Linear Differential Equation

What is a Differential Equation?

A differential equation is an equation involving one or more terms and the derivatives of one dependent variable with respect to the other independent variable.

Differential equation: $\frac{d y}{d x} =f(x)$
Where " $x$ " is an independent variable and " $y$ " is a dependent variable

Example of differential equation: $x \frac{d y}{d x}+2 y=0$
The above-written equation involves variables as well as the derivative of the dependent variable $\mathrm{y}$ with respect to the independent variable $\mathrm{x}$. Therefore, it is a differential equation.
The following relations are some of the examples of differential equations:
(i) $\frac{d y}{d x}=\sin 2 x+\cos x$
(ii) $\mathrm{k} \frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^2\right]^{3 / 2}$

What is a Linear Differential Equation?

The linear differential equation is a linear equation that involves one or more terms consisting of derivatives of the dependent variable concerning one or more independent variables.

The general equation of the first-order differential equation is the form of

$
\frac{d y}{d x}+P(x) \cdot y=Q(x)
$
Where P(x) and Q(x) are functions of x only or constants.

Non-Linear Differential Equation

When the equation is not linear in an unknown function and its derivatives, then it is said to be a nonlinear differential equation. It provides diverse solutions which can be seen for chaos.

Solving Linear Differential Equations

To solve these types of equations, various methods can be used:

1. Separation of Variables: Applicable to those types of equations which is very simple in which variables can be separated easily on each side of the equation.
2. Integrating Factor: For first-order(degree) linear equations of the form dy/dx+P(x)y=Q(x), we multiply by an integrating factor to simplify.
3. Characteristic Equation: For constant coefficient linear differential equations, we solve the characteristic polynomial to find the solutions of the equations.

Solve Linear Differential Equation using Integrating Factor

Integrating factor: A term, which when multiplied by an expression, converts it to an exact differential i.e. a function which is the derivative of another function.

We have, $\frac{d y}{d x}+P(x) \cdot y=Q(x)$
multiply both sides of Eq (i) by $\int e^{P(x) d x}$, we get
i.e. $\quad e^{\int P(x) d x} \cdot \frac{d y}{d x}+y \cdot P(x) \frac{d}{d x}\left(e^{\int P(x) d x}\right)=Q e^{\int P(x) d x}$
or $\quad \frac{d}{d x}\left(\mathrm{ye}^{\int P(x) d x}\right)=\mathrm{e}^{\int P(x) d x} \cdot Q(x)$
Integrating both sides, we get
or $\quad \int \mathrm{d}\left(y e^{\int P(x) d x}\right)=\int\left(e^{\int P(x) d x} \cdot Q(x)\right) d x$
$\Rightarrow \quad y \mathrm{e}^{\int P(x) d x}=\int \mathrm{Q}(\mathrm{x}) \mathrm{e}^{\int \mathrm{P}(\mathrm{x}) d \mathrm{dx}} \mathrm{dx}+\mathrm{C}$

Which is the required solution of the given differential equation.

The term $\mathrm{e}^{/ \mathrm{P}(\mathrm{x}) \mathrm{dx}}$ which convert the left hand expression of the equatio into a perfect differential is called an Integrating factor (IF).

Thus, we remember the solution of the above equation as

$
y(\mathrm{IF})=\int Q(\mathrm{IF}) d x+C
$
Note : Sometimes a given differential equation can be made linear if we take x as the dependent variable and y as the independent variable. So, we can check the equation with respect to both x and y.

Recommended Video Based on Linear Differential Equation

Solved Examples Based On Linear Differential Equation

Example 1: If the solution curve of the differential equation
$\left(y-2 \log _e x\right) d x+\left(x \log _e x^2\right) d y=0, x>1 {\text { passes through the points }}\left(e, \frac{4}{3}\right)$ and $\left(e^4, \alpha\right)$, then $\alpha$ is equal to $\qquad$
[JEE Main 2023]
Solution:

$
\begin{aligned}
& (y-2 \ln x) d x+(2 x \ln x) d y=0 \\
& d y(2 x \ln x)=[(2 \ln x)-y] d x \\
& \frac{d y}{d x}=\frac{1}{x}-\frac{y}{2 x \ln x} \\
& \frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{2 \mathrm{x} \ln \mathrm{x}}=\frac{1}{\mathrm{x}} \\
& \mathrm{I} . \mathrm{F}=e^{\int \frac{1}{2 x \ln x} d x} \\
& =\mathrm{e}^{\frac{1}{2} \int \frac{\mathrm{dt}}{\mathrm{t}}}=\mathrm{e}^{\frac{1}{2} \ln (\ln x)} \\
& \Rightarrow \mathrm{I} \cdot \mathrm{F}=(\ln x)^{1 / 2} \\
& \therefore \mathrm{y} \sqrt{\ln \mathrm{x}}=\int \frac{\sqrt{\ln \mathrm{x}}}{\mathrm{x}} \mathrm{dx} \quad\left(\text { Let }, \ln \mathrm{x}=\mathrm{u}^2\right) \\
& =2 \int u^2 d u
\end{aligned}
$

