Question

The solution of $(y+x+5)dy=(y-x+1)dx$ is
(where $C$ is integration constant)

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

Answer 1

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

The intersection of $y-x+1=0$ and $y+x+5=0$ is $(-2,-3).$ Put $x=X-2,y=Y-3.$
The given equation reduces to $\frac{dY}{dX}=\frac{Y-X}{Y+X}.$
Putting $Y=vX,$ we get
$X\frac{dv}{dX}=-\frac{v^2+1}{v+1}$
$\Rightarrow (-\frac{v}{v^2+1}-\frac{1}{v^2+1})dv=\frac{dX}{X}$
$\Rightarrow -\frac{1}{2}\ln |v^2+1|-{\tan }^{-1}v=\ln \left|X\right|+k,k$is constant of integration
$\Rightarrow \ln |Y^2+X^2|+2{\tan }^{-1}\frac{Y}{X}=C,$Here $C=-2k$
$\Rightarrow \ln |{(y+3)}^2+{(x+2)}^2|+2{\tan }^{-1}\frac{y+3}{x+2}=C$