In the previous section, we completed a discussion on the method to solve homogeneous differential equations. In this section, we will see linear differential equations.
Some basic details can be written in 3 steps:
Type I:
1. Consider the differential equation $\small{\frac{dy}{dx}~+~\rm{P}y~=~\rm{Q}}$
2. Suppose that P and Q 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
3. Then the equation written in (1), is a first order linear differential equation.
Let us see some examples:
•
$\small{\frac{dy}{dx}~+~y~=~\sin x}$
♦ Condition (iii) is satisfied
•
$\small{\frac{dy}{dx}~+~\left(\frac{1}{x} \right)y~=~e^x}$
♦ Condition (ii) is satisfied
•
$\small{\frac{dy}{dx}~+~\left(\frac{1}{x\,\log x} \right)y~=~\frac{1}{x}}$
♦ Condition (ii) is satisfied
Type II:
1. Consider the differential equation $\small{\frac{dx}{dy}~+~\rm{P}_1 x~=~\rm{Q}_1}$
2. Suppose that P1 and Q1 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
3. Then the equation written in (1), is a first order linear differential equation.
Let us see some examples:
•
$\small{\frac{dx}{dy}~+~x~=~\cos y}$
♦ Condition (iii) is satisfied
•
$\small{\frac{dx}{dy}~+~\left(\frac{-2}{y} \right)x~=~y^2\;e^y}$
♦ Condition (ii) is satisfied
So now we know what a first order linear differential equation is. Let us try to solve such equations:
First we will solve type I:
1. We have: $\small{\frac{dy}{dx}~+~\rm{P}y~=~\rm{Q}}$
2. Multiply both sides by a function of x, say $\small{g(x)}$.
•
We get:
$\small{g(x)\,\frac{dy}{dx}~+~\rm{P}\,g(x)\,y~=~{\rm{Q}}\,g(x)}$
3. The function $\small{g(x)}$ should be such that:
$\small{{\rm{Q}}\,g(x)~=~\frac{d}{dx}\left[g(x)\,y \right]}$
4. Based on (2) and (3), we get:
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{g(x)\,\frac{dy}{dx}~+~{\rm{P}}\,g(x)\,y} & {~=~} &{{\rm{Q}}\,g(x)} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{g(x)\,\frac{dy}{dx}~+~{\rm{P}}\,g(x)\,y} & {~=~} &{\frac{d}{dx}\left[g(x)\,y \right]} \\
{~\color{magenta} 3} &{{\Rightarrow}} &{g(x)\,\frac{dy}{dx}~+~{\rm{P}}\,g(x)\,y} & {~=~} &{g(x)\,\frac{dy}{dx}~+~y\;g'(x)} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{{\rm{P}}\,g(x)} & {~=~} &{g'(x)} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{{\rm{P}}} & {~=~} &{\frac{g'(x)}{g(x)}} \\
{~\color{magenta} 6 } &{{\Rightarrow}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\int{\left[\frac{g'(x)}{g(x)} \right]dx}} \\
{~\color{magenta} 7 } &{{\Rightarrow}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\log\left[g(x) \right]} \\
{~\color{magenta} 8 } &{{\Rightarrow}} &{g(x)} & {~=~} &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}} \\
\end{array}}$
5. So we have obtained $\small{g(x)}$.
•
It is called the integrating factor (abbreviated as I.F).
•
Now the equation in (2) becomes:
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{g(x)\,\frac{dy}{dx}~+~{\rm{P}}\,g(x)\,y} & {~=~} &{{\rm{Q}}\,g(x)} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}\,\frac{dy}{dx}~+~{\rm{P}}\,e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}\,y} & {~=~} &{{\rm{Q}}\,e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}} \\
{~\color{magenta} 3} &{{\Rightarrow}} &{\frac{d}{dx}\left[y\,e^{\left(\int{\left[{\rm{P}} \right]dx} \right)} \right]} & {~=~} &{{\rm{Q}}\,e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{y\,e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}} & {~=~} &{\int{\left[{\rm{Q}}\,e^{\left(\int{\left[{\rm{P}} \right]dx} \right)} \right]dx}} \\
\end{array}}$
• From this, we can write y.
