The solution of the equation $x{\int }_0^x \mathrm{y}\left(\mathrm{t}\right)\mathrm{dt}=(x+1){\int }_0^x \mathrm{ty}\left(\mathrm{t}\right)\mathrm{dt},\mathrm{x}>0$ is

$x{\int }_0^x y\left(t\right)dt=(x+1){\int }_0^x ty\left(t\right)dt$ 

             Diff w.r.t x 

$$\begin{gathered} \therefore xy\left(x\right)+{\int }_0^x y\left(t\right)dt=(x+1)xy\left(x\right)+1{\int }_0^x ty\left(t\right)dt \\ \Rightarrow {\int }_0^x y\left(t\right)dt =x^{2}y\left(x\right) +{\int }_0^x ty\left(t\right)dt \\ again differentiate w.r.t x \\ \therefore y=x^2\frac{dy}{dx}+y 2x+xy \end{gathered}$$

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