Question

The solution of the differential equation $\frac{dy}{dx}=\frac{2(y+2{)}^2}{(x+y-1{)}^2}$ is :
(where $C$ is integration constant)

• A. $\ln |y+2|+2{\tan }^{-1}\left(\frac{y+2}{x-3}\right)=C$
• B. $\ln |y-2|+2{\tan }^{-1}\left(\frac{y-2}{x-3}\right)=C$
• C. $\ln |y+2|+{\tan }^{-1}\left(\frac{y+2}{x-3}\right)=C$
• D. $\ln |y-2|+{\tan }^{-1}\left(\frac{y-2}{x-3}\right)=C$

Answer 1

The correct option is A $\ln |y+2|+2{\tan }^{-1}\left(\frac{y+2}{x-3}\right)=C$

Given:
$\frac{dy}{dx}=\frac{2(y+2{)}^2}{(x+y-1{)}^2}$
Put $x=X+h$ and $y=Y+k$
$\frac{dY}{dX}=\frac{2(Y+k+2{)}^2}{(X+h+Y+k-1{)}^2}$
For homogenous eqn such that, $h+k-1=0$ and $k+2=0$
$\therefore k=-2\text{and}h=3$
$\therefore \frac{dY}{dX}=\frac{2Y^2}{(X+Y{)}^2}=\frac{2{\left(\frac{Y}{X}\right)}^2}{{(1+\frac{Y}{X})}^2}$
Putting $Y=vX\text{and}\frac{dY}{dX}=v+X\frac{dv}{dX}$
We have
$v+X\frac{dv}{dX}=\frac{2v^2}{(1+v{)}^2}$
$\Rightarrow X\frac{dv}{dX}=\frac{2v^2-v(1+v^2+2v)}{(1+v{)}^2}=-\frac{v(1+v^2)}{(1+v{)}^2}$
$\Rightarrow \int -\frac{(1+v{)}^2}{v(1+v^2)}dv=\int \frac{dX}{X}$
$\Rightarrow -\int \frac{1+v^2+2v}{v(1+v^2)}dv=\ln |X|+c$
$\Rightarrow -\int (\frac{1}{v}+\frac{2}{(1+v^2)})dv=\ln |X|+c$
$\Rightarrow -\ln |v|-2{\tan }^{-1}v=\ln |X|+c$
$\Rightarrow \ln |Y|+2{\tan }^{-1}\frac{Y}{X}=C(\text{let}C=-c)$
$\Rightarrow \ln |y+2|+2{\tan }^{-1}\left(\frac{y+2}{x-3}\right)=C$