In the previous section, we saw linear differential equations and the method to solve them. We saw some solved examples also. In this section, we will see a few more solved examples.
Solved example 25.62
Find the general solution of the differential equation $\boldsymbol{\frac{dy}{dx}~-~y~=~\cos x}$.
Solution:
1. The given differential equation is:
$\small{\frac{dy}{dx}~-~y~=~\cos x}$
•
Here P = −1 and Q = cos x
• P and Q must satisfy any one of the following three conditions:
(i) Both P and Q are constants
(ii) Both P and Q are functions of x
(iii) P is a constant and Q is a function of x
• Here, P is a constant and Q is a function of x. So it is a first order linear differential equation.
2. Finding I.F = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta}
1 } &{{}} &{\int{\left[{\rm{P}} \right]dx}} &
{~=~} &{\int{\left[{\rm{-1}} \right]dx}~=~-x} \\
{~\color{magenta}
2 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}}
\right]dx} \right)}} & {~=~} &{e^{\left(-x \right)}} \\
\end{array}}$
3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$
We get: $\small{y\left[e^{-x} \right]~=~\int{\left[\cos x~\times~e^{\left(-x \right)} \right]dx}}$
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{\frac{(\sin x~-~\cos x)e^{-x}}{2}~+~\rm{C}}$
• The reader must write all the steps involved in this integration.
5. Multiply the above result in (4), by the reciprocal of the I.F
$\small{\begin{array}{ll}{~\color{magenta}
1 } &{{}} &{y} & {~=~} &{\left[\frac{(\sin
x~-~\cos x)e^{-x}}{2}~+~\rm{C} \right]\frac{1}{e^{-x}}} \\
{~\color{magenta}
2 } &{{\Rightarrow}} &{y} & {~=~}
&{\frac{(\sin x~-~\cos x)}{2}~+~{\rm{C}}\,e^{x}} \\
\end{array}}$
Solved example 25.63
Find the general solution of the differential equation $\boldsymbol{\frac{dy}{dx}~+~2y~=~\sin x}$.
Solution:
1. The given differential equation is:
$\small{\frac{dy}{dx}~+~2y~=~\sin x}$
• Here P = 2 and Q = sin x
• P and Q must satisfy any one of the following three conditions:
(i) Both P and Q are constants
(ii) Both P and Q are functions of x
(iii) P is a constant and Q is a function of x
• Here, P is a constant and Q is a function of x. So it is a first order linear differential equation.
2. Finding I.F = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\int{\left[{\rm{2}} \right]dx}~=~2x} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}} & {~=~} &{e^{\left(2x \right)}} \\
\end{array}}$
3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$
We get: $\small{y\left[e^{2x} \right]~=~\int{\left[\sin x~\times~e^{\left(2x \right)} \right]dx}}$
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{\frac{(2\sin x~-~\cos x)e^{2x}}{5}~+~\rm{C}}$
• The reader must write all the steps involved in this integration.
5. Multiply the above result in (4), by the reciprocal of the I.F
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\left[\frac{(2\sin x~-~\cos x)e^{2x}}{5}~+~\rm{C} \right]\frac{1}{e^{2x}}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{y} & {~=~} &{\frac{2\sin x~-~\cos x}{5}~+~{\rm{C}}\,e^{-2x}} \\
\end{array}}$
Solved example 25.64
Find the general solution of the differential equation $\boldsymbol{\cos^2 x \frac{dy}{dx}~+~y~=~\tan x~~\left(0\le x < \frac{\pi}{2} \right)}$.
Solution:
1. The given differential equation is:
$\small{\cos^2 x \frac{dy}{dx}~+~y~=~\tan x~~\left(0\le x\le\frac{\pi}{2} \right)}$
Dividing both sides by $\small{\cos^2 x}$,we get:
$\small{\frac{dy}{dx}~+~\left(\sec^2 x \right) \,y~=~\left(\sec^2 x \right)\tan x~~\left(0\le x\le\frac{\pi}{2} \right)}$
• Here P = $\small{\sec^2 x}$ and Q = $\small{\left(\sec^2 x \right)\tan x}$
• P and Q must satisfy any one of the following three conditions:
(i) Both P and Q are constants
(ii) Both P and Q are functions of x
(iii) P is a constant and Q is a function of x
• Here, both P and Q are functions of x. So it is a first order linear differential equation.
2. Finding I.F = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\int{\left[\sec^2 x \right]dx}~=~\tan x} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}} & {~=~} &{e^{\left(\tan x \right)}} \\
\end{array}}$
3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$
We get: $\small{y\left[e^{\tan x} \right]~=~\int{\left[\left(\sec^2 x \right)\tan x~\times~e^{\left(\tan x \right)} \right]dx}}$
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{e^{\left(\tan x \right)}[\tan x~-~1]~+~\rm{C}}$
• The reader must write all the steps involved in this integration.
