# FIRST ORDER DIFFERENTIAL EQUATION

### Introduction:

$\hspace{3cm}\ Since\ the\ time\ of\ Newton,\ physical\ problems\ have\ been$$investigated\ by\ formulating\ them\ mathematically\ as\ differential\ equations.$$Many\ mathematical\ models\ in\ engineering\ employ\ differential\ equations.\ extensively.$
$\hspace{3cm}\ In\ the\ first\ order\ differential\ equation,\ say \frac{dy}{dx}\ =\ f (x, y),$$it\ is\ sometimes\ possible\ to\ group\ function\ of\ x\ with\ dx\ on\ one\ side$$and\ function\ of\ y\ with\ dy\ on\ the\ other\ side.$$This\ type\ of\ equation\ is\ called\ variables\ separable\ differential\ equations.$$The\ solution\ can\ be\ obtained\ by\ integrating\ both\ sides\ after\ separating\ the\ variables.$

Example  1:

$1.\ \color{red}{Solve:\ xdy\ +\ ydx\ =\ 0}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$xdy\ =\ -\ ydx$
$\frac{dy}{y}\ =\ -\ \frac{dx}{x}$
$Integrating\ on\ both\ sides$
$\int \frac{dy}{y}\ =\ -\ \int \frac{dx}{x}$
$log\ y\ =\ -\ log\ x\ +\ log\ c$
$log\ y\ +\ log\ x\ =\ log\ c$
$\boxed{xy\ =\ c}$

Example  2:

$\color{red}{Solve:\ xdx\ +\ ydy\ =\ 0}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$xdx\ =\ -\ ydy$
$Integrating\ on\ both\ sides$
$\int x\ dx\ =\ -\ \int y\ dy$
$\boxed{\frac{x^2}{2}\ =\ -\frac{y^2}{2}\ +\ c}$

#### Example  3:

$\color{red}{Solve:\ \frac{dy}{dx}\ =\ e^x}\ \hspace{18cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\frac{dy}{dx}\ =\ e^x$
$dy\ =\ e^x\ dx$
$Integrating\ on\ both\ sides$
$\int dy\ =\ \int e^x\ dx$
$\boxed{y\ =\ e^x\ +\ c}$

Example  4:

$\color{red}{Solve:\ \frac{dy}{dx}\ =\ y\ e^x}\ \hspace{18cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\frac{dy}{dx}\ =\ y\ e^x$
$\frac{dy}{y}\ =\ e^x\ dx$
$Integrating\ on\ both\ sides$
$\int \frac{dy}{y}\ =\ \int e^x\ dx$
$\boxed{log\ y\ =\ e^x\ +\ c}$

#### Example  5:

$\color{red}{Solve:\ \frac{dy}{dx}\ =\ \frac{x}{1\ +\ x^2}}\ \hspace{18cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\frac{dy}{dx}\ =\ \frac{x}{1\ +\ x^2}$
$(1\ +\ x^2)dy\ =\ x\ dx$
$dy\ =\ \frac{x}{1\ +\ x^2}\ dx$
$Integrating\ on\ both\ sides$
$\int dy\ =\ \int \frac{x}{1\ +\ x^2}\ dx$
$put\ u\ =\ 1\ +\ x^2$
$du\ =\ 2x\ dx$
$\frac{1}{2}du\ =\ x\ dx$
$\int dy\ =\ \int \frac{1}{2} \frac{du}{u}$
$y\ =\ \frac{1}{2} \int \frac{du}{u}$
$=\ \frac{1}{2} log\ u\ +\ c$
$\boxed{y\ =\ \frac{1}{2} log\ (1\ +\ x^2)\ +\ c}$

Example  6:

$\color{red}{Solve:\ \frac{dy}{dx}\ =\ \frac{1\ +\ y^2}{1\ +\ x^2}}\ \hspace{18cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\frac{dy}{dx}\ =\ \frac{1\ +\ y^2}{1\ +\ x^2}$
$(1\ +\ x^2)dy\ =\ (1\ +\ y^2)dx$
$\frac{dy}{1\ +\ y^2}\ =\ \frac{dx}{1\ +\ x^2}$
$Integrating\ on\ both\ sides$
$\int \frac{dy}{1\ +\ y^2}\ =\ \int \frac{dx}{1\ +\ x^2}$
$\boxed{tan^{-1}\ y\ =\ tan^{-1}\ x\ +\ c}$

Example  7:

