Q:

An integral equation is an equation that contains an unknown function y(x) and an integral that involves y(x). Solve the given integral equation. [Hint: Use an initial condition obtained from the integral equation.] y(x) = 36 + x 2t y(t) dt 0

Accepted Solution

A:
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]