Question

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

Answer 1

It looks like the integral equation is

`y(x)=36+\displaystyle\int_0^x2ty(t)\,\mathrm dt`

We can get an initial condition right away by setting
`x=0`, for which we get

`y(0)=36+\displaystyle\int_0^02ty(t)\,\mathrm dt=36`

Now, differentiating both sides of the integral equation gives

`(\mathrm dy(x))/(\mathrm dx)=0+2xy(x)`

so that
`y(x)` solves the differential equation,

`y'-2xy=0`

This ODE is linear, and multiplying both sides by
`e^(-x^2)` lets us condense the left side into the derivative of a product:

`e^(-x^2)y'-2xe^(-x^2)y=0`

`\left(e^(-x^2)y\right)'=0`

Integrate both sides to get

`e^(-x^2)y=C`

and solve for
`y(x)`:

`y(x)=Ce^(x^2)`

Knowing that
`y(0)=36`, we find
`C=36`, so that the integral equation has the particular solution,

`\boxed{y(x)=36e^(x^2)}`