For $x\in R,x\ne 0,$ if y(x) is a differentiable function such that $x\underset{1}{\overset{x}{\int }}y\left(t\right)dt=(x+1)\underset{1}{\overset{x}{\int }}ty\left(t\right)dt,$ then y(x) equals: (where C is a constant)

$$\begin{gathered} \mathrm{x} {\int }_1^{\mathrm{x}} \mathrm{y}\left(\mathrm{t}\right) \mathrm{dt}=(\mathrm{x}+1) {\int }_1^{\mathrm{x}} \mathrm{t} \mathrm{y}\left(\mathrm{t}\right) \mathrm{dt} \\ \mathrm{differentiate} \mathrm{w}.\mathrm{r}.\mathrm{t} \text{'}\mathrm{x}\text{'} \\ \mathrm{x} \mathrm{y}+{\int }_1^{\mathrm{x}} \mathrm{y}\left(\mathrm{t}\right) \mathrm{dt} \\ =(\mathrm{x}+1) \left(\mathrm{xy}\right)+{\int }_1^{\mathrm{x}} \mathrm{t} \mathrm{y}\left(\mathrm{t}\right) \mathrm{dt} \\ {\int }_1^{\mathrm{x}} \mathrm{y}\left(\mathrm{t}\right) \mathrm{dt}={\mathrm{x}}^2\mathrm{y}+{\int }_1^{\mathrm{x}} \mathrm{t} \mathrm{y} \left(\mathrm{t}\right) \mathrm{dt} \\ \mathrm{differentiate} \mathrm{w}.\mathrm{r}.\mathrm{t} \text{'}\mathrm{x}\text{'} \\ \mathrm{y}={\mathrm{x}}^2 \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}\left(2\mathrm{x}\right)+\mathrm{xy} \end{gathered}$$

$$\begin{gathered} {\mathrm{x}}^2 \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}(1-3\mathrm{x}) \\ \frac{\mathrm{dy}}{\mathrm{y}}=\frac{(1-3\mathrm{x}) \mathrm{dx}}{{\mathrm{x}}^2} \\ \log \mathrm{y}=\int \left(\frac{1}{{\mathrm{x}}^2}-\frac{3}{\mathrm{x}}\right) \mathrm{dx} \\ =\frac{-1}{\mathrm{x}}-3\log \mathrm{x}+\log \mathrm{c} \\ \log \frac{\mathrm{y} {\mathrm{x}}^3}{\mathrm{c}}=\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right. \\ \mathrm{y}=\frac{\mathrm{c}}{{\mathrm{x}}^2} {\mathrm{e}}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$\mathrm{x}$}\right.} \end{gathered}$$