Differential Equations

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

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

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 $=y\times I.F={\displaystyle \int e^{-2x}\left(I.F\right).dx+c}$

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

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 $y\times I.F={\displaystyle \int Q\times \left(I.F\right)dx+c}$

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

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

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

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

The given D.E.is

$\begin{array}{l}\left[xsi{n}^2\left(\frac{y}{x}\right)-y\right]dx+xdy=0\\ =\left[xsi{n}^2\left(\frac{y}{x}\right)-y\right]dx=-xdy\\ =\frac{dy}{dx}=\frac{\left[xsi{n}^2\left(\frac{y}{x}\right)-y\right]}{-x}\\ =-\left[si{n}^2\left(\frac{y}{x}\right)-\frac{y}{x}\right]=f\left(\frac{y}{x}\right)\end{array}$

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

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

$\begin{array}{l}=v+\frac{dv}{dx}=-\left[si{n}^{2}v-v\right]=v-si{n}^{2}v\\ =\frac{dv}{dx}=-si{n}^{2}v\\ =\frac{dv}{si{n}^{2}v}=-dx\end{array}$

Integrating both sides we get,

$\begin{array}{l}={\displaystyle \int cose{c}^{2}vdv={\displaystyle \int -dx}}\\ =-cotv=-log\left|x\right|+c\\ =cotv=log\left|x\right|-c\end{array}$

Putting back $v=\frac{y}{x}$ we have,

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

Then, $y=\frac{\pi }{4}$ when, $x=1$

$\begin{array}{l}cot\frac{\pi }{4}=log\left|1\right|-c\\ =c=-1\end{array}$

$\therefore$ The required particular solution is,

$\begin{array}{l}cot\frac{y}{x}=log\left|x\right|+1=log\left|x\right|+logc\left\{?loge=1\right\}\\ =cot\frac{y}{x}=log\left|ex\right|\end{array}$

The given D.E. is

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

i.e, the D.E is homogenous.

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

Then, $v+x\frac{dv}{dx}=-\left[v+v^2\right]=-v-v^2$

$\begin{array}{l}=x\frac{dv}{dx}=-2v-v^2=-v\left(2+v\right)\\ =\frac{dv}{v\left(2+v\right)}=-\frac{dx}{x}\end{array}$

Integrating both sides we get,

$\begin{array}{l}{\displaystyle \int \frac{dv}{v\left(2+v\right)}=-{\displaystyle \int \frac{dx}{x}}}\\ =\frac{1}{2}{\displaystyle \int \frac{2dv}{2\left(v+2\right)}=}-{\displaystyle \int \frac{dx}{x}}\\ =\frac{1}{2}{\displaystyle \int \frac{v+2-v}{v\left(v+2\right)}}dv={\displaystyle \int -\frac{dx}{x}}\\ =\frac{1}{2}\left\{{\displaystyle \int \frac{1}{v}dv-\frac{1}{v+2}dv}\right\}={\displaystyle \int -\frac{dx}{x}}\end{array}$

$\begin{array}{l}=\frac{1}{2}\left[logv-log\left|v+2\right|\right]=-logx+logc\\ =\frac{1}{2}log\left(\frac{v}{v+2}\right)=log\frac{c}{x}\\ =log{\left(\frac{v}{v+2}\right)}^{\frac{1}{2}}=log\frac{c}{x}\\ ={\left(\frac{v}{v+2}\right)}^{\frac{1}{2}}=\frac{c}{x}\\ =\frac{v}{v+2}={\left(\frac{c}{x}\right)}^2\end{array}$

Putting back $v=\frac{y}{x}$ we get,

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

Given, y = 1 when x = 1

So, $=\frac{1}{1+2}=c^2=c^2=\frac{1}{3}$

Hence, the required particular solution is,

$\begin{array}{l}=\frac{x^{2}y}{y+2x}=\frac{1}{3}\\ =3x^{2}y=y+2x\end{array}$

The given D.E. is

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

i.e, homogenous

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

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

$\begin{array}{l}=x\frac{dv}{dx}=\frac{v-1}{v+1}-v=\frac{v-1-v^2-v}{v+1}=-\frac{\left(v^2+1\right)}{v+1}\\ =\left[\frac{v+1}{v^2+1}\right]dv=-\frac{dx}{x}\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int \frac{v+1}{v^2+1}dv={\displaystyle \int -\frac{dx}{x}}}\\ =\frac{1}{2}{\displaystyle \int \frac{2v}{v^2+1}}dv+{\displaystyle \int \frac{1}{v^2+1}dv=-logx+c}\\ =\frac{log\left|v^2+1\right|}{2}+ta{n}^{-1}v=-logx+c\end{array}$

Putting back $v=\frac{y}{x}$ we get,

$\begin{array}{l}=\frac{1}{2}log\left|\frac{y^2}{x^2}+1\right|+ta{n}^{-1}\frac{y}{x}=-logx+c\\ =\frac{1}{2}\left[log\left(y^2+x^2\right)-log{x}^2\right]+ta{n}^{-1}\frac{y}{x}+logx=c\\ =\frac{1}{2}log\left(y^2+x^2\right)-\frac{1}{2}log{x}^2+logx+ta{n}^{-1}\frac{y}{x}=c\\ =\frac{1}{2}log\left(y^2+x^2\right)-logx+logx+ta{n}^{-1}\frac{y}{x}=c\\ =\frac{1}{2}log\left(y^2+x^2\right)+ta{n}^{-1}\frac{y}{x}=c\end{array}$

Given, $y=1,when,x=1$

So, $=\frac{1}{2}log\left(1^2+1^2\right)+ta{n}^{-1}\frac{1}{1}=c$

$=\frac{1}{2}log2+\frac{\pi }{4}=c$

Hence, the particular solution is

$\frac{1}{2}log\left(1^2+1^2\right)+ta{n}^{-1}\frac{y}{x}=\frac{1}{2}log2+\frac{\pi }{4}$