In the previous section, we saw some solved examples on the solution of linear differential equations. In this section, we will see a few more solved examples.
Solved example 25.71
Find the general solution
of the differential equation $\boldsymbol{y dx~+~\left(x - y^2 \right)dy~=~0}$.
Solution:
1. The given differential equation is:
$\small{y dx~+~\left(x - y^2 \right)dy~=~0}$
This can be rearranged as: $\small{\frac{dy}{dx}~=~\frac{y}{y^2 - x}}$
It is not a linear form. So we rearrange in another way: $\small{\frac{dx}{dy}~=~\frac{y^2 - x}{y}}$
$\small{\Rightarrow \frac{dx}{dy}~=~y~-~\left(\frac{1}{y} \right)x}$
$\small{\Rightarrow \frac{dx}{dy}~+~\left(\frac{1}{y} \right)x~=~y}$
• Here P1 = $\small{\frac{1}{y}}$ 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, both P1 and Q1 are functions of y. So it is a first order linear differential equation.
2. Finding I.F = $\small{e^{\left(\int{\left[{\rm{P}_1} \right]dy} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{\int{\left[{\rm{P}_1} \right]dy}} & {~=~} &{\int{\left[\frac{1}{y} \right]dy}~=~\log \left|y \right|} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}_1} \right]dx} \right)}} & {~=~} &{e^{\left(\log \left|y \right| \right)} ~=~\left|y \right|} \\
\end{array}}$
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: $\small{x\left[y \right]~=~\int{\left[y~\times~y \right]dy}~=~\int{\left[y^2 \right]dy}}$
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{\frac{y^3}{3}~+~\rm{C}}$
5. Multiply the above result in (4), by the reciprocal of the I.F
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{x} & {~=~} &{\left[\frac{y^3}{3}~+~\rm{C} \right]\frac{1}{y}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{x} & {~=~} &{\frac{y^2}{3}~+~\frac{{\rm{C}}}{y}} \\
\end{array}}$
Solved example 25.72
Find the general solution of the differential equation $\boldsymbol{\left(x + 3y^2 \right)\frac{dy}{dx}~=~y~(y>0)}$.
Solution:
1. The given differential equation is:
$\small{\left(x + 3y^2 \right)\frac{dy}{dx}~=~y~(y>0)}$
This can be rearranged as: $\small{\frac{dy}{dx}~=~\frac{y}{x + 3y^2}}$
It is not a linear form. So we rearrange in another way: $\small{\frac{dx}{dy}~=~\frac{x + 3y^2}{y}}$
$\small{\Rightarrow \frac{dx}{dy}~=~\frac{x}{y}~+~3y}$
$\small{\Rightarrow \frac{dx}{dy}~-~\left(\frac{1}{y} \right)x~=~3y}$
• Here P1 = $\small{\frac{-1}{y}}$ and Q1 = $\small{3y}$
• 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 = $\small{e^{\left(\int{\left[{\rm{P}_1} \right]dy} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{\int{\left[{\rm{P}_1} \right]dy}} & {~=~} &{\int{\left[\frac{-1}{y} \right]dy}~=~-\log \left|y \right|~=~\log \left|y \right|^{-1}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}_1} \right]dx} \right)}} & {~=~} &{e^{\left(\log \left|y \right| \right)} ~=~\left|y \right|^{-1}~=~\frac{1}{\left|y \right|}} \\
\end{array}}$
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: $\small{x\left[\frac{1}{y} \right]~=~\int{\left[3y~\times~\frac{1}{y} \right]dy}~=~\int{\left[3 \right]dy}}$
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{3y~+~\rm{C}}$
5. Multiply the above result in (4), by the reciprocal of the I.F
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{x} & {~=~} &{\left[3y~+~\rm{C} \right]y} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{x} & {~=~} &{3y^2~+~{\rm{C}}\,y} \\
\end{array}}$
Solved example 25.73
Find the general solution
of the differential equation $\boldsymbol{\left(1 + x^2 \right)dy~+~2xy\,dx~=~\cot x\,dx~~\left(x\ne0 \right)}$.