5. Multiply the above result in (4), by the reciprocal of the I.F
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\left[e^{\left(\tan x \right)}[\tan x~-~1]~+~\rm{C} \right]\frac{1}{e^{\tan x}}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{y} & {~=~} &{\tan x ~-~1~+~{\rm{C}}\,e^{-\tan x}} \\
\end{array}}$
Solved example 25.65
Find the general solution
of the differential equation $\boldsymbol{\frac{dy}{dx}~+~(\sec x)y~=~\tan x~~\left(0\le x < \frac{\pi}{2} \right)}$.
Solution:
1. The given differential equation is:
$\small{\frac{dy}{dx}~+~(\sec x)y~=~\tan x~~\left(0\le x\le\frac{\pi}{2} \right)}$
• Here P = $\small{\sec x}$ and Q = $\small{\tan x}$
• P and Q must satisfy any one of the following three conditions:
(i) Both P and Q are constants
(ii) Both P and Q are functions of x
(iii) P is a constant and Q is a function of x
• Here, both P and Q are functions of x. So it is a first order linear differential equation.
2. Finding I.F = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\int{\left[\sec x \right]dx}~=~\log \left|\sec x~+~\tan x \right|} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}} & {~=~} &{e^{\left(\log \left|\sec x~+~\tan x \right| \right)}~=~\left|\sec x~+~\tan x \right|} \\
\end{array}}$
3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$
We get: $\small{y\left[\sec x~+~\tan x \right]~=~\int{\left[\tan x~\times~(\sec x~+~\tan x) \right]dx}}$
$\small{~=~\int{\left[\sec x\,\tan x~+~\tan^2 x) \right]dx}}$
•
Given that $\small{\left(0\le x\le\frac{\pi}{2} \right)}$. This is the first quadrant. Cosine and tangent are +ve. So we can discard the absolute value symbol.
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{\sec x~+~\tan x~-~x~+~\rm{C}}$
• The reader must write all the steps involved in this integration.
5. Multiply the above result in (4), by the reciprocal of the I.F
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\left[\sec x~+~\tan x~-~x~+~\rm{C} \right]\frac{1}{\sec x +\tan x}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{y(\sec x +\tan x)} & {~=~} &{\sec x~+~\tan x~-~x~+~\rm{C}} \\
\end{array}}$
Solved example 25.66
Find the general solution
of the differential equation $\boldsymbol{x\,\frac{dy}{dx}~+~y~-~x~+~(xy)\cot x~=~0~~\left(x\ne0 \right)}$.
Solution:
1. The given differential equation is:
$\small{x\,\frac{dy}{dx}~+~y~-~x~+~(xy)\cot x~=~0~~\left(x\ne0 \right)}$
Dividing by x and rearranging, we get:
$\small{\frac{dy}{dx}~+~\left(\frac{1}{x} \right)y~-~1~+~(y)\cot x~=~0}$
$\small{\Rightarrow\frac{dy}{dx}~+~\left(\frac{1}{x}~+~\cot x \right)y~-~1~=~0}$
$\small{\Rightarrow \frac{dy}{dx}~+~\left(\frac{1}{x}~+~\cot x \right)y~=~1}$
• Here P = $\small{\left(\frac{1}{x}~+~\cot x \right)}$ and Q = $\small{1}$
• P and Q must satisfy any one of the following three conditions:
(i) Both P and Q are constants
(ii) Both P and Q are functions of x
(iii) P is a constant and Q is a function of x
• Here, both P and Q are functions of x. So it is a first order linear differential equation.
2. Finding I.F = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\int{\left[\frac{1}{x}~+~\cot x \right]dx}~=~\log \left|x \right|+\log \left|\sin x \right|} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\log\left[\left|x \right|~\times~\left|\sin x \right| \right]~=~\log\left[ \left|x\,\sin x \right| \right]} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}} & {~=~} &{e^{\left(\log\left[ x\,\sin x \right] \right)}~=~x\,\sin x } \\
\end{array}}$
3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$
We get: $\small{y\left[x\,\sin x \right]~=~\int{\left[1~\times~x\,\sin x \right]dx}}$
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{\sin\left(x\right) - x \cos\left(x\right)~+~\rm{C}}$
• The reader must write all the steps involved in this integration.
