**#26,pg.350:** Solve the integro-differential equation
$\phi'(t) + \displaystyle\int_0^t (t-\xi)\phi(\xi) d\xi = t; \phi(0)=0$ by using the Laplace transform. By differentiating the equation a sufficient number of times, convert it into an initial value problem. Solve the initial value problem and verify the solution is the same as the one found by using the Laplace transform.

**Solution:** To solve this using Laplace transforms, recognize the integral as the convolution $f*\phi$ where $f(t)=t$. Then we take the Laplace transform (using the convolution theorem on the second term) to get
$$(s\mathscr{L}\{\phi\}-\phi(0))+\mathscr{L}\{f\}\mathscr{L}\{\phi\}=\dfrac{1}{s^2},$$
and since $\mathscr{L}\{f\}=\dfrac{1}{s^2}$ and $\phi(0)=0$ we have
$$\left( s + \dfrac{1}{s^2} \right) \mathscr{L}\{\phi\}= \dfrac{1}{s^2},$$
or
$$\mathscr{L}\{\phi\} = \dfrac{1}{s^2 (s+\frac{1}{s^2})} = \dfrac{s^2}{s^2(s^3+1)}=\dfrac{1}{s^2(s^3+1)}$$
by factoring the sum of cubes we see $s^3+1=(s+1)(s^2-s+1)$.
Hence
$$\mathscr{L}\{\phi\}=\dfrac{1}{(s+1)(s^2-s+1)}.$$
The quadratic does not factor nicely, so we complete the square to see $s^2-s+1=\left(s - \dfrac{1}{2} \right)^2+\dfrac{3}{4}$ so we have
$$\mathscr{L}\{\phi\} = \dfrac{1}{(s+1)((s-\frac{1}{2})^2+\frac{3}{4})}.$$
Partial fractions shows that
$$\begin{array}{ll}
\dfrac{1}{(s+1)((s-\frac{1}{2})^2+\frac{3}{4})} &= \dfrac{\frac{1}{3}}{s+1} + \dfrac{(-\frac{1}{3})s+\frac{2}{3}}{(s-\frac{1}{2})^2+\frac{3}{4}} \\
&=\dfrac{1}{3} \left[ \dfrac{1}{s+1} - \dfrac{s-2}{(s-\frac{1}{2})^2+\frac{3}{4}} \right] \\
&= \dfrac{1}{3} \left[ \dfrac{1}{s+1} - \dfrac{s-\frac{1}{2}}{(s-\frac{1}{2})^2+\frac{3}{4}} - \dfrac{-\frac{3}{2}}{(s-\frac{1}{2})^2+\frac{3}{4}} \right].
\end{array}$$
Hence we have derived
$$\mathscr{L}\{\phi\}=\dfrac{1}{3} \left[ \dfrac{1}{s+1} - \dfrac{s-\frac{1}{2}}{(s-\frac{1}{2})^2+\frac{3}{4}} - \dfrac{-\frac{3}{2}}{(s-\frac{1}{2})^2+\frac{3}{4}} \right].$$
Therefore
$$\phi(t) = \dfrac{e^{-t}}{3} - e^{\frac{t}{2}} \cos \left( \sqrt{\dfrac{3}{4}} t \right) +\sqrt{3} e^{\frac{t}{2}} \sin \left( \sqrt{\dfrac{3}{4}} t \right).$$

Recall the fundamental theorem of calculus says
$$\dfrac{d}{dt} \displaystyle\int_0^t f(\xi) d\xi = f(t).$$
You should note that you may not apply this to the integral in the integro-differential equation because the variable $t$ is not allowed to appear in the integrand when applying the fundamental theorem. We will get out of this by factoring it out of the integral.

To derive a differential equation, note that the integro-differential equation can be written as
$$(*) \hspace{35pt} \phi'(t) + t \displaystyle\int_0^t \phi(\xi) d\xi - \displaystyle\int_0^t \xi\phi(\xi) d\xi=t;\phi(0)=0.$$
*Note: the $t$ can be pulled out of the integral because it is not the variable of integration, $\xi$ -- this occurs frequently in calculus 3*

If we differentiate this expression (using product rule on the second term) we get
$$\phi''(t) + \displaystyle\int_0^t \phi(\xi) d\xi + t\phi(t) - t\phi(t) = 1,$$
or simplified
$$(**) \hspace{35pt} \phi''(t) + \displaystyle\int_0^t \phi(\xi) d\xi = 1.$$
This expression still has an integral, so we differentiate again to get
$$\phi'''(t) + \phi(t)=0.$$
The original problem tells us the initial condition $\phi(0)=0$. Since this is a third order differential equation, we need two more initial conditions. To find one of them plug $t=0$ into $(*)$ to get $\phi'(0)=0$ and plug $t=0$ into $(**)$ to get $\phi''(0)=1$. Thus we have shown that the integro-differential equation is equivalent to the following differential equation:
$$\phi'''(t)+\phi(t)=0; \phi(0)=0,\phi'(0)=0,\phi''(0)=1.$$
This equation may be solved using the standard techniques we learned earlier in the class and we see from calculation that the solution matches the one we found earlier.