Differential Equations

The given D.E. is

$x\frac{dy}{dx}+2y=x^{2}logx$

$\Rightarrow \frac{dy}{dx}+\frac{2}{x}.y=xlogx$ which is of form $\frac{dy}{dx}+Py=Q$

So, $P=\frac{2}{x}\&Q=xlogx$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int \frac{2}{x}dx}}=e^{2logx}=e^{log{x}^2}=x^2$

Thus, the general solution is of the form.

$y*x^2={\displaystyle \int xlogx.x^{2}dx+c}$

$={\displaystyle \int logx}{x}^{3}dx+c$

$\begin{array}{l}=logx{\displaystyle \int x^{3}dx-{\displaystyle \int \frac{d}{dx}logx{\displaystyle \int x^{3}dxdx+c}}}\\ =\frac{logx.x^4}{4}-{\displaystyle \int \frac{1}{4}*\frac{x^4}{4}dx+c}\end{array}$

$\begin{array}{l}=y{x}^2=\frac{x^4}{4}logx-\frac{x^4}{16}+c\\ =y=\frac{x^2}{4}logx-\frac{x^2}{16}+\frac{c}{x^2}\end{array}$

The given D.E.is

$\begin{array}{l}co{s}^{2}x\frac{dy}{dx}+y=tanx\\ =\frac{dy}{dx}+\frac{1}{co{s}^{2}x}y=\frac{tanx}{co{s}^{2}x}\end{array}$

$=\frac{dy}{dx}+se{c}^{2}xy=se{c}^{2}xtanx$ Which is of form $\frac{dy}{dx}+Py=Q$

So, $P=se{c}^{2}x\&Q=se{c}^{2}xtanx$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int se{c}^{2}dx}}=e^{tanx}$

Thus, the general solution is of the form.

$y.e^{tanx}={\displaystyle \int se{c}^{2}xtanx.}{e}^{tanx}dx+c$

Let, $tanx=t=se{c}^{2}xdx=dt$

$\begin{array}{l}=y{e}^t={\displaystyle \int t.}{e}^{t}dt+c\\ =t{\displaystyle \int e^{t}dt-{\displaystyle \int \frac{d}{dt}t{\displaystyle \int e^{t}dt.dt+c}}}\\ =t{e}^t-{\displaystyle \int e^{t}dt+c}\\ =t{e}^t-e^t+c\\ =e^t\left(t-1\right)+c\end{array}$

$\begin{array}{l}\Rightarrow y{e}^{tanx}=e^{tanx}\left(tanx-1\right)+c\\ y=\left(tanx-1\right)+c{e}^{-tanx}\end{array}$

The given D.E.is

$\frac{dy}{dx}+\left(secx\right)y=tanx$ Which is in the form $\frac{dy}{dx}+Py=Q$

So, $P=secx\&Q=tanx$

$\therefore$ $\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int secxdx}}=e^{log\left|secx+tanx\right|}=secx+tanx$

Thus, the general solution is ,

$\begin{array}{l}y*I.F={\displaystyle \int Q*I.Fdx+c}\\ =y*\left(secx+tanx\right)={\displaystyle \int tanx}\left(secx+tanx\right)dx+c\end{array}$

$\begin{array}{l}={\displaystyle \int \left(tanxsecx+ta{n}^{2}x\right)dx+c}\\ ={\displaystyle \int \left(tanxsecx+se{c}^2-1\right)}dx+c\\ =sec+tanx-x+c\end{array}$

$=\left(secx+tanx\right)y=secx+tanx-x+c\left\{?se{c}^{2}x=ta{n}^{2}x+1\right\}$

The given D.E. is

$\frac{dy}{dx}+\frac{y}{x}=x^2$ Which is in the form $\frac{dy}{dx}+Py=Q$

So,  $P=\frac{1}{x}=\&Q=x^2$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int \frac{1}{2}dx}}=e^{logx}=x\left\{?e^{logx}=x\right\}$

Thus, the general solution is

$\begin{array}{l}y*I.F={\displaystyle \int Q*I.Fdx+c}\\ y.x={\displaystyle \int x^2.xdx+c}\\ =xy={\displaystyle \int x^{3}dx+c}\\ =xy=\frac{x^4}{4}+c\end{array}$

Given, D.E. is

$\frac{dy}{dx}+3y=e^{-2x}$ which is of the form

$\frac{dy}{dx}+Py=Q$

Where $\begin{array}{l}P=3\\ \&Q=e^{-2x}\end{array}$

So, I.F $=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int 3dx}}=e^{3x}$

So, the solution is $=y*I.F={\displaystyle \int e^{-2x}\left(I.F\right).dx+c}$

$\begin{array}{l}=y*e^{3x}={\displaystyle \int e^{-2x}.}{e}^{3x}dx+c\\ =e^{3x}y={\displaystyle \int e^{x}dx+c}\\ =e^{3x}y=e^x+c\\ =y=\frac{e^x}{e^{3x}}+\frac{c}{e^{3x}}\\ =y=e^{-2x}+c{e}^{-3x}\end{array}$

Is the required general solution.

