LINEAR TYPE DIFFERENTIAL EQUATION (Excercise Problems with Solutions)

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