Question

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

Answer 1

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

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

$\Rightarrow \frac{dx}{dy}-xy=x^2{y}^3.....\left(1\right)$

hence $P=-y,Q=y^3,x^n=x^2$

Dividing by $\frac{1}{n^2}$ both side of eqn $\left(1\right)$

$\Rightarrow \frac{1}{n^2}\frac{dx}{dy}=\frac{1}{x}y=y^3.......\left(2\right)$

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

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

$\Rightarrow \frac{1}{x^2}\frac{dx}{dy}=-\frac{du}{dy}.....\left(3\right)$

Putting eqn $\left(3\right)$ in $\left(2\right)$ we have

$-\frac{du}{dy}-uy=y^2\Rightarrow \frac{du}{dx}+uy=-y^3$

$I.F=e^{\int ydy}=e^{y^2/2}dy$ Let $y^2/2=t2tdy=dt$

$u.e^{y^2/2}=\int -y^3.e^{y^2/2}dy$

$\frac{e^{y^2/2}}{x}=-\int 2t.e^{t}dt$

$e^{y^2/2}=-2x\left(\int t\int e^{t}dt-\int (\frac{dt}{dt}\int e^{f}dt)dt\right)$

$e^{y^2/2}=-2x(t{e}^1-e^t+c)$

$\Rightarrow e^{y^2/2}=-2x(y^2/2.e^{y^2/2}+C)$ (Ans)