Differential Equations

Given, $e^{x}tanydx+\left(1-e^x\right)se{c}^{2}ydy=0$

Dividing throughout by $\left(1-e^x\right)tany$ we get,

$\begin{array}{l}\frac{e^{x}tany}{\left(1-e^x\right)tany}dx+\frac{\left(1-e^x\right)se{c}^{2}y}{\left(1-e^x\right)tany}dy=0\\ =\frac{e^x}{1-e^x}dx+\frac{se{c}^{2}y}{tany}dy=0\end{array}$

Integrating both sides

$\begin{array}{l}={\displaystyle \int \frac{-e^x}{1-e^x}dx+{\displaystyle \int \frac{se{c}^{2}y}{tany}dy=clogc}}\\ =-log\left|1-e^x\right|+log\left|tany\right|=clogc\\ =log\frac{tany}{1-e^x}=logc\\ =\frac{tany}{1-e^x}=c\end{array}$

$=tany=\left(1-e^x\right)c$ is the general solution.

Given,  $\frac{dy}{dx}=si{n}^{-1}x$

$\Rightarrow dy=si{n}^{-1}xdx$

Integrating

Given,  $x^5\frac{dy}{dx}=-y^5$

$\Rightarrow \frac{dy}{y^5}=-\frac{dx}{x^5}$

Integrating both sides

$\begin{array}{l}{\displaystyle \int \frac{dy}{y^5}=-{\displaystyle \int \frac{dx}{x^5}}}\\ \Rightarrow {\displaystyle \int y^{-5}dy=-{\displaystyle \int x^{-5}dx}}\end{array}$

$\begin{array}{l}\Rightarrow \frac{y^{-5+1}}{\left(-5+1\right)}=-\frac{x^{-5+1}}{\left(-5+1\right)}+c\\ \Rightarrow \frac{1}{-4y^4}=\frac{1}{4x^4}+c\end{array}$

$\Rightarrow \frac{1}{y^4}=\frac{1}{x^4}+4c$ is the general solution.

Given, $ylogydx-xdy=0$

$\begin{array}{l}\Rightarrow ylogydx=xdy\\ \Rightarrow \frac{dy}{ylogy}=\frac{dx}{x}\end{array}$

Integration both sides,

${\displaystyle \int \frac{dy}{ylogy}={\displaystyle \int \frac{dx}{x}}}$

Put log $y=t\Rightarrow \frac{1}{y}=\frac{dt}{dy}\Rightarrow \frac{dy}{y}=dt$

Hence, ${\displaystyle \int \frac{dt}{t}={\displaystyle \int \frac{dx}{x}}}$

$\begin{array}{l}\Rightarrow log\left|t\right|=log\left|x\right|+log\left|c\right|=log\left|xc\right|\\ \Rightarrow t=\pm xc\end{array}$

$\Rightarrow logy=ax$ where $a=\pm c$

$\Rightarrow y=e^{ax}$ is the general solution.

Given,  $\frac{dy}{dx}=\left(1+x^2\right)\left(1+y^2\right)$

$\Rightarrow \frac{dy}{\left(1+y^2\right)}=\left(1+x^2\right)dx$

Integrating both sides

$\begin{array}{l}{\displaystyle \int \frac{dy}{\left(1+y^2\right)}dy={\displaystyle \int \left(x^2+1\right)dx}}\\ \Rightarrow ta{n}^{-1}\frac{y}{1}=\frac{x^3}{3}+x+c\end{array}$

$\Rightarrow ta{n}^{-1}y=\frac{x^3}{3}+x+c$ is the general solution.

Given,  $\left(e^x+e^{-x}\right)dy-\left(e^x-e^{-x}\right)dx=0$

$\begin{array}{l}\Rightarrow \left(e^x+e^{-x}\right)dy=\left(e^x-e^{-x}\right)dx\\ \Rightarrow dy=\frac{e^x-e^{-x}}{e^x+e^{-x}}dx\end{array}$

Integrating both sides

$\Rightarrow {\displaystyle \int dy=}\frac{e^x-e^{-x}}{e^x+e^{-x}}dx\left\{?{\displaystyle \int \frac{f^{|}\left(x\right)}{f\left(x\right)}}dx=log\left|x\right|\right\}$

$\Rightarrow y=log\left|e^x+e^{-x}\right|+c$ is the required general solution.

Given, $se{c}^{2}xtanydx+se{c}^{2}ytanxdy=0$

Dividing throughout by ' $tanxtany$ ' we get,

$\begin{array}{l}\frac{se{c}^{2}xtany}{tanxtany}dx+\frac{se{c}^{2}ytanx}{tanxtany}dy=0\\ \Rightarrow \frac{se{c}^{2}x}{tanx}dx+\frac{se{c}^{2}y}{tany}dy=0\end{array}$

Integrating both sides we get,

$\begin{array}{l}{\displaystyle \int \frac{se{c}^{2}x}{tanx}dx+{\displaystyle \int \frac{se{c}^{2}y}{tany}dy=logc}}\\ \Rightarrow log\left|tanx\right|+log\left|tany\right|=logc\left\{{\displaystyle \int \frac{f^{|}\left(x\right)}{f\left(x\right)}}dx-log\left|f\left(x\right)\right|\right\}\\ \Rightarrow log\left|\left(tanx+tany\right)\right|=logc\end{array}$

$\Rightarrow tanxtany=\pm c$ is the required general solution.

Given,  $\frac{dy}{dx}+y=1$

$\Rightarrow \frac{dy}{dx}=1-y=-\left(y-1\right)$

By separable of variable,

$\frac{dy}{\left(y-1\right)}=-dx$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{dy}{\left(y-1\right)}=-{\displaystyle \int dx}}\\ \Rightarrow log\left|y-1\right|=-x+c\\ \Rightarrow \left|y-1\right|=e^{-x+c}\\ \Rightarrow y-1=\pm e^{-x}.e^c\end{array}$

$\Rightarrow y=1+Ac$ where $A=\pm e^c$

Is the general solution.

Kindly go through the solution

The given D.E is

$\frac{dy}{dx}=\frac{1-cosx}{1+cosx}$

By separable of variable,

$\begin{array}{l}\Rightarrow dy=\frac{1-cosx}{1+cosx}dx\left\{\begin{array}{l}cos2x=1-2si{n}^{2}x\\ =2si{n}^{2}x=1-cos2x\\ =2si{n}^2\frac{x}{2}=1-cosx\\ cos2x=2co{s}^{2}x-1\end{array}\right\}\\ \Rightarrow dy=\frac{2si{n}^2\frac{x}{2}}{2co{s}^2\frac{x}{2}}dx\\ \Rightarrow dy=ta{n}^2\frac{x}{2}dx\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int dy={\displaystyle \int ta{n}^2\frac{x}{2}dx\left\{se{c}^{2}x=1+ta{n}^2\right\}}}\\ \Rightarrow y={\displaystyle \int \left(se{c}^2\frac{x}{2}-1\right)}dx\end{array}$

$\Rightarrow y=\frac{tan\frac{x}{2}}{\frac{1}{2}}-x+c$ c = constant

$\Rightarrow y=2tan\frac{x}{2}-x+c$ is the general solution.