Question

The solution of the equation $\frac{dy}{dx}=\frac{x+y}{x-y}$ is

• A. $c{(x^2+y^2)}^{1/2}+e^{{\tan }^{-1}\left(y/x\right)}=0$
• B. $c{(x^2+y^2)}^{1/2}=e^{{\tan }^{-1}\left(y/x\right)}$
• C. $c{(x^2-y^2)}^{}=e^{{\tan }^{-1}\left(2y/x\right)}$
• D. None of these

Answer 1

Answer: (B)

$c{(x^2+y^2)}^{1/2}=e^{{\tan }^{-1}\left(y/x\right)}$

$\frac{dy}{dx}=\frac{x+y}{x-y}$

$\frac{dy}{dx}=\frac{1+y/x}{1-y/x}$

Let $y=Vx$

$\frac{dy}{dx}=V+\frac{xdv}{dx}$

$V+\frac{xdv}{dx}=\frac{1+V}{1-V}$

$\frac{xdv}{dx}=\frac{1+V}{1-V}-V=\frac{1+V-V+V^2}{1-V}=\frac{1+V^2}{1-V}$

$\frac{xdv}{dx}=\frac{1+V^2}{1-V}\Rightarrow \frac{1-V}{1+V^2}dv=\frac{dx}{x}$

$\int \frac{dv}{1+v^2}-\int \frac{1}{2}\left(\frac{2V}{1+V^2}\right)dv=\int \frac{dx}{x}$

${\tan }^{-1}V-\frac{1}{2}\int \frac{2v}{1+v^2}dv=\ln x+e\ln c$

${\tan }^{-1}\frac{y}{x}-\frac{1}{2}\ln (x^2+y^2)+\frac{1}{2}2x\ln x=\ln x+e\ln c$

${\tan }^{-1}\frac{y}{x}=e\ln c(x^2+y^2{)}^{1/2}$

$C(x^2+y^2{)}^{1/2}=e^{{\tan }^{-1}\left(y/x\right)}$