# LINEAR TYPE DIFFERENTIAL EQUATION (Excercise Problems with Solutions)

$\underline{PART\ -\ A}$
$1.\ \color{red}{Find\ the\ integrating\ factor\ of\ \frac{dy}{dx}\ +\ \frac{2}{x}\ y\ =\ x}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$Compare\ with\ \frac{dy}{dx}\ +\ P\ y\ =\ Q$
$Here\ P\ =\ \frac{2}{x}$
$I.F\ =\ e^{\int P\ dx}$
$=\ e^{ \int \frac{2}{x}\ dx}$
$=\ e^{2\int \frac{1}{x}\ dx}$
$=\ e^{2\ log\ x}$
$=\ e^{log\ x^{2}}$
$=\ x^{2}$
$\boxed{I.F\ =\ x^2}$
$\underline{PART\ -\ B}$
$2.\ \color{red}{Find\ the\ integrating\ factor\ of\ \frac{dy}{dx}\ +\ y\ tan\ x\ =\ x^2}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$Compare\ with\ \frac{dy}{dx}\ +\ P\ y\ =\ Q$
$Here\ P\ =\ tan\ x$
$I.F\ =\ e^{\int P\ dx}$
$=\ e^{ \int \frac{sin\ x}{cos\ x}\ dx}$
$put\ u\ =\ cos\ x$
$du\ =\ -\ sin\ x\ dx$
$I.F\ =\ e^{ \int \frac{-du}{u}}$
$=\ e^{-\ log\ u}$
$=\ e^{log\ u^{-\ 1}}$
$=\ u^{-\ 1}$
$=\ \frac{1}{u}$
$\boxed{I.F\ =\ \frac{1}{cos\ x}}$
$3.\ \color{red}{Find\ the\ integrating\ factor\ of\ \frac{dy}{dx}\ -\ y\ cot\ x\ =\ sin\ x}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$Compare\ with\ \frac{dy}{dx}\ +\ P\ y\ =\ Q$
$Here\ P\ =\ -\ cot\ x$
$I.F\ =\ e^{\int P\ dx}$
$=\ e^{ -\ \int \frac{cos\ x}{sin\ x}\ dx}$
$put\ u\ =\ sin\ x$
$du\ =\ cos\ x\ dx$
$I.F\ =\ e^{-\ \int \frac{du}{u}}$
$=\ e^{-\ log\ u}$
$=\ e^{log\ u^{-\ 1}}$
$=\ u^{-\ 1}$
$=\ \frac{1}{u}$
$\boxed{I.F\ =\ \frac{1}{sin\ x}}$
$\underline{PART\ -\ C}$
$4.\ \color{red}{Solve:\ \frac{dy}{dx}\ -\ \frac{2}{x}\ y\ =\ x^2\ sin\ x}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$Compare\ with\ \frac{dy}{dx}\ +\ P\ y\ =\ Q$
$Here\ P\ =\ -\ \frac{2}{x}\ \hspace{2cm}\ Q\ =\ x^2\ sin\ x$
$I.F\ =\ e^{\int P\ dx}$
$=\ e^{-\ \int \frac{2}{x}\ dx}$
$=\ e^{-\ 2\int \frac{1}{x}\ dx}$
$=\ e^{-\ 2\ log\ x}$
$=\ e^{log\ x^{-2}}$
$=\ x^{-2}$
$\boxed{I.F\ =\ \frac{1}{x^2}}$
$The\ required\ solution\ is$
$y\ (I.F)\ =\ \int Q\ (I.F)\ dx\ +\ c$
$y\ (\frac{1}{x^2})\ =\ \int x^2\ sin\ x\ (\frac{1}{x^2})\ dx\ +\ c$
$=\ \int sin\ x\ dx\ +\ c$
$\boxed{\frac{y}{x^2}\ =\ -\ cos\ x\ +\ c}$
$5.\ \color{red}{Solve:\ \frac{dy}{dx}\ +\ y\ tan\ x\ =\ e^x\ cos\ x}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$Compare\ with\ \frac{dy}{dx}\ +\ P\ y\ =\ Q$
$Here\ P\ =\ tan\ x\ \hspace{2cm}\ Q\ =\ e^x\ cos\ x$
$I.F\ =\ e^{\int P\ dx}$
$=\ e^{ \int \frac{sin\ x}{cos\ x}\ dx}$
$put\ u\ =\ cos\ x$
