Question

Solve the differential equation: $(x+y)dx+(y-x)dy=0$

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

Answer 1

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

Given differential eqn
$(x+y)dx+(y-x)dy=0$
$\frac{dy}{dx}=\frac{x+y}{x-y}$
Put $y=vx$
$\frac{dy}{dx}=v+x\frac{dv}{dx}$
So, the given differential equation becomes
$v+x\frac{dv}{dx}=\frac{1+v}{1-v}$
$x\frac{dv}{dx}=\frac{1+v^2}{1-v}$
$\Rightarrow \frac{(1-v)dv}{1+v^2}=\frac{dx}{x}$
Integrating both sides, we get
$\Rightarrow \int \frac{dv}{1+v^2}-\int \frac{vdv}{1+v^2}=\int \frac{dx}{x}$
Put $1+v^2=t$ in second integral
$\Rightarrow 2vdv=dt$
$\Rightarrow {\tan }^{-1}v--\frac{1}{2}\int \frac{dt}{t}=\log x+\log C$
$\Rightarrow {\tan }^{-1}\frac{y}{x}-\frac{1}{2}\log (1+v^2)=\log x+\log C$
$\Rightarrow {\tan }^{-1}\frac{y}{x}-\frac{1}{2}\log (x^2+y^2)+\log x=\log x+\log C$