Solution:
1. The given differential equation is:
$\small{\left(1 + x^2 \right)dy~+~2xy\,dx~=~\cot x\,dx~~\left(x\ne0 \right)}$
This can be rearranged as shown below:
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{\left(1 + x^2 \right)dy~+~2xy\,dx } & {~=~} &{\cot x\,dx} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\left(1 + x^2 \right)dy} & {~=~} &{\left(\cot x~-~2xy \right)\,dx} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\frac{dy}{dx}} & {~=~} &{\frac{\cot x~-~2xy}{1 + x^2}} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\frac{dy}{dx}~+~\left(\frac{2x}{1+x^2} \right)y} & {~=~} &{\frac{\cot x}{1 + x^2}} \\
\end{array}}$
• Here P = $\small{\frac{2x}{1+x^2}}$ and Q = $\small{\frac{\cot x}{1 + x^2}}$
• 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 = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\int{\left[\frac{2x}{1+x^2} \right]dx}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\log\left[\left|1 + x^2 \right| \right]} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}} & {~=~} &{e^{\left(\log\left[ 1 + x^2 \right] \right)}~=~1 + x^2 } \\
\end{array}}$
3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$
We get: $\small{y\left[1 + x^2 \right]~=~\int{\left[\frac{\cot x}{1 + x^2}~\times~1 + x^2 \right]dx}~=~\int{\left[\cot x \right]dx}}$
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{\log \left(\left|\sin x \right| \right)~+~\rm{C}}$
• 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
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\left[\log \left(\left|\sin x \right| \right)~+~\rm{C} \right]\frac{1}{1 + x^2}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{y} & {~=~} &{(1 + x^2)^{-1}\,\log \left(\left|\sin x \right| \right)~+~\rm{C}\,(1 + x^2)^{-1}} \\
\end{array}}$
Solved example 25.74
For the differential equation $\boldsymbol{\frac{dy}{dx}~+~y\,\cot
x~=~2x~+~x^2\cot x~~\left(x\ne0 \right)}$; y = 0 when x = $\small{\frac{\pi}{2}}$, find the
particular solution satisfying the given conditions.
Solution:
1. The given differential equation is:
$\small{\frac{dy}{dx}~+~y\,\cot x~=~2x~+~x^2\cot x~~\left(x\ne0 \right)}$
• Here P = $\small{\cot x}$ and Q = $\small{2x~+~x^2\cot x}$
• 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 = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\int{\left[\cot x \right]dx}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\log\left[\left|\sin x \right| \right]} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}} & {~=~} &{e^{\left(\log\left[\sin x \right] \right)}~=~\sin x } \\
\end{array}}$
3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$
We get: $\small{y\left[\sin x \right]~=~\int{\left[\left(2x~+~x^2\cot x \right)~\times~\sin x \right]dx}}$
$\small{\Rightarrow y\left[\sin x \right]~=~\int{\left[2x \sin x~+~x^2\,\cos x \right]dx}}$
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{2\left(\sin x~-~x\,\cos x \right)~+~\left(x^{2} - 2\right) \sin\left(x\right) + 2x \cos\left(x\right)~+~\rm{C}}$
$\small{\Rightarrow 2 \sin x - 2x \,\cos x+ x^2 \sin x - 2 \sin x + 2x \cos x~+~\rm{C}}$
$\small{\Rightarrow x^2 \sin x ~+~\rm{C}}$
• 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
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\left[x^2 \sin x~+~\rm{C} \right]\frac{1}{\sin x}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{y \sin x} & {~=~} &{x^2 \sin x~+~\rm{C}} \\
\end{array}}$
6. Now we can find the particular solution.
• Given that: y = 0 when x = $\small{\frac{\pi}{2}}$
• Substituting in the general solution, we get:
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{y \sin x} & {~=~} &{x^2 \sin x~+~\rm{C}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{0} & {~=~} &{\left(\frac{\pi^2}{4} \right)\sin \left(\frac{\pi}{2} \right)~+~\rm{C}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{0} & {~=~} &{\left(\frac{\pi^2}{4} \right) \left(1 \right)~+~\rm{C}} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\rm{C}} & {~=~} &{(-1)\frac{\pi^2}{4}} \\
\end{array}}$
• So the particular solution is:
$\small{y \sin x ~=~x^2\,\sin x~-~\frac{\pi^2}{4}}$
Solved example 25.75
For the differential equation $\boldsymbol{\frac{dy}{dx}~+~y\,\cot
x~=~4x\,\csc x~~\left(x\ne0 \right)}$; y = 0 when x = $\small{\frac{\pi}{2}}$, find the
particular solution satisfying the given conditions.
Solution:
1. The given differential equation is:
$\small{\frac{dy}{dx}~+~y\,\cot x~=~4x\,\csc x~~\left(x\ne0 \right)}$
• Here P = $\small{\cot x}$ and Q = $\small{4x\,\csc x}$
• 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 = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\int{\left[\cot x \right]dx}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\log\left[\left|\sin x \right| \right]} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}} & {~=~} &{e^{\left(\log\left[\sin x \right] \right)}~=~\sin x } \\
\end{array}}$
3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$
We get: $\small{y\left[\sin x \right]~=~\int{\left[\left(4x\,\csc x \right)~\times~\sin x \right]dx}}$
$\small{\Rightarrow y\left[\sin x \right]~=~\int{\left[4x \right]dx}}$
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{2x^2~+~\rm{C}}$
• 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
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\left[2x^2~+~\rm{C} \right]\frac{1}{\sin x}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{y \sin x} & {~=~} &{2x^2~+~\rm{C}} \\