◼ Remarks:
•
The L.H.S of [(2) magenta color], is the derivative of $y\,e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}$, w.r.t x.
•
In [(4) magenta color], we integrate both sides of [(3) magenta color]
The above five steps can be rewritten in an "easy to remember" form:
1. Write the given differential equation in the form:
$\small{\frac{dy}{dx}~+~{\rm{P}}y~=~\rm{Q}}$
•
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
2. Find I.F:
I.F =$\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$
3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$
4. Do the integration in the R.H.S, and add the constant of integration.
5. Multiply the above result in (4), by the reciprocal of the I.F
Next, we will solve type II.
We can directly write the "easy to remember" five steps:
1. Write the given differential equation in the form:
$\small{\frac{dx}{dy}~+~{\rm{P}_1}y~=~\rm{Q}_1}$
• 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
2. Find I.F:
I.F =$\small{e^{\left(\int{\left[{\rm{P}_1} \right]dy} \right)}}$
3. Write the equation with independent x:
$\small{x\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q_1(I.F)}} \right]dy}}$
4. Do the integration in the R.H.S, and add the constant of integration.
5. Multiply the above result in (4), by the reciprocal of the I.F
Now we will see some solved examples:
Solved example 25.56
Find the general solution of the differential equation $\boldsymbol{\frac{dy}{dx}~+~3y~=~e^{-2x}}$.
Solution:
1. The given differential equation is:
$\small{\frac{dy}{dx}~+~3y~=~e^{-2x}}$
• Here P = 3 and Q = $\small{e^{-2x}}$
• 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, P is a constant and Q is a function of x. So it is a first order linear differential equation.
2. Finding I.F = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\int{\left[3 \right]dx}~=~ 3x} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}} & {~=~} &{e^{\left(3x \right)}} \\
\end{array}}$
3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$
We get: $\small{y\left[e^{\left(3x \right)} \right]~=~\int{\left[e^{-2x}~\left(e^{\left(3x \right)} \right) \right]dx}~=~\int{\left[e^{x} \right]dx}}$
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{e^x~+~\rm{C}}$
5. Multiply the above result in (4), by the reciprocal of the I.F
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\left[e^x~+~\rm{C} \right]\frac{1}{e^{3x}}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{y} & {~=~} &{e^{\left(-2x \right)}~+~{\rm{C}}e^{\left(-3x \right)}} \\
\end{array}}$
Solved example 25.57
Find the general solution of the differential equation $\boldsymbol{\frac{dy}{dx}~+~\frac{y}{x}~=~x^2}$.
Solution:
1. The given differential equation is:
$\small{\frac{dy}{dx}~+~\frac{y}{x}~=~x^2}$
• Here P = 1/x and Q = x2
• 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 = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta}
1 } &{{}} &{\int{\left[{\rm{P}} \right]dx}} &
{~=~} &{\int{\left[\frac{1}{x} \right]dx}~=~ \log x} \\
{~\color{magenta}
2 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}}
\right]dx} \right)}} & {~=~} &{e^{\left(\log x
\right)}} \\
{~\color{magenta} 3 } &{{\Rightarrow}}
&{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}} &
{~=~} &{x~~\left({\rm{since}}~~e^{\log u}~=~u \right)} \\
\end{array}}$
3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$
We get: $\small{y\left[x \right]~=~\int{\left[x^2~\left(x \right) \right]dx}~=~\int{\left[x^3 \right]dx}}$
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{\frac{x^4}{4}~+~\rm{C}}$
• 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
$\small{\begin{array}{ll}{~\color{magenta}
1 } &{{}} &{y} & {~=~}
&{\left[\frac{x^4}{4}~+~\rm{C} \right]\frac{1}{x}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{xy} & {~=~} &{\frac{x^4}{4}~+~{\rm{C}}} \\
\end{array}}$
Solved example 25.58
Find the general solution of the differential equation $\boldsymbol{x\frac{dy}{dx}~+~2y~=~x^2~~(x\ne 0)}$.
