It looks like the integral equation is
[tex]y(x)=36+\displaystyle\int_0^x2ty(t)\,\mathrm dt[/tex]
We can get an initial condition right away by setting [tex]x=0[/tex], for which we get
[tex]y(0)=36+\displaystyle\int_0^02ty(t)\,\mathrm dt=36[/tex]
Now, differentiating both sides of the integral equation gives
[tex]\dfrac{\mathrm dy(x)}{\mathrm dx}=0+2xy(x)[/tex]
so that [tex]y(x)[/tex] solves the differential equation,
[tex]y'-2xy=0[/tex]
This ODE is linear, and multiplying both sides by [tex]e^{-x^2}[/tex] lets us condense the left side into the derivative of a product:
[tex]e^{-x^2}y'-2xe^{-x^2}y=0[/tex]
[tex]\left(e^{-x^2}y\right)'=0[/tex]
Integrate both sides to get
[tex]e^{-x^2}y=C[/tex]
and solve for [tex]y(x)[/tex]:
[tex]y(x)=Ce^{x^2}[/tex]
Knowing that [tex]y(0)=36[/tex], we find [tex]C=36[/tex], so that the integral equation has the particular solution,
[tex]\boxed{y(x)=36e^{x^2}}[/tex]