5. Multiply the above result in (4), by the reciprocal of the I.F
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\left[\sin\left(x\right) - x \cos\left(x\right)~+~\rm{C} \right]\frac{1}{x\,\sin x}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{y} & {~=~} &{\frac{1}{x}~-~\cot x~+~\frac{\rm{C}}{x\,\sin x}} \\
\end{array}}$
Solved example 25.67
For the differential equation $\boldsymbol{\frac{dy}{dx}~+~2y\,\tan x~=~\sin x}$; y = 0 when x = $\small{\frac{\pi}{3}}$, find the particular solution satisfying the given conditions.
Solution:
1. The given differential equation is:
$\small{\frac{dy}{dx}~+~2y\,\tan x~=~\sin x}$
• Here P = $\small{2\tan x}$ and Q = $\small{\sin x}$
• P and Q must satisfy any one of the following three conditions:
(i) Both P and Q are constants
(ii) Both P and Q are functions of x
(iii) P is a constant and Q is a function of x
• Here, both P and Q are functions of x. So it is a first order linear differential equation.
2. Finding I.F = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\int{\left[2\tan x \right]dx}~=~(-2)\log \left|\cos x \right|} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\log\left[\sec^2 x \right]} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}} & {~=~} &{e^{\left(\log\left[\sec^2 x \right] \right)}~=~\sec^2 x} \\
\end{array}}$
3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$
We get: $\small{y\left[\sec^2 x \right]~=~\int{\left[\sin x~\times~\sec^2 x \right]dx}}$
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{\sec x~+~\rm{C}}$
• The reader must write all the steps involved in this integration.
5. Multiply the above result in (4), by the reciprocal of the I.F
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\left[\sec x~+~\rm{C} \right]\frac{1}{\sec^2 x}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{y} & {~=~} &{\cos x~+~\rm{C}\,\cos^2x} \\
\end{array}}$
6. Now we can find the particular solution.
• Given that: y = 0 when x = $\small{\frac{\pi}{3}}$
• Substituting in the general solution, we get:
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\cos x~+~\rm{C}\,\cos^2x} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{0} & {~=~} &{\cos \left(\frac{\pi}{3} \right)~+~\rm{C}\,\cos^2\left(\frac{\pi}{3} \right)} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{0} & {~=~} &{\frac{1}{2}~+~\rm{C}\,\left(\frac{1}{4} \right)} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{0} & {~=~} &{2~+~\rm{C}} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{\rm{C}} & {~=~} &{-2} \\
\end{array}}$
• So the particular solution is:
$\small{y~=~\cos x~-~2\cos^2x}$
Solved example 25.68
For the differential
equation $\boldsymbol{\frac{dy}{dx}~-~3y\,\cot x~=~\sin (2x)}$; y = 2 when x
= $\small{\frac{\pi}{2}}$, find the particular solution satisfying the
given conditions.
Solution:
1. The given differential equation is:
$\small{\frac{dy}{dx}~-~3y\,\cot x~=~\sin (2x)}$
• Here P = $\small{-3\cot x}$ and Q = $\small{\sin (2x)}$
• P and Q must satisfy any one of the following three conditions:
(i) Both P and Q are constants
(ii) Both P and Q are functions of x
(iii) P is a constant and Q is a function of x
• Here, both P and Q are functions of x. So it is a first order linear differential equation.
2. Finding I.F = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\int{\left[-3\cot x \right]dx}~=~(-3)\log \left|\sin x \right|} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\log\left[\csc^3 x \right]} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}} & {~=~} &{e^{\left(\log\left[\csc^3 x \right] \right)}~=~\csc^3 x} \\
\end{array}}$
3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$
We get: $\small{y\left[\csc^3 x \right]~=~\int{\left[\sin (2x)~\times~\csc^3 x \right]dx}}$
$\small{\Rightarrow y\left[\csc^3 x \right]~=~\int{\left[\frac{2 \cos x}{\sin^2 x} \right]dx}}$
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{-2\csc x~+~\rm{C}}$
• The reader must write all the steps involved in this integration.
5. Multiply the above result in (4), by the reciprocal of the I.F
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\left[-2\csc x~+~\rm{C} \right]\frac{1}{\csc^3 x}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{y} & {~=~} &{-2\sin^2 x~+~\rm{C}\,\sin^3 x} \\
\end{array}}$
6. Now we can find the particular solution.
• Given that: y = 2 when x = $\small{\frac{\pi}{2}}$
• Substituting in the general solution, we get:
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{y} & {~=~} &{-2\sin^2 x~+~\rm{C}\,\sin^3 x} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{2} & {~=~} &{-2 \sin^2 \left(\frac{\pi}{2} \right)~+~\rm{C}\,\sin^3\left(\frac{\pi}{2} \right)} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{2} & {~=~} &{-2~+~\rm{C}} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\rm{C}} & {~=~} &{4} \\
\end{array}}$
• So the particular solution is:
$\small{y~=~-2\sin^2 x~+~4\sin^3 x}$
Solved example 25.69
Find the general solution
of the differential equation $\boldsymbol{(x+y)\frac{dy}{dx}~=~1}$.
