Obtaining the Fredholm alternative kernel

Question

I am reading through Greenberg's Foundations of Applied Mathemtics, and I am having an issue understanding the Fredholm's Alternative. Consider the initial value problem $$x'' + 3x = 6t, x(0)=x'(0)=0$$ Integrating from $0$ to $t$ and using the condition $x'(0)=0$, we obtain $$x'(t) + 3\int_{0}^t x(\tau) d\tau = 3t^2$$ Integrating from $0$ to $t$ and using the condition $x(0)=0$, we obtain $$x(t) = t^3 - 3\int_{0}^{t} \int_{0}^{t'} x(\tau) d\tau dt'$$ Now, the book makes a leap I don't understand. Greenberg says "inverting the order of integration and integrating on $t'$", $$x(t)= t^3 + 3\int_{0}^t (\tau - t)x(\tau) d\tau$$ How did they get the $(\tau - t)$ term in the integrand? Shouldn't it become $$x(t)= t^3 + 3\int_{0}^t t'x(\tau) d\tau$$ Why is $t' = \tau -t$, if it even is?

Answer 1

Draw a picture with the $x$-axis being $\tau$ and the $y$-axis being $t'$. Draw the line $t'=\tau$. You should find that the integrating region is the upper right triangle above the line $t'=\tau$. If you switch the integrating order, you are still integrating over this triangle. But it should be obvious that the integral becomes $$ \int_0^t\int_{\tau}^tx(\tau)\,\mathrm{d}t'\mathrm{d}\tau=\int_0^t x(\tau)\mathrm{d}\tau\int_{\tau}^t\,\mathrm{d}t'=\int_0^t x(\tau)(t-\tau)\,\mathrm{d}\tau. $$