A curve passes through (2,3) and satisfying the D.E ∫0x ty(t)dt=x2y(x),x>0 is
x2+y2=13
y2=92x
x28+y218=1
xy=C
∫0x ty(t)dt=x2y(x) Differentiating w.r.t. x, we getxy(x)+x2y1(x)=0→xdydx+y=0⇒logx+logy=logC⇒xy=C