# FIRST ORDER DIFFERENTIAL EQUATION (Excercise Problems with Solutions)

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