Question

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

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

Answer 1

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

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
$X=\frac{dv}{dX}=-\frac{v^2+1}{v+1}$
$\Rightarrow (-\frac{v}{v^2+1}-\frac{1}{v^2+1})dv=\frac{dX}{X}$
$\Rightarrow \frac{1}{2}log(v^2+1)-ta{n}^{-1}v=log|X|+constant\Rightarrow log(Y^2+X^2)+ta{n}^{-1}\frac{Y}{X}=constant\Rightarrow log(Y^2+X^2)+2ta{n}^{-1}\frac{Y}{X}=constant\Rightarrow log(y+3{)}^2)+(x+2{)}^{)}+2ta{n}^{-1}\frac{y+3}{x+2}=C$