$
\begin{aligned}
& (y-2 \ln x) d x+(2 x \ln x) d y=0 \\
& d y(2 x \ln x)=[(2 \ln x)-y] d x \\
& \frac{d y}{d x}=\frac{1}{x}-\frac{y}{2 x \ln x} \\
& \frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{y}}{2 \mathrm{x} \ln \mathrm{x}}=\frac{1}{\mathrm{x}} \\
& \mathrm{I} . \mathrm{F}=e^{\int \frac{1}{2 x \ln x} d x} \\
& =\mathrm{e}^{\frac{1}{2} \int \frac{\mathrm{dt}}{\mathrm{t}}}=\mathrm{e}^{\frac{1}{2} \ln (\ln x)} \\
& \Rightarrow \mathrm{I} \cdot \mathrm{F}=(\ln x)^{1 / 2} \\
& \therefore \mathrm{y} \sqrt{\ln \mathrm{x}}=\int \frac{\sqrt{\ln \mathrm{x}}}{\mathrm{x}} \mathrm{dx} \quad\left(\text { Let }, \ln \mathrm{x}=\mathrm{u}^2\right) \quad \frac{1}{x} d x=2 \mathrm{udu} \\
& =2 \int u^2 d u
\end{aligned}
$

$
\begin{aligned}
& \mathrm{y} \sqrt{\ln x}=\frac{2}{3}(\ln x)^{3 / 2}+c \leftarrow\left(e, \frac{4}{3}\right) \\
& \frac{4}{3}=\frac{2}{3}+c \Rightarrow c=\frac{2}{3} \\
& y \sqrt{\ln x}=\frac{2}{3}(\ln x)^{3 / 2}+\frac{2}{3} \leftarrow\left(e^4, \alpha\right) \\
& \alpha \cdot 2=\frac{2}{3} \times 8+\frac{2}{3} \\
& \alpha=3
\end{aligned}
$

Hence, the answer is 3 .
Example 2: If $y=y(x)$ is the solution of the differential equation $\frac{d y}{d x}+\frac{4 x}{\left(x^2-1\right)} y=\frac{x+2}{\left(x^2-1\right)^{\frac{5}{2}}}, x>1 \quad$ such that $y(2)=\frac{2}{9} \log _c(2+\sqrt{3})$ and $y(\sqrt{2})=\alpha \log _{\mathrm{c}}(\sqrt{\alpha}+\beta)+\beta-\sqrt{\gamma}, \alpha, \beta, \gamma, \in \mathrm{N}$, then $\alpha \beta \gamma$ is equal to $\qquad$ .
[JEE Main 2023]
Solution:
Given differential equation $\frac{d y}{d x}+\frac{4 x}{\left(x^2-1\right)} y=\frac{x+2}{\left(x^2-1\right)^{5 / 2}}$ is linear D.E.

$
\begin{aligned}
& \text { I.F. }=\int_e \frac{4 x}{x^2-1} d x=e_e 2 \ln \left(x^2-1\right)={ }_e \ln \left(x^2-1\right)^2=\left(x^2-1\right)^2 \\
& y\left(x^2-1\right)^2=\int \frac{x+2}{\left(x^2-1\right)^{5 / 2}}\left(x^2-1\right)^2 d x \\
& =\int \frac{x}{\sqrt{x^2-1}} d x+\int \frac{2 d x}{\sqrt{x^2-1}} \\
& =\sqrt{x^2-1}+2 \ln \left[x+\sqrt{x^2-1}\right]+C \\
& \text { put } y(2)=\frac{2}{9} \ln (2+\sqrt{3}) \\
& \frac{2}{9} \ln (2+\sqrt{3})(9)=\sqrt{3}+2 \ln [2+\sqrt{3}]+C \\
& =C=-\sqrt{3} \\
& \text { put } x=\sqrt{2} \\
& y=1+2 \ln [\sqrt{2}+1]-\sqrt{3}
\end{aligned}
$

$
\begin{aligned}
& \frac{2}{9} \ln (2+\sqrt{3})(9)=\sqrt{3}+2 \ln [2+\sqrt{3}]+C \\
& =C=-\sqrt{3} \\
& \text { put } x=\sqrt{2} \\
& y=1+2 \ln [\sqrt{2}+1]-\sqrt{3} \\
& \alpha=2, \beta=1=\gamma=3 \\
& \alpha \beta \gamma=2(1)(3)=6
\end{aligned}
$

Hence, the answer is 6 .
Example 3: If $y=y(x)$ is the solution curve of the differential equation $\frac{d y}{d x}+y \tan x=x \sec x, 0 \leq x \leq \frac{\pi}{3}, y(0)=1$, then $y\left(\frac{\pi}{6}\right)$ is equal to [JEE Main 2023]

Solution:
Given D.E. is linear D.E.