The given D.E. is

$\frac{dy}{dx}+2y=2sinx$ which is of form $\frac{dy}{dx}+Px=Q$

We have, P = 2

$Q=sinx$

So, I.F. $=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int 2dx}}=e^{2x}$

The solution is $y*I.F={\displaystyle \int Q*\left(I.F\right)dx+c}$

$\begin{array}{l}y{e}^{2x}={\displaystyle \int sinx{e}^{2x}dx+c}\\ =y{e}^{2x}=I+c---------(1)\\ Where,I={\displaystyle \int sinx{e}^{2x}}\\ =sinx{\displaystyle \int e^{2x}-{\displaystyle \int \left(\frac{d}{dx}sinx{\displaystyle \int e^{2x}dx}\right)dx}}\\ =sinx\frac{e^{2x}}{2}-\frac{1}{2}\left[{\displaystyle \int cosx{e}^{2x}dx}\right]\\ =\frac{e^{2x}sinx}{2}-\frac{1}{2}\left[cos{\displaystyle \int e^{2x}dx-{\displaystyle \int \left(\frac{d}{dx}cosx{\displaystyle \int e^{2x}dx}\right)dx}}\right]\\ =\frac{e^{2x}sinx}{2}-\frac{1}{2}\left[cosx\frac{e^{2x}}{2}\frac{1}{2}{\displaystyle \int \left(-cosx\right)e^{2x}dx}\right]\\ =\frac{e^{2x}sinx}{2}-\frac{e^{2x}cosx}{4}-\frac{1}{4}{\displaystyle \int sinx{e}^{2x}dx}\end{array}$

$\begin{array}{l}=I=e^{2x}\frac{sinx}{2}-e^{2x}\frac{cosx}{4}-\frac{1}{4}1\\ =I+\frac{1}{4}I=2e^{2x}\frac{sinx-e^{2x}cosx}{4}\\ =\frac{5}{4}I=e^{2x}\frac{\left(2sinx-cosx\right)}{4}\\ =I=e^{2x}\frac{\left(2sinx-cosx\right)}{5}\end{array}$

Hence, equation, (1) becomes,

$\begin{array}{l}y{e}^{2x}=\frac{e^{2x}}{5}\left(2sinx-cosx\right)+c\\ =y=\frac{\left(2sinx-cosx\right)}{5}+c{e}^{2x}\end{array}$

Is the required solution.

In (D), each of the terms has a degree 2.

Hence, (D) is homogenous

$\therefore$ Option (D) is correct.

For a homogenous D.E. of the formula $f\left(\frac{y}{x}\right)$

We put,  $\frac{x}{y}=0=x=vy$

$\therefore$ Option (c) is correct.

The given D.E. is

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

i.e, the given is homogenous.

Let, $y=vx$ $=\frac{y}{x}=v$ so that $\frac{dy}{dx}=v+x\frac{dv}{dx}$ is the D.E.

Then, $v+x\frac{dv}{dx}=v+\frac{1}{2}{v}^2$

$\begin{array}{l}=x\frac{dv}{dx}=\frac{1}{2}{v}^2\\ =\frac{dv}{v^2}=\frac{dx}{2x}\end{array}$

Now, $={\displaystyle \int \frac{dv}{v^2}={\displaystyle \int \frac{dx}{2x}}}$

$\begin{array}{l}=\frac{v-2+1}{-2+1}=\frac{1}{2}log\left|x\right|+c\\ =\frac{-1}{v}=\frac{1}{2}log\left|x\right|+c\end{array}$

Putting back $\frac{y}{x}=v$ we get,

$=-\frac{x}{y}=\frac{1}{2}log\left|x\right|+c$

$Given\frac{y}{x}=vwhenx=1$ and y= 2

$\begin{array}{l}=-\frac{1}{2}=\frac{1}{2}log\left|1\right|+c\\ =c=-\frac{1}{2}\end{array}$

$\therefore$ The particular solution is,

$\begin{array}{l}=-\frac{x}{y}=\frac{1}{2}log\left|x\right|-\frac{1}{2}\\ =-\frac{2x}{y}=log\left|x\right|-1\\ =y=\frac{-2x}{log\left|x\right|-1}=\frac{2x}{1-log\left|x\right|}\end{array}$

The given D.E.is

$\begin{array}{l}\frac{dy}{dx}-\frac{y}{x}+cosec\left(\frac{y}{x}\right)=0\\ =\frac{dy}{dx}=\frac{y}{x}-cosec\left(\frac{y}{x}\right)=f\left(\frac{y}{x}\right)\end{array}$

i.e, the given D.E. is homogenous.

Let, $y=vx=\frac{y}{x}=v$ So that, $\frac{dy}{dx}=v+x\frac{dv}{dx}$ in the D.E

Then, $v+x\frac{dv}{dx}=v-cosecv$

$\begin{array}{l}=x\frac{dv}{dx}=-cosecv\\ =\frac{dv}{cosecv}=-\frac{dx}{x}\\ =sinvdv=-\frac{dx}{x}\end{array}$

Integrating both sides we get,

$\begin{array}{l}{\displaystyle \int sinvdv=-{\displaystyle \int \frac{dx}{x}}}\\ =-cosv=-log\left|x\right|+c\\ =cosv=log\left|x\right|-c\end{array}$

Putting back $v=\frac{y}{x}$ we get,

$=cos\frac{y}{x}=log\left|x\right|-c$

Given, $y=0,when,x=1$

$\begin{array}{l}=cos0=log1-c\\ =c=-1\end{array}$

$\therefore$ The required particular solution is

$\begin{array}{l}cos\left(\frac{y}{x}\right)=log\left|x\right|+1=log\left|x\right|+log\left|c\right|\\ =cos\left(\frac{y}{x}\right)=log\left|cx\right|\end{array}$