LINEAR TYPE DIFFERENTIAL EQUATION

\[Differential\ equations\ of\ the\ form\ \frac{dy}{dx}\ +\ P\ y\ =\ Q,\]\[Where\ P\ and\ Q\ are\ functions\ of\ x\ are\ called\ Linear\ Differential\ Equations\]
\[The\ solution\ of\ linear\ differential\ equation\ is\ given\ by\]
\[y\ (I.F)\ =\ \int Q\ (I.F)\ dx\ +\ c\]
\[Where\ I.F\ (Integrating\ Factor)\ =\ e^{\int P\ dx}\]

Example  1:

\[1.\ \color{red}{Find\ the\ integrating\ factor\ of\ \frac{dy}{dx}\ +\ \frac{y}{x}\ =\ x}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[Compare\ with\ \frac{dy}{dx}\ +\ P\ y\ =\ Q\]
\[Here\ P\ =\ \frac{1}{x}\]
\[I.F\ =\ e^{\int P\ dx}\]
\[=\ e^{\int \frac{1}{x}\ dx}\]
\[=\ e^{log\ x}\]
\[\boxed{I.F\ =\ x}\]

Example  2:

\[2.\ \color{red}{Find\ the\ integrating\ factor\ of\ \frac{dy}{dx}\ -\ \frac{3}{x}\ y\ =\ x^2}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[Compare\ with\ \frac{dy}{dx}\ +\ P\ y\ =\ Q\]
\[Here\ P\ =\ -\ \frac{3}{x}\]
\[I.F\ =\ e^{\int P\ dx}\]
\[=\ e^{-\ \int \frac{3}{x}\ dx}\]
\[=\ e^{-\ 3\int \frac{1}{x}\ dx}\]
\[=\ e^{-\ 3\ log\ x}\]
\[=\ e^{log\ x^{-3}}\]
\[=\ x^{-3}\]
\[\boxed{I.F\ =\ \frac{1}{x^3}}\]

Example  3:

\[3.\ \color{red}{Find\ the\ integrating\ factor\ of\ \frac{dy}{dx}\ +\ y\ cot\ x\ =\ x}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[Compare\ with\ \frac{dy}{dx}\ +\ P\ y\ =\ Q\]
\[Here\ P\ =\ cot\ x\]
\[I.F\ =\ e^{\int P\ dx}\]
\[=\ e^{ \int \frac{cos\ x}{sin\ x}\ dx}\]
\[put\ u\ =\ sin\ x\]
\[du\ =\ cos\ x\ dx\]
\[I.F\ =\ e^{ \int \frac{du}{u}}\]
\[=\ e^{log\ u}\]
\[=\ e^{log\ sin\ x}\]
\[=\ sin\ x\]
\[\boxed{I.F\ =\ sin\ x}\]

Example  4:

\[4.\ \color{red}{Solve:\ \frac{dy}{dx}\ +\ y\ tan\ x\ =\ 4x\ cos\ x}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[Compare\ with\ \frac{dy}{dx}\ +\ P\ y\ =\ Q\]
\[Here\ P\ =\ tan\ x\ \hspace{2cm}\ Q\ =\ 4x\ cos\ x\]
\[I.F\ =\ e^{\int P\ dx}\]
\[=\ e^{ \int \frac{sin\ x}{cos\ x}\ dx}\]
\[put\ u\ =\ cos\ x\]
\[du\ =\ -\ sin\ x\ dx\]
\[I.F\ =\ e^{ \int \frac{-du}{u}}\]
\[=\ e^{-\ log\ u}\]
\[=\ e^{log\ u^{-\ 1}}\]
\[=\ u^{-\ 1}\]
\[=\ \frac{1}{u}\]
\[\boxed{I.F\ =\ \frac{1}{cos\ x}}\]
\[The\ required\ solution\ is\]
\[y\ (I.F)\ =\ \int Q\ (I.F)\ dx\ +\ c\]
\[y\ (\frac{1}{cos\ x})\ =\ \int 4x\ cos\ x\ (\frac{1}{cos\ x})\ dx\ +\ c\]
\[y\ sec\ x\ =\ \int 4x\ \ dx\ +\ c\]
\[=\ 4\ \frac{x^2}{2}\ dx\ +\ c\]
\[\boxed{y\ sec\ x\ =\ 2\ x^2\ +\ c}\]

Example  5:

\[5.\ \color{red}{Solve:\ \frac{dy}{dx}\ -\ \frac{2x}{1\ +\ x^2}\ y\ =\ (1\ +\ x^2)}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[Compare\ with\ \frac{dy}{dx}\ +\ P\ y\ =\ Q\]
\[Here\ P\ =\ -\ \frac{2x}{1\ +\ x^2}\ \hspace{2cm}\ Q\ =\ 1\ +\ x^2\]
\[I.F\ =\ e^{\int P\ dx}\]
\[=\ e^{ -\ \int \frac{2\ x}{1\ +\ x^2}\ dx}\]
\[put\ u\ =\ 1\ +\ x^2\]
\[du\ =\ 2\ x\ dx\]
\[I.F\ =\ e^{ \int \frac{-du}{u}}\]
\[=\ e^{-\ log\ u}\]
\[=\ e^{log\ u^{-\ 1}}\]
\[=\ u^{-\ 1}\]
\[=\ \frac{1}{u}\]
\[\boxed{I.F\ =\ \frac{1}{1\ +\ x^2}}\]
\[The\ required\ solution\ is\]
\[y\ (I.F)\ =\ \int Q\ (I.F)\ dx\ +\ c\]
\[y\ (\frac{1}{1\ +\ x^2})\ =\ \int (1\ +\ x^2)\ (\frac{1}{1\ +\ x^2})\ dx\ +\ c\]
\[\frac{y}{1\ +\ x^2}\ =\ \int 1\ dx\ +\ c\]
\[\boxed{\frac{y}{1\ +\ x^2}\ =\ x\ +\ c}\]