FIRST ORDER DIFFERENTIAL EQUATION (Excercise Problems with Solutions)

\[\underline{PART\ -\ A}\]
\[1.\ \color{red}{Solve:\ sec\ x\ tan\ x\ dx\ +\ sec\ y\ tan\ y\ dy\ =\ 0}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[sec\ x\ tan\ x\ dx\ =\ -\ sec\ y\ tan\ y\ dy\]
\[Integrating\ on\ both\ sides\]
\[\int sec\ x\ tan\ x\ dx\ =\ -\ \int sec\ y\ tan\ y\ dy\]
\[\boxed{sec\ x\ =\ – sec\ y\ +\ c}\]
\[\underline{PART\ -\ B}\]
\[2.\ \color{red}{Solve:\ x\ \frac{dy}{dx}\ =\ y}\ \hspace{18cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[x\ \frac{dy}{dx}\ =\ y\]
\[\frac{dy}{y}\ =\ \frac{dx}{x}\]
\[Integrating\ on\ both\ sides\]
\[\int \frac{dy}{y}\ =\ \int \frac{dx}{x}\]
\[log\ y\ =\ log\ x\ +\ log\ c\]
\[log\ y\ =\ log\ x\ c\]
\[\boxed{y\ =\ x\ c}\]
\[3.\ \color{red}{Solve:\ xdx\ -\ ydy\ =\ 0}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[x\ dx\ =\ \ y\ dy\]
\[Integrating\ on\ both\ sides\]
\[\int x\ dx\ =\ \int y\ dy \]
\[\boxed{\frac{x^2}{2}\ =\ \frac{y^2}{2}\ +\ c}\]
\[4.\ \color{red}{Solve:\ \frac{dy}{dx}\ =\ \frac{1}{1\ +\ x^2}}\ \hspace{18cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[\frac{dy}{dx}\ =\ \frac{1}{1\ +\ x^2}\]
\[(1\ +\ x^2)dy\ =\ dx\]
\[dy\ =\ \frac{1}{1\ +\ x^2}\ dx\]
\[Integrating\ on\ both\ sides\]
\[\int dy\ =\ \int \frac{1}{1\ +\ x^2}\ dx\]
\[\boxed{y\ =\ tan^{-1}\ x\ +\ c}\]
\[\underline{PART\ -\ C}\]
\[5.\ \color{red}{Solve:\ (1\ +y^2)\ dx\ +\ (1\ +\ x^2)\ dy\ =\ 0}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[(1\ +y^2)\ dx\ +\ (1\ +\ x^2)\ dy\ =\ 0\]
\[(1\ +\ x^2)\ dy\ =\ -\ (1\ +\ y^2)\ dx\]
\[\frac{dy}{1\ +\ y^2}\ =\ -\ \frac{dx}{1\ +\ x^2}\]
\[Integrating\ on\ both\ sides\]
\[\int \frac{dy}{1\ +\ y^2}\ =\ -\ \int \frac{dx}{1\ +\ x^2}\]
\[\boxed{tan^{-1}\ y\ =\ -\ tan^{-1}\ x\ +\ c}\]
\[6.\ \color{red}{Solve:\ \frac{dy}{dx}\ =\ \sqrt{\frac{1\ -\ y^2}{1\ -\ x^2}}}\ \hspace{18cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[\frac{dy}{dx}\ =\ \sqrt{\frac{1\ -\ y^2}{1\ -\ x^2}}\]
\[\frac{dy}{dx}\ =\ {\frac{\sqrt{1\ -\ y^2}}{\sqrt{1\ -\ x^2}}}\]
\[\sqrt{1\ -\ x^2}\ dy\ =\ \sqrt{1\ -\ y^2}\ dx\]
\[\frac{dy}{\sqrt{1\ +\ y^2}}\ =\ \frac{dx}{\sqrt{1\ +\ x^2}}\]
\[Integrating\ on\ both\ sides\]
\[\int \frac{dy}{\sqrt{1\ +\ y^2}}\ =\ \int \frac{dx}{\sqrt{1\ +\ x^2}}\]
\[\boxed{sin^{-1}\ y\ =\ sin^{-1}\ x\ +\ c}\]
\[7.\ \color{red}{Solve:\ tan\ x\ sec^2\ y\ dy\ +\ tan\ y\ sec^2\ x\ dx\ =\ 0}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[tan\ x\ sec^2\ y\ dy\ +\ tan\ y\ sec^2\ x\ dx\ =\ 0\]
\[tan\ x\ sec^2\ y\ dy\ =\ -\ tan\ y\ sec^2\ x\ dx\]
\[\frac{sec^2\ y}{tan\ y}\ dy\ =\ -\ \frac{sec^2\ x}{tan\ x}\ dx\]
\[Integrating\ on\ both\ sides\]
\[\int \frac{sec^2\ y}{tan\ y}\ dy\ =\ -\ \int \frac{sec^2\ x}{tan\ x}\ dx\]
\[put\ u\ =\ tan\ y\ \hspace{5cm}\ put\ z\ =\ tan\ x\]
\[du\ =\ sec^2\ y\ dy\ \hspace{5cm}\ dz\ =\ sec^2\ x\ dx\]
\[\int \frac{du}{u}\ =\ -\ \int \frac{dz}{z}\]
\[log\ u\ =\ -\ log\ z\ +\ log\ c\]
\[log\ (tan\ y)\ =\ -\ log\ (tan\ x)\ +\ log\ c\]
\[log\ (tan\ y)\ +\ log\ (tan\ x)\ =\ log\ c\]
\[log\ (tan\ y)\ log\ (tan\ x)\ =\ log\ c\]
\[\boxed{tan\ y\ tan\ x\ =\ c}\]
%d bloggers like this: