Maths

106. Kindly go through the solution

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

105. Given,  $y=5cosx-3sinx$

Differentiating w r t x we get,

$\frac{dy}{dx}=5\frac{d}{dx}cosx-3\frac{d}{dx}sinx$

$-5sinx-3cosx.$

Differentiating again w r t. 'x' we get,

$\frac{d^{2}y}{d{x}^2}=-5\frac{d}{dx}sinx-3\frac{d}{dx}cosx$

$=-5cosx+3sinx$

$=-\left[5cosx-3sinx\right]$

$=-y$

$\Rightarrow \frac{d^{2}y}{d{x}^2}+y=0$ . Hence proved.

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.

104. Let $y=sin\left(logx\right)$

so, $\frac{dy}{dx}=\frac{d}{dx}sin\left(logx\right)=cos\left(logx\right)\frac{d}{dx}logx=\frac{cos\left(logx\right)}{x}$

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{x\frac{d}{dx}cos\left(logx\right)-cos\left(logx\right)\frac{dx}{dx}}{x^2}$

$=\frac{x\left[-sin\left(logx\right)\right]\frac{d}{dx}logx-cos\left(logx\right)}{x^2}$

$=\frac{-\left[x\cdot sin\left(logx\right)*\frac{1}{x}+cos\left(logx\right)\right]}{x^2}$

$=\frac{-\left[sin\left(logx\right)+cos\left(logx\right)\right]}{x^2}$

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.

103. Let $y=log\left(logx\right)$

So,  $\frac{dy}{dx}=\frac{1}{logx}\frac{d}{dx}logx=\frac{1}{xlogx}$

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{xlogx\frac{d}{dx}\left(1\right)-1\cdot \frac{d}{dx}\left(xlogx\right)}{{\left(xlogx\right)}^2}$

$=\frac{-\left[x\frac{d}{dx}logx+logx\frac{dx}{dx}\right]}{{\left[xlogx\right]}^2}$

$=\frac{-\left(x*\frac{1}{x}+logx\right)}{{\left[xlogx\right]}^2}$

$=\frac{-\left(1+logx\right)}{{\left(xlogx\right)}^2}$