Question

Solve the differential equation: $ydx-xdy+3x^2{y}^2{e}^{x^3}dx$

• A. $\frac{x}{y}=e^{x^3}+c.$
• B. $-\frac{x}{y}=e^{x^{-3}}+c.$
• C. $-\frac{x}{y}=e^{x^3}+c.$
• D. $\frac{x}{y}=e^{x^{-3}}+c.$

Answer 1

The correct option is C $-\frac{x}{y}=e^{x^3}+c.$

Given, $ydx=xdy+3x^2{y}^2{e}^{x^3}dx$
$\Rightarrow x\frac{dy}{dx}-y=3x^2{y}^2{e}^{x^3}$

$\Rightarrow \frac{1}{y^2}\frac{dy}{dx}-\frac{1}{x}\frac{1}{y}=3e^{x^3}$
Put $-\frac{1}{y}=v\Rightarrow \frac{+1}{y^2}dy=dv$
$\therefore \frac{dv}{dx}+\frac{v}{x}=3e^{x^3}$ ...(1)
Here, $P=\frac{1}{x}$

$\Rightarrow \int Pdx=\int \frac{1}{x}dx=\log x$
$\therefore$ $I.F.=e^{\log x}=x$
Multiplying (1) by $I.F.$, we get,
$x\frac{dv}{dx}+v=3x^2{e}^{x^3}$
Integrating both sides we get,
$xv=e^{x^3}+C$

$\Rightarrow -\frac{x}{y}=e^{x^3}+C$