The solution of the equationx∫0xy(t)dt=(x+1)∫x0ty(t)dt,x>0 is

[d] The equation is$x\int x0y\left(t\right)dt=\left(x+1\right)\int x0ty\left(t\right)dt,?\left(i\right)$Differentiating both the sides with respect to x, we get$xy\left(x\right)+\int x0y\left(t\right)dt=\left(x+1\right)xy\left(x\right)+\int x0ty\left(t\right)dt$⇒∫x0y(t)dt=x2y(x)+∫x0ty(t)dt                    ? (ii)$D\iff erentiat\in gaga\in withrespect\to x,we\ge t$y(x)=2xy(x)+x2y′(x)+xy(x)$\Rightarrow x2dydx=\left(1-3x\right)y\left[writ\in gy\left(x\right)=y\right]$⇒dyy=(1−3xx2)dx=(1x2−3x)dx$Integrat\in gwe\ge t,\log y=-1x-3\log x+a,aiscons\tan t$⇒logy+3logx=a−1x$\Rightarrow \log \left(yx3\right)=a-1x\Rightarrow yx3=ea-1x=c.e-1x$where c=ea∴y=cx3e−1x