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

$c\left(x^2+y^2\right)^{1 / 2}+e^{\tan ^{-1}(y / x)}=0$

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

JEE MATHEMATICS MCQ: DIFFERENTIAL EQUATIONS

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Application +4 | -1

The solution of the equation

\frac{d y}{d x}=\frac{x+y}{x-y}

View answer and explanation

Correct option: B

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

Option B (

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

) is the correct solution, obtained by solving the homogeneous differential equation using the substitution

y=vx

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