$\color{red}{Solve:\ \frac{dy}{dx}\ +\ \frac{1\ +\ x^2}{1\ +\ y^2}\ =\ 0}\ \hspace{18cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\frac{dy}{dx}\ =\ -\ \frac{1\ +\ x^2}{1\ +\ y^2}$
$(1\ +\ y^2)dy\ =\ -\ (1\ +\ x^2)dx$
$Integrating\ on\ both\ sides$
$\int (1\ +\ y^2)dy\ =\ -\ \int (1\ +\ x^2)dx$
$\boxed{y\ +\ \frac{y^3}{3}\ =\ -\ x\ -\ \frac{x^3}{3}\ +\ c}$

Example  8:

$\color{red}{Solve:\ \frac{dy}{dx}\ =\ e^{x\ -\ y}\ +\ 3\ x^2\ e^{-y}}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$\frac{dy}{dx}\ =\ e^{x\ -\ y}\ +\ 3\ x^2\ e^{-y}$
$\frac{dy}{dx}\ =\ e^x\ e^{-y}\ +\ 3\ x^2\ e^{-y}$
$\frac{dy}{dx}\ =\ e^{-y}(e^x\ +\ 3\ x^2)$
$\frac{dy}{e^{-y}}\ =\ (e^x\ +\ 3\ x^2)dx$
$e^{y}\ dy\ =\ (e^x\ +\ 3\ x^2)dx$
$Integrating\ on\ both\ sides$
$\int e^y\ dy\ =\ \int (e^x\ +\ 3\ x^2)\ dx$
$e^y\ =\ e^x\ +\ 3\ \frac{x^3}{3}\ +\ c$
$\boxed{e^y\ =\ e^x\ +\ x^3\ +\ c}$

Example  9:

$\color{red}{Solve:\ (1\ +\ e^x)\ sec^2\ y\ dy\ -\ e^x\ tan\ y\ dx\ =\ 0}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$(1\ +\ e^x)\ sec^2\ y\ dy\ -\ e^x\ tan\ y\ dx\ =\ 0$
$(1\ +\ e^x)\ sec^2\ y\ dy\ =\ \ e^x\ tan\ y\ dx$
$\frac{sec^2\ y}{tan\ y}\ dy\ =\ \frac{e^x}{1\ +\ e^x}\ dx$
$Integrating\ on\ both\ sides$
$\int \frac{sec^2\ y}{tan\ y}\ dy\ =\ \int \frac{e^x}{1\ +\ e^x}\ dx$
$put\ u\ =\ tan\ y\ \hspace{5cm}\ put\ z\ =\ 1\ +\ e^x$
$du\ =\ sec^2\ y\ dy\ \hspace{5cm}\ dz\ =\ e^x\ dx$
$\int \frac{du}{u}\ =\ \int \frac{dz}{z}$
$log\ u\ =\ log\ z\ +\ log\ c$
$log\ (tan\ y)\ =\ log\ (1\ +\ e^x)\ +\ log\ c$
$log\ (tan\ y)\ =\ log\ (1\ +\ e^x)\ c$
$tan\ y\ =\ (1\ +\ e^x)\ c$

Example  9:

$\color{red}{Solve:\ (1\ +\ e^y)\ sec^2\ x\ dx\ +\ 5\ e^y\ tan\ x\ dy\ =\ 0}\ \hspace{15cm}$
$\color {blue}{Soln:}\ \hspace{20cm}$
$(1\ +\ e^y)\ sec^2\ x\ dx\ +\ 5\ e^y\ tan\ x\ dy\ =\ 0$
$(1\ +\ e^y)\ sec^2\ x\ dx\ =\ -\ 5\ e^y\ tan\ x\ dy$
$-\ 5\ e^y\ tan\ x\ dy\ =\ (1\ +\ e^y)\ sec^2\ x\ dx$
$\frac{e^y}{1\ +\ e^y}\ dy\ =\ -\ \frac{1}{5} \frac{sec^2\ x}{tan\ x}\ dx$
$Integrating\ on\ both\ sides$
$\int \frac{e^y}{1\ +\ e^y}\ dy\ =\ -\ \frac{1}{5} \int \frac{sec^2\ x}{tan\ x}\ dx$
$put\ u\ =\ 1\ +\ e^y\ \hspace{5cm}\ put\ z\ =\ tan\ x$
$du\ =\ e^y\ dy\ \hspace{5cm}\ dz\ =\ sec^2\ x\ dx$
$\int \frac{du}{u}\ =\ -\ \frac{1}{5} \int \frac{dz}{z}$
$log\ u\ =\ -\ \frac{1}{5}\ log\ z\ +\ log\ c$
$log\ u\ =\ \ log\ z^\frac{-1}{5}\ +\ log\ c$
$log\ u\ =\ \ log\ z^\frac{-1}{5}\ c$
$u\ =\ \ z^\frac{-1}{5}\ c$
$1\ +\ e^y\ =\ (tan\ x)^\frac{-1}{5}\ c$