Solution:
1. The given differential equation is:
$\small{x\frac{dy}{dx}~+~2y~=~x^2~~(x\ne 0)}$
Dividing both sides by x, we get:
$\small{\frac{dy}{dx}~+~\left(\frac{2}{x} \right)y~=~x~~(x\ne 0)}$
• Here P = 2/x and Q = 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
• Both P and Q are functions of x. So it is a first order linear differential equation.
2. Finding I.F = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta}
1 } &{{}} &{\int{\left[{\rm{P}} \right]dx}} &
{~=~} &{\int{\left[\frac{2}{x} \right]dx}~=~2 \log x} \\
{~\color{magenta}
2 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}}
\right]dx} \right)}} & {~=~} &{e^{\left(2 \log x
\right)}~=~e^{\left( \log x^2 \right)}} \\
{~\color{magenta} 3
} &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}}
\right]dx} \right)}} & {~=~}
&{x^2~~\left({\rm{since}}~~e^{\log u}~=~u \right)} \\
\end{array}}$
3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$
We get: $\small{y\left[x^2 \right]~=~\int{\left[x~\left(x^2 \right) \right]dx}}$
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{\frac{x^4}{4}~+~\rm{C}}$
• 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
$\small{\begin{array}{ll}{~\color{magenta}
1 } &{{}} &{y} & {~=~}
&{\left[\frac{x^4}{4}~+~\rm{C} \right]\frac{1}{x^2}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{y} & {~=~} &{\frac{x^2}{4}~+~{\rm{C}}\,x^{-2}} \\
\end{array}}$
Solved example 25.59
Find the general solution of the differential equation $\boldsymbol{x\frac{dy}{dx}~+~2y~=~x^2\,\log x}$.
Solution:
1. The given differential equation is:
$\small{x\frac{dy}{dx}~+~2y~=~x^2\,\log x}$
Dividing both sides by x, we get:
$\small{\frac{dy}{dx}~+~\left(\frac{2}{x} \right)y~=~x\,\log x}$
• Here P = $\small{\frac{2}{x}}$ and Q = $\small{x\,\log x}$
• 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 = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\int{\left[\frac{2}{x} \right]dx}~=~ 2\,\log x~=~\log\left(x^2 \right)} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}} & {~=~} &{e^{\log\left(x^2 \right)}~=~x^2} \\
\end{array}}$
3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$
We get: $\small{y\left[x^2 \right]~=~\int{\left[x \log x~\left(x^2 \right) \right]dx}~=~\int{\left[x^3~\times~\log x \right]dx}}$
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{\frac{x^4 \log \left|x \right|}{4}~-~\frac{x^4}{16}~+~\rm{C}}$
5. Multiply the above result in (4), by the reciprocal of the I.F
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\left[\frac{x^4 \log \left|x \right|}{4}~-~\frac{x^4}{16}~+~\rm{C} \right]\frac{1}{x^2}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{y} & {~=~} &{\frac{x^2 \log \left|x \right|}{4}~-~\frac{x^2}{16}~+~{\rm{C}}x^{-2}} \\
\end{array}}$
Solved example 25.60
Find the general solution of the differential equation $\boldsymbol{x\,\log x \frac{dy}{dx}~+~y~=~\left(\frac{2}{x} \right)\log x}$.
