Question

The solution of differential equation $\frac{dy}{dx}-\frac{1}{x}+e^y\cdot x=0,y(1)=0$

• A. $e^y\left(x^3+3\right)=4x$
• B. $e^y\left(x^3+2\right)=3x$
• C. $e^y\left(x^2+2\right)=3x$
• D. $e^y\left(x^2+3\right)=4x$

Answer 1

Answer: (B)

$\frac{dy}{dx}-\frac{1}{x}=-e^{y}x\Rightarrow e^{-y}\frac{dy}{dx}-\frac{1}{x}{e}^{-y}=-x$
$\text{ put }{e}^{-y}=z\Rightarrow \frac{dz}{dx}+\frac{1}{x}\cdot z=x$
$\text{ I.F. }=x$
solution $z\cdot x=\int x^{2}dx+C\Rightarrow zx=\frac{x^3}{3}+c$
$e^{-y}\cdot x=\frac{x^3}{3}+c$
$e^0\cdot 1=\frac{1}{3}+C\Rightarrow C=\frac{2}{3}$
$e^{-y}\cdot x=\frac{x^3}{3}+\frac{2}{3}\Rightarrow 3\cdot e^{-y}x=x^3+2$
$3x=\left(x^3+2\right)e^y$