Question

$(x^3-y^3)dx+x{y}^{2}dy=0.$ Solving this we get $\frac{k}{x}=e^{y^m/n{x}^r}$.Find $m+n+r$ ?

Answer 1

$(x^3-y^3)dx+x{y}^{2}dy=0.$
$\frac{dy}{dx}=\left(\frac{y}{x}\right)-{\left(\frac{x}{y}\right)}^2$
Let $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\Rightarrow v+x\frac{dv}{dx}=v-\frac{1}{v^2}$
$\Rightarrow v^{2}dv=-\frac{dx}{x}$
Integrating we get,
$\frac{v^3}{3}=-\ln x+\ln k=\ln \frac{k}{x}=\frac{y^3}{3x^3}\therefore \frac{k}{x}=e^{\frac{y^3}{3x^3}}$