Question

Solve: $\frac{dy}{dx}(x^2{y}^3+xy)=1$

Answer 1

Given,

$\frac{dy}{dx}(x^2{y}^3+xy)=1$

$\frac{dy}{dx}=\frac{1}{x^2{y}^3+xy}$

$\frac{dx}{dy}=x^2{y}^3+xy$

$\frac{dx}{dy}-xy=x^2{y}^3$

$\frac{1}{x^2}\frac{dx}{dy}-\frac{y}{x}=y^3$

substitute $\frac{1}{x}=u$

$\frac{du}{dy}=-\frac{1}{x^2}\frac{dx}{dy}$

$-\frac{du}{dy}-uy=y^3$

$\frac{du}{dy}+uy=-y^3$

$I.F=e^{\int ydy}=e^{\frac{y^2}{2}}$

$u\times e^{\frac{y^2}{2}}=-\int y^3{e}^{\frac{y^2}{2}}dy$

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

$=(2-y^2)e^{\frac{y^2}{2}}+c$

$\Rightarrow x(2-y^2)+cx{e}^{-\frac{y^2}{2}}=1$