Question

The general solution of differential equation $y(x^{2}y+e^x)dx-e^{x}dy=0$ is:

• A. $x^{3}y-3e^x=cy$
• B. $x^{3}y+3e^x=cy$
• C. $y^{3}x-3e^y=cx$
• D. $y^{3}x+3e^y=cx$

Answer 1

The correct option is B $x^{3}y+3e^x=cy$

We have $y(x^{2}y+e^x)dx-e^{x}dy=0$

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

Dividing by $y^2{e}^x$, we get $\frac{1}{y^2}\frac{dy}{dx}-\frac{1}{y}=x^2{e}^{-x}$

Substitute $\frac{1}{y}=V$ So that $\frac{-1}{y^2}\frac{dy}{dx}=\frac{dV}{dx}$ .

We thus have $\frac{dV}{dx}+V=-x^2{e}^{-x}$, which is linear.

$\therefore I.F.=e^{\int 1dx}=e^x$

Hence the solution is
$V.e^x=-\int x^2{e}^{-x}.e^{x}dx+\frac{c}{3}$
or $\frac{1}{y}{e}^x=-\frac{x^3}{3}+\frac{c}{3}$ or $x^{3}y+3e^x=cy$