Question

$(x^2-y^2)\frac{dy}{dx}=xy$. Solve the above equation.

• A. $e^{-x^2/2y^2}=kx$
• B. $e^{-x^2/2y^2}=ky$
• C. $e^{x^2/2y^2}=kx$
• D. $e^{x^2/2y^2}=ky$

Answer 1

The correct option is B $e^{-x^2/2y^2}=ky$

Given, $\frac{dy}{dx}=\frac{xy}{x^2-y^2}$
Let $y=vx$
Then $\frac{dy}{dx}=v+\frac{xdv}{dx}$
Hence, $v+x\frac{dv}{dx}=\frac{v{x}^2}{x^2-v{x}^2}$
$\frac{xdv}{dx}=\frac{v}{1-v^2}-v$
$\frac{xdv}{dx}=\frac{v-v+v^3}{1-v^2}$
$\frac{xdv}{dx}=\frac{v^3}{1-v^2}$
$\frac{1-v^2}{v^3}.dv=\frac{dx}{x}$
$v^{-3}-\frac{1}{v}.dv=\frac{1}{x}dx$
$\int v^{-3}-\frac{1}{v}=\int \frac{dx}{x}$
$\frac{-v^{-2}}{2}-\ln v=\ln x+\ln C$
$\frac{-v^{-2}}{2}=\ln v+\ln x+\ln C$
$\frac{-v^{-2}}{2}=\ln cvx$
$e^{-\frac{1}{2v^2}}=cvx$
Now $y=vx$
Hence, $e^{-\frac{x^2}{2y^2}}=cy$.