$
\begin{aligned}
& \text { I.F. }=e^{\int \tan x d x} \\
& =e^{\ell \operatorname{nsec} x}=\operatorname{secx} \\
& y \sec x=\int x \sec ^2 x d x \\
& =\mathrm{x} \tan \mathrm{x}-\int \tan x d x \\
& \Rightarrow \quad y \sec x=x \tan x-\ell \mathrm{n} \operatorname{sect}+c \\
& \text { Put } y(0)=1 \\
& 1=0-0+\mathrm{c} \Rightarrow \mathrm{c}=1 \\
& \mathrm{Y}(\mathrm{x})=\frac{x \tan x}{\sec x}-\frac{\ell \operatorname{nsec} x}{\sec x}+\frac{1}{\sec x} \\
& y\left(\frac{\pi}{6}\right)=\frac{\left(\frac{\pi}{6}\right)\left(\frac{1}{\sqrt{3}}\right)}{\left(\frac{2}{\sqrt{3}}\right)}-\frac{\ln \left(\frac{2}{\sqrt{3}}\right)}{\left(\frac{2}{\sqrt{3}}\right)}+\frac{\sqrt{3}}{2} \\
& =\frac{\pi}{12}-\frac{\sqrt{3}}{2} \ell n\left(\frac{2}{\sqrt{3}}\right)+\frac{\sqrt{3}}{2} \ell n e \\
& =\frac{\pi}{12}-\frac{\sqrt{3}}{2} \ell n\left(\frac{2}{e \sqrt{3}}\right)
\end{aligned}
$

Hence, the required answer is $\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e\left(\frac{2}{e \sqrt{3}}\right)$

Example 4: Let $\mathrm{y}=\mathrm{y}(\mathrm{x})$ be the solution of the differential equation
$x\left(1-x^2\right) \frac{d y}{d x}+\left(3 x^2 y-y-4 x^3\right)=0, x>1$, with $y(2)=-2$. Then $y(3)$ equal to
[JEE Main 2022] is

Solution:

$
\begin{aligned}
& \frac{d y}{d x}+\frac{\left(3 x^2-1\right)}{x\left(1-x^2\right)} y=\frac{4 x^3}{x\left(1-x^2\right)} \\
& \text { I.F }=e^{\int\left(\frac{x^2-1}{x\left(1-x^2\right)}+\frac{2 x^2}{x\left(1-x^2\right)}\right) d x}=e^{-\ln x-\ln \left(1-x^2\right)}=\frac{1}{x\left(1-x^2\right)} \\
& \Rightarrow y \times \frac{1}{x\left(1-x^2\right)}=\int \frac{4 x^3}{x^2\left(1-x^2\right)^2} d x=\frac{+2}{1-x^2}+c \\
& \text { at } x=2, y=-2 \Rightarrow c=1 \\
& \Rightarrow y=2 x+x\left(1-x^2\right) \\
& \text { at } x=3 ; y=6+3(1-9)=-18
\end{aligned}
$

Hence, the required answer is -18 .
Example 5: Let $\mathrm{y}=\mathrm{y}(\mathrm{x})$ be the solution curve of the differential equation

$
\begin{aligned}
& \frac{d y}{d x}+\frac{1}{x^2-1} y=\left(\frac{x-1}{x+1}\right)^{1 / 2}, x>1 \\
& \left(2, \sqrt{\frac{1}{3}}\right) . \text { Then } \sqrt{7} y(8) \\
& \text { is equal to: } \\
& \text { [JEE Main 2022] }
\end{aligned}
$

Solution:

$
\begin{aligned}
& \text { For } I \cdot F \\
& =\int \frac{1}{(x-1)(x+1)} d x \\
& =\frac{1}{2}\left[\int \frac{1}{x-1} d x-\int \frac{1}{x+1} d x\right] \\
& =\frac{1}{2} \ln \left(\frac{x-1}{x+1}\right) \\
& =\ln \sqrt{\frac{x-1}{x+1}} \\
& \therefore I \cdot F \cdot=\sqrt{\frac{x-1}{x+1}}
\end{aligned}
$

$\therefore$ Solution is

$
\begin{aligned}
& y \cdot \sqrt{\frac{x-1}{x+1}}=\int \frac{x-1}{x+1} d x \\
& y \sqrt{\frac{x-1}{x+1}}=x-2 \ln (x+1)+c
\end{aligned}
$

Using $\left(2, \sqrt{\frac{1}{3}}\right)$

$
\begin{aligned}
& \sqrt{\frac{1}{3}} \cdot \sqrt{\frac{1}{3}}=2-2 \ln 3+c \\
& c=2 \ln 3-\frac{5}{3} \\
& \text { put } x=8 \\
& y \sqrt{\frac{7}{9}}=8-2 \ln 9+2 \ln 3-\frac{5}{3} \\
& \sqrt{7} y=3\left[\frac{19}{3}-2 \ln 3\right] \\
& =19-6 \ln 3
\end{aligned}
$

Hence, the required answer is $19-6 \ln 3$