Question

Let $x \frac{d y}{d x}-y=x^2\left(x e^x+e^x-1\right) \forall x \in R-\{0\}$ such that $y(1)=e-1$. If $y(2)=k$ $y(1)(y(1)+2)$, then the value of $k$ is not less than

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (A), (B)

$\Theta x \frac{d y}{d x}-y=x^2\left(x e^x+e^x-1\right)$
$\Rightarrow \frac{ dy }{ dx }-\frac{1}{ x } \cdot y = x \left( xe ^{ x }+ e ^{ x }-1\right)$
$\text { I.F. }=e^{\int \frac{-1}{x} d x}=e^{-\ln x}=\frac{1}{x}$
$\therefore$ Solution is
$y y \cdot \frac{1}{x}=\int\left(x e^x+e^x-1\right) d x=x e^x-x+c $
$\Theta y(1) =e-1 \Rightarrow e-1=e-1+c \Rightarrow c=0 $
$\therefore y=x^2 \left(e^x-1\right)$
$\therefore y(2) =4\left(e^2-1\right)$
$=4(e-1)(e+1)$
$ =4 y(1)(y(1)+2)$
$\therefore k =4$.