$du\ =\ -\ sin\ x\ dx$
$I.F\ =\ e^{ \int \frac{-du}{u}}$
$=\ e^{-\ log\ u}$
$=\ e^{log\ u^{-\ 1}}$
$=\ u^{-\ 1}$
$=\ \frac{1}{u}$
$\boxed{I.F\ =\ \frac{1}{cos\ x}}$
$The\ required\ solution\ is$
$y\ (I.F)\ =\ \int Q\ (I.F)\ dx\ +\ c$
$y\ (\frac{1}{cos\ x})\ =\ \int e^x\ cos\ x\ (\frac{1}{cos\ x})\ dx\ +\ c$
$y\ sec\ x\ =\ \int e^x\ \ dx\ +\ c$
$\boxed{y\ sec\ x\ =\ e^x\ +\ c}$
$6.\ \color{red}{Solve:\ \frac{dy}{dx}\ -\ y\ tan\ x\ =\ e^x\ sec\ x}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$Compare\ with\ \frac{dy}{dx}\ +\ P\ y\ =\ Q$
$Here\ P\ =\ -\ tan\ x\ \hspace{2cm}\ Q\ =\ e^x\ sec\ x$
$I.F\ =\ e^{\int P\ dx}$
$=\ e^{ \int -\ \frac{sin\ x}{cos\ x}\ dx}$
$put\ u\ =\ cos\ x$
$du\ =\ -\ sin\ x\ dx$
$I.F\ =\ e^{ \int \frac{du}{u}}$
$=\ e^{log\ u}$
$=\ u$
$\boxed{I.F\ =\cos\ x}$
$The\ required\ solution\ is$
$y\ (I.F)\ =\ \int Q\ (I.F)\ dx\ +\ c$
$y\ cos\ x\ =\ \int e^x\ sec\ x\ cos\ x\ dx\ +\ c$
$y\ cos\ x\ =\ \int e^x\ dx\ +\ c$
$\boxed{y\ cos\ x\ =\ e^x\ +\ c}$
$7.\ \color{red}{Solve:\ \frac{dy}{dx}\ +\ y\ tan\ x\ =\ cos^3\ x}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$Compare\ with\ \frac{dy}{dx}\ +\ P\ y\ =\ Q$
$Here\ P\ =\ tan\ x\ \hspace{2cm}\ Q\ =\ cos^3\ x$
$I.F\ =\ e^{\int P\ dx}$
$=\ e^{ \int \frac{sin\ x}{cos\ x}\ dx}$
$put\ u\ =\ cos\ x$
$du\ =\ -\ sin\ x\ dx$
$I.F\ =\ e^{ \int \frac{-du}{u}}$
$=\ e^{-\ log\ u}$
$=\ e^{log\ u^{-\ 1}}$
$=\ u^{-\ 1}$
$=\ \frac{1}{u}$
$\boxed{I.F\ =\ \frac{1}{cos\ x}}$
$The\ required\ solution\ is$
$y\ (I.F)\ =\ \int Q\ (I.F)\ dx\ +\ c$
$y\ (\frac{1}{cos\ x})\ =\ \int cos^3\ x\ (\frac{1}{cos\ x})\ dx\ +\ c$
$=\ \int cos^2\ x\ dx\ +\ c$
$=\ \int (\frac{1\ +\ cos\ 2\ x}{2})\ dx\ +\ c$
$\boxed{\frac{y}{cos\ x}\ =\ \frac{1}{2} \big [x\ +\ \frac{sin\ 2x}{2}\ \big ]\ +\ c}$
$8.\ \color{red}{Solve:\ \frac{dy}{dx}\ +\ y\ cot\ x\ =\ 2\ cos\ x}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$Compare\ with\ \frac{dy}{dx}\ +\ P\ y\ =\ Q$
$Here\ P\ =\ cot\ x\ \hspace{2cm}\ Q\ =\ 2 \cos\ x$
$I.F\ =\ e^{\int P\ dx}$
$=\ e^{ \int \frac{cos\ x}{sin\ x}\ dx}$
$put\ u\ =\ sin\ x$
$du\ =\ cos\ x\ dx$
$I.F\ =\ e^{ \int \frac{du}{u}}$
$=\ e^{log\ u}$
$=\ e^{log\ sin\ x}$
$=\ sin\ x$
$\boxed{I.F\ =\ sin\ x}$
$The\ required\ solution\ is$
$y\ (I.F)\ =\ \int Q\ (I.F)\ dx\ +\ c$
$y\ sin\ x\ =\ \int 2\ cos\ x\ sin\ x\ dx\ +\ c$
$y\ sin\ x\ =\ \int sin\ 2x\ dx\ +\ c$
$\boxed{y\ sin\ x\ =\ -\ \frac{cos\ 2x}{2}\ +\ c}$