# True-or-False: The solution of $\large\frac{dy}{dx}$$=\bigg(\large\frac{y}{x}\bigg)^\frac{1}{3} is y^{\large\frac{2}{3}}-x^{\large\frac{2}{3}}=c. ## 1 Answer Toolbox: • Differential equation of the type \large\frac{dy}{dx}$$=f(x)$ can be solved by seperating the variables and then integrating on both sides.
• $\int x^n dx=x^{n+1}/n+1+c$
$\large\frac{dy}{dx}=\bigg(\large\frac{y}{x}\bigg)^{1/3}$
$(ie) \large\frac{dy}{dx}=\large\frac{y^{\Large\frac{1}{3}}}{x^{\Large\frac{1}{3}}}$
Seperating the variables we get,
$=> \large\frac{dy}{y^{\Large\frac{1}{3}}}=\large\frac{dx}{x^{\Large\frac{1}{3}}}$
$y^{\Large\frac{-1}{3}}.dy=x^{\Large\frac{-1}{3}}.dx$
Integrating on both sides we get,
$\int y^{\Large\frac{-1}{3}}.dy=\int x^{\Large\frac{-1}{3}}.dx$
$\large\frac{x^{\Large\frac{-1}{3}+1}}{\Large\frac{-1}{3}+1}=\large\frac {y^{\Large\frac{-1}{3}+1}}{\Large\frac{-1}{3}+1}+c$
$=>\Large\frac{x^{\Large\frac{2}{3}}}{\frac{2}{3}}=\large\frac{y^{\Large\frac{2}{3}}}{\Large\frac{2}{3}}+c$
$=>x^{\Large\frac{2}{3}}-y^{\Large\frac{2}{3}}$$=c or =>y^{\Large\frac{2}{3}}-x^{\Large\frac{2}{3}}$$=c$
Hence the statment