A curve passing through $(2,3)$  and satisfying the differential equation $\underset{0}{\overset{x}{\int }}ty\left(t\right)dt=x^{2}y\left(x\right)(x>0)$  is 

$\underset{0}{\overset{x}{\int }}ty\left(t\right)dt=x^{2}y\left(x\right)$

diff w r t x         $xy\left(x\right)=x^2{y}^{\text{'}}\left(x\right)+2xy\left(x\right)$    $xy\left(x\right)+x^2{y}^{\text{'}}\left(x\right)=0$$÷x$   $y\left(x\right)+x{y}^{\text{'}}\left(x\right)=0$   $y+x\frac{dy}{dx}=0$  $y=-x\frac{dy}{dx}\Rightarrow \int \frac{dx}{x}=-\int \frac{dy}{y}$$\mathcal{l}nx+\mathcal{l}ny=\mathcal{l}nc$

At point  $(2,3)\Rightarrow 2\times 3=c\Rightarrow c=6$  $\therefore xy=6$