy =x+y+ 1
x−y−1dx
dy =x+ 1
x−y−1dx +y
x−y−1dx
Step 3: Integrate both sides.
Zdy =Zx+ 1
x−y−1dx +Zy
x−y−1dx
Step 4: For the first integral, substitute u=x−y−1.
Zx+ 1
x−y−1dx =Zu+ 3
udu =Z(1 + 3
u)du
Step 5: Integrate the above expression.
Z(1 + 3
u)du =u+ 3 ln |u|+C
=x−y−1 + 3 ln |x−y−1|+C1
Step 6: For the second integral, substitute v=x−y−1.
Zy
x−y−1dx =Zv+ 1 −y
vdv =Z(1 + 1−y
v)dv
Step 7: Integrate the above expression.
Z(1 + 1−y
v)dv =v+ ln |v| − y+C
=x−y−1 + ln |x−y−1| − y+C2
Step 8: Putting it all together,
y=x−y−1 + 3 ln |x−y−1|+C1+x−y−1 + ln |x−y−1| − y+C2
Step 9: Simplifying the expression gives the general solution:
2y= 2x−2 + 4 ln |x−y−1|+C
22
Question 23
Question
Find the general solution to the following first-order differential equation:
dy
dx =2y+x−1
x2
Solution
Step 1: Rewrite the equation in standard form. To do this, we can multiply
both sides by x2to get:
x2dy
dx = 2y+x−1
Step 2: Rewrite in terms of dy and dx separately by multiplying both sides
by dx:
x2dy = (2y+x−1)dx
Step 3: Rearrange terms:
x2dy −2ydx −xdx =−dx
Step 4: Divide through by x2:
dy
dx −2y
x−1 = −1
x2
Step 5: Rewrite as a homogeneous first-order linear differential equation:
dy
dx −2y
x= 1 −1
x2
Step 6: Find the integrating factor µ(x):
µ(x) = eR−2
xdx =e−2 ln |x|=eln 1
x2=1
x2
Step 7: Multiply both sides of the differential equation by the integrating
factor: 1
x2
dy
dx −2
x3y=1
x2−1
Step 8: Rewrite the equation incorporating the integrating factor:
d
dx y
x2=1
x2−1
Step 9: Integrate both sides:
Zd
dx y
x2dx =Z1
x2−1dx
23
Step 10: Solve the integrals:
y
x2=−1
x−x+C
Step 11: Multiply through by x2to find the general solution:
y=−1−x3+Cx2
where Cis the constant of integration.
Question 24
Question
Find the particular solution to the differential equation y′=ex−ythat satisfies
the initial condition y(0) = 2.
Solution
Step 1: First, let’s rewrite the differential equation in the form dy
dx +y=ex.
Step 2: This is a first-order linear differential equation. To solve it, we can
use an integrating factor µ(x) = eR1dx =ex.
Step 3: Multiply both sides of the differential equation by the integrating
factor:
exdy
dx +exy=e2x
Step 4: Recognize that the left-hand side is the derivative of the product
exy, so we have:
d
dx (exy) = e2x
Step 5: Integrate both sides with respect to x:
Zd
dx (exy)dx =Ze2xdx
Step 6: Solving the integrals gives us:
exy=1
2e2x+C
Step 7: Now, apply the initial condition y(0) = 2 to find the constant:
e0·2 = 1
2e0+C
2 = 1
2+C
C=3
2
24
Step 8: Substitute the value of C back into our equation to find the particular
solution:
exy=1
2e2x+3
2
y=1
2ex+3
2e−x
Therefore, the particular solution to the differential equation with the initial
condition is y=1
2ex+3
2e−x.
Question 25
Question
Solve the following first-order differential equation:
dy
dx =x2−1
y
Solution
Step 1: Rewrite the differential equation in a more convenient form by separating
the variables.
y dy = (x2−1) dx
Step 2: Integrate both sides with respect to their respective variables.
Zy dy =Z(x2−1) dx
Step 3: Simplify the integrals and evaluate them.
1
2y2=1
3x3−x+C
Step 4: Solve for yto obtain the general solution.
y2=2
3x3−2x+ 2C
y=±r2
3x3−2x+ 2C
Therefore, the general solution for the given first-order differential equation
is y=±q2
3x3−2x+ 2C.
25
Question 26
Question
Solve the following first-order differential equation:
dy
dx =2x+y−1
x−2y+ 4
Solution
Step 1: First, let’s rewrite the given differential equation in a more standard
form by multiplying both sides by (x−2y+ 4):
(x−2y+ 4)dy = (2x+y−1)dx
Step 2: Next, we integrate both sides of the equation:
Z(x−2y+ 4)dy =Z(2x+y−1)dx
Step 3: Integrating the left side gives us:
Z(x−2y+ 4)dy =xy −y2+ 4y+C1
Step 4: Integrating the right side gives us:
Z(2x+y−1)dx =x2+1
2y−x+C2
Step 5: So the equation now becomes:
xy −y2+ 4y+C1=x2+1
2y−x+C2
Step 6: Rearranging terms gives us:
y2−xy +1
2y−x+ 4y−x=C2−C1
Step 7: Simplifying further:
y2−3
2y−2x=C
Therefore, the solution to the given first-order differential equation is y2−
3
2y−2x=C, where Cis the constant of integration.