\end{array}}$
6. Now we can find the particular solution.
• Given that: y = 0 when x = $\small{\frac{\pi}{2}}$
• Substituting in the general solution, we get:
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{y \sin x} & {~=~} &{2x^2~+~\rm{C}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{0} & {~=~} &{2\left(\frac{\pi^2}{4} \right)~+~\rm{C}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\rm{C}} & {~=~} &{(-1)\frac{\pi^2}{2}} \\
\end{array}}$
• So the particular solution is:
$\small{y \sin x ~=~2x^2~-~\frac{\pi^2}{2}}$
Solved example 25.76
Find the general solution
of the differential equation $\boldsymbol{\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}~-~\frac{y}{\sqrt{x}} \right]\frac{dx}{dy}~=~1~~\left(x\ne0 \right)}$.
Solution:
1. The given differential equation is:
$\small{\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}~-~\frac{y}{\sqrt{x}} \right]\frac{dx}{dy}~=~1~~\left(x\ne0 \right)}$
This can be rearranged as shown below:
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{\frac{dy}{dx}} & {~=~} &{\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}~-~\frac{y}{\sqrt{x}}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\frac{dy}{dx}~+~\frac{y}{\sqrt{x}}} & {~=~} &{\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\frac{dy}{dx}~+~\left(\frac{1}{\sqrt{x}} \right)y} & {~=~} &{\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}} \\
\end{array}}$
• Here P = $\small{\frac{1}{\sqrt{x}}}$ and Q = $\small{\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}}$
• 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 = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\int{\left[\frac{1}{\sqrt{x}} \right]dx}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{2 \sqrt{x}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}} & {~=~} &{e^{\left(2 \sqrt{x} \right)} } \\
\end{array}}$
3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$
We get: $\small{y\left[e^{\left(2 \sqrt{x} \right)} \right]~=~\int{\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}~\times~e^{\left(2 \sqrt{x} \right)} \right]dx}~=~\int{\left[\frac{1}{\sqrt{x}} \right]dx}}$
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{2 \sqrt{x}~+~\rm{C}}$
• 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
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\left[2 \sqrt{x}~+~\rm{C} \right]e^{\left(-2 \sqrt{x} \right)}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{y\,e^{\left(2 \sqrt{x} \right)}} & {~=~} &{2 \sqrt{x}~+~\rm{C}} \\
\end{array}}$
Solved example 25.77
Find the equation of a curve passing through the point (0,1). If the slope of the tangent to the curve at any point (x,y) is equal to the sum of the x coordinate (abscissa) and the product of the x coordinate and y coordinate of that point.
Solution:
1. The slope of the tangent to the curve at any point (x,y) is $\small{\frac{dy}{dx}}$.
• Given that: This slope is equal to the sum of the x coordinate (abscissa) and the product of the x coordinate and y coordinate of that point.
• So we can write: $\small{\frac{dy}{dx}~=~x + xy}$
• This is the differential equation that we need to solve.
This can be rearranged as shown below:
$\small{\frac{dy}{dx}~-~(x)y~=~x}$
• Here P = $\small{-x}$ and Q = $\small{x}$
• 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 = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\int{\left[-x \right]dx}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{(-1)\frac{x^2}{2}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}} & {~=~} &{e^{\left(\frac{-x^2}{2} \right)} } \\
\end{array}}$
3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$
We get: $\small{y\left[e^{\left(\frac{-x^2}{2} \right)} \right]~=~\int{\left[x~\times~e^{\left(\frac{-x^2}{2} \right)} \right]dx}}$
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{-e^{\left(\frac{-x^2}{2} \right)}~+~\rm{C}}$
• 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
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\left[-e^{\left(\frac{-x^2}{2} \right)}~+~\rm{C} \right]\frac{1}{e^{\left(\frac{-x^2}{2} \right)}}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{y} & {~=~} &{\left[-e^{\left(\frac{-x^2}{2} \right)}~+~\rm{C} \right]e^{\left(\frac{x^2}{2} \right)}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{y} & {~=~} &{-1~+~{\rm{C}}\,e^{\left(\frac{x^2}{2} \right)} } \\
\end{array}}$
6. Now we can find the particular solution.
• Given that: y = 1 when x = 0 because, the curve passes through (0,1)
• Substituting in the general solution, we get:
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{y} & {~=~} &{-1~+~{\rm{C}}\,e^{\left(\frac{x^2}{2} \right)}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{1} & {~=~} &{-1~+~{\rm{C}}\,e^{\left(\frac{0^2}{2} \right)}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{2} & {~=~} &{\rm{C}} \\
\end{array}}$
• So the particular solution is:
$\small{y ~=~-1~+~{2}\,e^{\left(\frac{x^2}{2} \right)}}$
• This is the equation of the required curve.
The link below gives a few more solved examples:
Exercise 9.5
In the next section, we will see linear differential equations.
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