Question

If $x = x ( y )$ is the solution of the differential equation $y \frac{d x}{d y}=2 x+y^{3}(y+1) e^{y}, x(1)=0$; then $x(e)$ is equal to :

• A. $e^{3}\left(e^{e}-1\right)$
• B. $e ^{ e }\left( e ^{3}-1\right)$
• C. $e ^{2}\left( e ^{ e }+1\right)$
• D. $e ^{ e }\left( e ^{2}-1\right)$

Answer 1

Answer: (A)

$y \frac{d x}{d y}=2 x+y^{3}(y+1) e^{y}, x(1)=0$
$\frac{d x}{d y}-\frac{2}{y} x=y^{2}(y+1) e^{y}$
I.f $=e^{\int \frac{-2}{y} d y}=\frac{1}{y^{2}}$
$x \cdot \frac{1}{y^{2}}=\int(y+1) e^{y} d y$
$\frac{x}{y^{2}}=(y+1) e^{y}-e^{y}+c=y \cdot e^{y}+c$
$x=y^{3} e^{y}+c y^{2}$
For $x =0, y =1 \Rightarrow c =- e$
$x=y^{3} e^{y}-e \cdot y^{2}$
$x(e)=e^{3}\left(e^{e}-1\right)$