Question

The solution of $(y+x+5)dy=(y-x+1)dx$ is:

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

Answer 1

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

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

$\frac{dY}{dX}=\frac{Y-X}{Y+X}$.

This is a homogeneous equation, so putting $Y=vX$, we get

$X\frac{dv}{dX}=-\frac{v^2+1}{v+1}\Rightarrow (-\frac{v}{v^2+1}-\frac{1}{v^2+1})dv=\frac{dX}{X}$

integrating on both sides

$\Rightarrow -\frac{1}{2}\log (v^2+1)-{\tan }^{-1}v=\log \left|X\right|+$ Const.

$\Rightarrow \log (Y^2+X^2)+2{\tan }^{-1}\frac{Y}{X}=$ Const.

$\Rightarrow \log ({(y+3)}^2+{(x+2)}^2)+2{\tan }^{-1}\frac{y+3}{x+2}=C$.