Solution:
1. The given differential equation is:
$\small{(x+y)\frac{dy}{dx}~=~1}$
•
This can be rearranged as: $\small{\frac{dy}{dx}~=~\frac{1}{x+y}}$
•
It is not a linear form. So we rearrange in another way: $\small{\frac{dx}{dy}~=~x+y}$
$\small{\Rightarrow \frac{dx}{dy}~-~x~=~y}$
• Here P1 = $\small{-1}$ and Q1 = $\small{y}$
• P1 and Q1 must satisfy any one of the following three conditions:
(i) Both P1 and Q1 are constants
(ii) Both P1 and Q1 are functions of y
(iii) P1 is a constant and Q1 is a function of y
• Here, P1 is a constant and Q1 is a function of y. So it is a first order linear differential equation.
2. Finding I.F = $\small{e^{\left(\int{\left[{\rm{P}_1} \right]dy} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{\int{\left[{\rm{P}_1} \right]dy}} & {~=~} &{\int{\left[-1 \right]dy}~=~-y} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}_1} \right]dx} \right)}} & {~=~} &{e^{\left(-y \right)} } \\
\end{array}}$
3. Write the equation with independent x:
$\small{x\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q_1(I.F)}} \right]dy}}$
We get: $\small{x\left[e^{\left(-y \right)} \right]~=~\int{\left[y~\times~e^{\left(-y \right)} \right]dy}}$
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{(-1)(y+1)e^{-y}~+~\rm{C}}$
• The reader must write all the steps involved in this integration.
5. Multiply the above result in (4), by the reciprocal of the I.F
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{x} & {~=~} &{\left[(-1)(y+1)e^{-y}~+~\rm{C} \right]\frac{1}{e^{-y}}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{x} & {~=~} &{(-1)(y+1)~+~{\rm{C}}\,e^y} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{x+y+1} & {~=~} &{{\rm{C}}\,e^y} \\
\end{array}}$
Solved example 25.70
Find the general solution
of the differential equation $\boldsymbol{y dx~-~\left(x + 2y^2 \right)dy~=~0}$.
Solution:
1. The given differential equation is:
$\small{y dx~-~\left(x + 2y^2 \right)dy~=~0}$
This can be rearranged as: $\small{\frac{dy}{dx}~=~\frac{y}{x + 2y^2}}$
It is not a linear form. So we rearrange in another way: $\small{\frac{dx}{dy}~=~\frac{x + 2y^2}{y}}$
$\small{\Rightarrow \frac{dx}{dy}~=~\left(\frac{1}{y} \right)x~+~2y}$
$\small{\Rightarrow \frac{dx}{dy}~-~\left(\frac{1}{y} \right)x~=~2y}$
• Here P1 = $\small{\frac{-1}{y}}$ and Q1 = $\small{2y}$
• P1 and Q1 must satisfy any one of the following three conditions:
(i) Both P1 and Q1 are constants
(ii) Both P1 and Q1 are functions of y
(iii) P1 is a constant and Q1 is a function of y
• Here, both P1 and Q1 are functions of y. So it is a first order linear differential equation.
2. Finding I.F = $\small{e^{\left(\int{\left[{\rm{P}_1} \right]dy} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{\int{\left[{\rm{P}_1} \right]dy}} & {~=~} &{\int{\left[\frac{-1}{y} \right]dy}~=~-\log \left|y \right|} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}_1} \right]dx} \right)}} & {~=~} &{e^{\left(-\log \left|y \right| \right)} ~=~\left|y \right|^{-1}~=~\frac{1}{\left|y \right|}} \\
\end{array}}$
3. Write the equation with independent x:
$\small{x\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q_1(I.F)}} \right]dy}}$
We get: $\small{x\left[\frac{1}{y} \right]~=~\int{\left[2y~\times~\frac{1}{y} \right]dy}~=~\int{\left[2 \right]dy}}$
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{2y~+~\rm{C}}$
5. Multiply the above result in (4), by the reciprocal of the I.F
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{x} & {~=~} &{\left[2y~+~\rm{C} \right]y} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{x} & {~=~} &{2y^2~+~{\rm{C}}y} \\
\end{array}}$
In the next section, we will see a few more solved examples.
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