2.7 Linear differential equation

1.1 Order and degree of a differential equation
1.2 Formation of a differential equation

2.1 Variable separable method
2.2 Equations reducible to variable separable form
2.3 Homogeneous differential equation
2.4 Equations reducible to homogeneous form
2.5 Exact differential equation
2.6 Equations reducible to exact form
2.7 Linear differential equation
2.8 Equations reducible to linear form

A differential equation is a linear differential equation if a dependent variable and its derivative occur only in the first degree, order may be more than one.

In general form, a linear differential equation of first order is written as $$\frac{{dy}}{{dx}} + P(x).y = Q(x)$$ where $y$ is a dependent variable, $P(x)$ and $Q(x)$ are functions of $x$ only or constant.

The general solution can be find out by multiplying both sides by an integrating factor (I.F.) i.e., ${e^{\int {Pdx} }}$ and solve as follows:

$$ {e^{\int {Pdx} }}.\frac{{dy}}{{dx}} + yP.{e^{\int {Pdx} }} = Q{e^{\int {Pdx} }} $$ $$ \Rightarrow {e^{\int {Pdx} }}.\frac{{dy}}{{dx}} + y\frac{{d\left( {{e^{\int {Pdx} }}} \right)}}{{dx}} = Q{e^{\int {Pdx} }} $$ $$ \Rightarrow \frac{{d\left( {y{e^{\int {Pdx} }}} \right)}}{{dx}} = Q.{e^{\int {Pdx} }} $$

Integrating, we get

$$y{e^{\int {Pdx} }} = \int {Q{e^{\int {Pdx} }}} dx + C$$

which is the solution of the first order linear differential equation.

Note: If differential equation becomes linear by taking $x$ as a dependent variable and $y$ as independent variable, then the general form is written as $$\frac{{dx}}{{dy}} + {P_1}(y)x = {Q_1}(y)$$ where ${P_1}$ and ${Q_1}$ are functions of $y$ and integrating factor (I.F.) becomes ${e^{\int {{P_1}dy} }}$.

Question 11. Solve $$\frac{{dy}}{{dx}} + \frac{y}{x} = \log x$$

Solution: It is a linear differential equation of form $$\frac{{dy}}{{dx}} + P(x).y = Q(x)$$ where $P(x) = \frac{1}{x}$ and $Q(x) = \log x$. The solution of linear differential equation is $$y{e^{\int {Pdx} }} = \int {Q{e^{\int {Pdx} }}} dx + C$$ Integrating factor is $${e^{\int {Pdx} }} = {e^{\int {\frac{{dx}}{x}} }} = {e^{\log x}} = x$$ The general solution is $$\begin{equation} \begin{aligned} yx = \int {\log x.xdx} + C \\ yx = \log x.\frac{{{x^2}}}{2} - \int {\frac{1}{x}} .\frac{{{x^2}}}{2}dx + C \\ yx = \frac{{{x^2}}}{2}(\log x) - \frac{{{x^2}}}{4} + C \\\end{aligned} \end{equation} $$