Question

The solution of the D.E $ydx-xdy+3x^2{y}^2{e}^{x^3}dx=0$ is

• A. $x+e^{x^3}=cy$
• B. $x+y{e}^{x^3}-cy=0$
• C. $x-y{e}^{x^3}+c=0$
• D. $x+y{e}^{x^3}+c{y}^2=0$

Answer 1

Answer: (B)

$x+y{e}^{x^3}-cy=0$

$ydx-xdy+3x^2{y}^2{e}^{x^3}dx=0$

$\Rightarrow 3x^2{y}^2{e}^{x^3}dx=xdy-ydx$

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

$\Rightarrow 3y^2{e}^{x^3}=\frac{1}{x}\frac{dy}{dx}{-}^y{/}_{x^2}$

$\Rightarrow 3y^2{e}^{x^3}=\frac{d}{dx}{}^y{/}_x$

$\Rightarrow 3x^2{e}^{x^3}(\frac{y}{x}{)}^2=\frac{d}{dx}{(}^y{/}_x)$

$\Rightarrow \int 3x^2{e}^{x^3}dx=\int \frac{d\left(\frac{y}{x}\right)}{(\frac{y}{x}{)}^2}+c$

$\Rightarrow e^{x^3}=\frac{-1}{{}^y{/}_x}+c$

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

$\Rightarrow x+y{e}^{x^3}-cy=0$