$\text{ Let }\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}={\mathrm{x}}^2\left({\mathrm{xe}}^{\mathrm{x}}+{\mathrm{e}}^{\mathrm{x}}-1\right)\mathrm{\forall }\mathrm{x}\in \mathrm{R}-\left\{0\right\}\text{ such that }\mathrm{y}\left(1\right)=\mathrm{e}-1.\text{ If }\mathrm{y}\left(2\right)=\mathrm{ky}\left(1\right)\left(\mathrm{y}\right(1)+2) \text{the k is}$

$\begin{array}{l}\frac{\text{ dy }}{dx}-\frac{1}{x}y=x\left(x{e}^x+e^x-1\right)\\ I.F=e^{-\log x}=\frac{1}{x}\end{array}$

$\begin{array}{l}\text{ Solution of the differential equation in }\mathrm{y}·\frac{1}{\mathrm{x}}=\int \left({\mathrm{e}}^{\mathrm{x}}(\mathrm{x}+1)-1\right)\mathrm{dx}\\ \frac{y}{x}=\int x{e}^{x}dx+\int e^{x}dx-\int 1dx\\ \Rightarrow \frac{\mathrm{y}}{\mathrm{x}}={\mathrm{xe}}^{\mathrm{x}}-\int e^{x}dx+\int e^{x}dx-\mathrm{x}+\mathrm{c}\\ \Rightarrow \frac{\mathrm{y}}{\mathrm{x}}={\mathrm{xe}}^{\mathrm{x}}-\mathrm{x}+\mathrm{c} ----\left(1\right)\\ \text{ Put }\mathrm{x}=1,\mathrm{y}=\mathrm{e}-1\\ e-1=\mathrm{e}-1+\mathrm{c}\Rightarrow \mathrm{c}=0\\ \left(1\right) becomes \\ \Rightarrow \mathrm{x}+\frac{\mathrm{y}}{\mathrm{x}}={\mathrm{xe}}^{\mathrm{x}}\\ x=2\end{array}$

$\begin{array}{l}\Rightarrow 2+\frac{y}{2}=2{\mathrm{e}}^2\\ \frac{y}{2}=2\left(e^2-1\right)\\ y=4e^2-4=4(e^2-1)\\ \text{ y(1)}=e-1\\ y\left(2\right)=4(e-1)(e+1)\\ =4y\left(1\right)\left(y\right(1)+2)\\ \Rightarrow k=4.\end{array}$