Question 27
Question
Find the general solution to the differential equation: dy
dx =y
x−xln x.
26
Solution
Step 1: Write the differential equation in standard form. Let’s multiply through
by dx to separate variables:
dy
dx =y
x−xln x
dy =y
x−xln xdx
Step 2: Rewrite the equation in a more convenient form. Let’s separate the
variables on the right-hand side and rewrite the equation as:
dy
y=dx
x−ln xdx
Step 3: Integrate both sides of the equation. Integrating the left-hand side
with respect to yand the right-hand side with respect to x, we have:
Z1
ydy =Z1
xdx −Zln xdx
ln |y|= ln |x| − xln x−x+C
Step 4: Clear the natural logarithm on the left-hand side. Exponentiate
both sides to get rid of the natural logarithms:
|y|=eln |x|−xln x−x+C
|y|=eln |x|e−xln xe−xeC
|y|=|x|e−xln xKe−x, where K=eC
Step 5: Consider the absolute value on the right-hand side. We remove
the absolute value by splitting the equation into two cases: y=xKe−xln xor
y=−xKe−xln x. Therefore, the general solution to the differential equation is:
y=±xKe−xln x
Question 28
Question
Consider the first-order differential equation:
y′+ (2x+ 3)y= 4x2+ 6x
Find the general solution of the given differential equation.
27
Solution
Step 1: First, we identify the integrating factor which is given by:
IF (x) = eR(2x+3) dx
Step 2: Compute the integrating factor:
IF (x) = eR(2x+3) dx =ex2+3x
Step 3: Multiply the integrating factor with the given differential equation:
ex2+3x(y′+ (2x+ 3)y) = ex2+3x(4x2+ 6x)
Step 4: Rewrite the left-hand side as the derivative of a product:
(ex2+3xy)′= (4x2+ 6x)ex2+3x
Step 5: Integrate both sides of the equation to find the general solution:
ex2+3xy=Z(4x2+ 6x)ex2+3xdx
Step 6: Solve the integral on the right-hand side:
Z(4x2+ 6x)ex2+3xdx = 2ex2+3x(2x2+ 3x+ 1) + C
Step 7: Multiply by the integrating factor to find the general solution for y:
y=e−(x2+3x)(2x2+ 3x+ 1) + Ce−(x2+3x)
Question 29
Question
Consider the first-order differential equation given by dy
dx =y2−x2
xy . Find the
general solution to this differential equation.
Solution
Step 1: Rewrite the differential equation in a more convenient form. We have
dy
dx =y2−x2
xy . This can be rewritten as 1
y
dy
dx =y
x−x
y.
Step 2: Integrate both sides with respect to x. Integrating the left side
with respect to x, we get:
Z1
y
dy
dx dx =Zy
x−x
ydx.
28
This simplifies to:
ln |y|=y2
2−x2
2+C,
where Cis the constant of integration.
Step 3: Solve for y. Exponentiating both sides, we get:
|y|=ey2
2−x2
2+C.
This further simplifies to:
y=±ey2
2−x2
2+C.
Therefore, the general solution to the differential equation is:
y=±ey2
2−x2
2+C.
Question 30
Question
Find the general solution to the first-order differential equation:
dy
dx =y
x+3x2
y
Solution
Step 1: Recognize that the given differential equation is a Bernoulli equation,
which can be transformed into a linear first-order differential equation by making
a suitable substitution. Let z=y1−n, where nis different from 0 and 1. In this
case, n=−1 will be used:
y=z−1and dy
dx =dz
dx (−z−2)
Step 2: Substitute y=z−1and dy
dx =dz
dx (−z−2) into the given differential
equation:
dz
dx (−z−2) = 1
xz−1+3x2
z−1
Step 3: Simplify the equation by multiplying through by −z2to get rid of
negative exponents:
dz
dx =−z
x−3x2z3
Step 4: This is now a linear first-order ordinary differential equation. We
can solve it using an integrating factor. First, find the integrating factor I(x)
by integrating the function in front of z:
I(x) = eR−1
xdx =e−ln |x|=1
|x|
29
Step 5: Multiply the integrating factor I(x) with the original differential
equation: 1
|x|
dz
dx =−1
xz−3x2z3
Step 6: Rearrange the equation and integrate both sides:
Z1
|x|dz =Z(−z−3x2z3)dx
Step 7: Integrate both sides with respect to their respective variables:
1
|x|z=−1
2z2−3
4x2z4+C
Step 8: Finally, solve for z:
z=4Cx2
4x2+|x|=⇒y=z−1=1
4Cx2
4x2+|x|
So, the general solution to the given differential equation is y=4x2+|x|
4Cx2.
30