Solution:
1. The given differential equation is:
$\small{x\,\log x \frac{dy}{dx}~+~y~=~\left(\frac{2}{x} \right)\log x}$
Dividing both sides by x log x, we get:
$\small{\frac{dy}{dx}~+~\left(\frac{1}{x\,\log x} \right)y~=~\frac{2}{x^2}}$
• Here P = $\small{\frac{1}{x\,\log x}}$ and Q = $\small{\frac{2}{x^2}}$
• 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 = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\int{\left[\frac{1}{x\,\log x} \right]dx}~=~\log \left|\log \left|x \right| \right|} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}} & {~=~} &{e^{\log \left|\log \left|x \right| \right|}~=~\log \left|x \right|} \\
\end{array}}$
3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$
We get: $\small{y\left[\log x \right]~=~\int{\left[\frac{2}{x^2}~\left(\log \left|x \right| \right) \right]dx}}$
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{\frac{-2 \log \left|x \right|~-~2}{x}~+~\rm{C}}$
• 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
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\left[\frac{-2 \log \left|x \right|~-~2}{x}~+~\rm{C} \right]\frac{1}{\log \left|x \right|}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{y\,\log \left|x \right|} & {~=~} &{\frac{-2 \log \left|x \right|~-~2}{x}~+~\rm{C}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{y\,\log \left|x \right|} & {~=~} &{\frac{-2 }{x}\left(\log \left|x \right|~+~1 \right)~+~\rm{C}} \\
\end{array}}$
Solved example 25.61
For the differential equation $\boldsymbol{\left(1 +x^2 \right)
\frac{dy}{dx}~+~2xy~=~\frac{1}{1+x^2}}$; y = 0 when x = 1, find the particular solution satisfying the given conditions.
Solution:
1. The given differential equation is:
$\small{\left(1 +x^2 \right) \frac{dy}{dx}~+~2xy~=~\frac{1}{1+x^2}}$
Dividing both sides by $\small{\left(1 +x^2 \right)}$, we get:
$\small{\frac{dy}{dx}~+~\left(\frac{2x}{1 +x^2 } \right)y~=~\frac{1}{\left(1 +x^2 \right)^2}}$
• Here P = $\small{\frac{2x}{1 +x^2 }}$ and Q = $\small{\frac{1}{\left(1 +x^2 \right)^2}}$
• 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 = $\small{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{\int{\left[{\rm{P}} \right]dx}} & {~=~} &{\int{\left[\frac{2x}{1 +x^2 } \right]dx}~=~\log (1+x^2)} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}} \right]dx} \right)}} & {~=~} &{e^{\log (1+x^2)}~=~1 + x^2} \\
\end{array}}$
3. Write the equation with independent y:
$\small{y\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q(I.F)}} \right]dx}}$
We get: $\small{y\left[1 + x^2 \right]~=~\int{\left[\frac{1}{\left(1 +x^2 \right)^2}~\left(1 + x^2 \right) \right]dx}~=~\int{\left[\frac{1}{1 +x^2} \right]dx}}$
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{\tan^{-1}x~+~\rm{C}}$
5. Multiply the above result in (4), by the reciprocal of the I.F
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\left[\tan^{-1}x~+~\rm{C} \right]\frac{1}{1 + x^2}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{y\,\left(1 + x^2 \right)} & {~=~} &{\tan^{-1}x~+~\rm{C}} \\
\end{array}}$
6. Now we can find the particular solution.
•
Given that: y = 0 when x = 1
•
Substituting in the general solution, we get:
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{y\,\left(1 + x^2 \right)} & {~=~} &{\tan^{-1}x~+~\rm{C}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{(0)\left(1 + 1^2 \right)} & {~=~} &{\tan^{-1}1~+~\rm{C}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{0} & {~=~} &{\tan^{-1}1~+~\rm{C}} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\rm{C}} & {~=~} &{(-1)\tan^{-1}1} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{\rm{C}} & {~=~} &{\frac{-\pi}{4}} \\
\end{array}}$
• So the particular solution is:
$\small{y\,\left(1 + x^2 \right)~=~\tan^{-1}x~-~\frac{\pi}{4}}$
In the next section, we will see a few more solved examples.
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