Solve $f(t)=e^t+e^t\int_0^te^{-\tau}f(\tau)\mathrm{d}\tau$

Question

The question:
$$\begin{equation*} f(t)=e^t+e^t\int_0^te^{-\tau}f(\tau)\mathrm{d}\tau \end{equation*}.$$

I believe we need to take the Laplace transform of all terms. I am getting stuck with this part:
$$e^t\int_0^te^{-\tau}f(\tau)\mathrm{d}\tau.$$

I believe we could utilize the convolution theorem

$$\begin{equation*} (f\ast g)(t)=\int_0^tf(\tau)g(t-\tau)\mathrm{d}\tau \end{equation*}$$

but the $e^t$ in front of the integral is causing problems for me.

Any help is appreciated, thanks.

Answer 1

Hint

Suppose that $f$ is a solution of $$f(t)=e^t+e^t\int_0^te^{-\tau}f(\tau)\,\mathrm d \tau.$$

In one hand, $f(0)=1$. Moreover, if you derivate, you get \begin{align*} f'(t)&=e^t+\underbrace{e^t\int_0^t e^{-\tau}f(\tau)\,\mathrm d \tau}_{=f(t)-e^t}+f(t)=2f(t). \end{align*}

I let you conclude.

Answer 2

You may produce a first-order ODE by differentiating $f$ with the help of Leibniz integral rule, as it is done in Surb's answer, and it turns out to be a quick method. If you really want to use the convolution theorem with Laplace transforms, you can follow the steps described below.

Note that $$ e^t \int_0^t e^{-\tau}f(\tau) \,\mathrm{d}\tau = \int_0^t f(\tau)g(t-\tau) \,\mathrm{d}\tau, $$ with $g(t) := e^t$, so that $$ f(t) = g(t) + (f*g)(t) $$ and $$ F(s) = G(s) + F(s)G(s), $$ thanks to the said convolution theorem, where $G(s) = \mathcal{L}[e^t](s) = \frac{1}{s-1}$, hence $$ F(s) = \frac{G(s)}{1-G(s)} = \frac{1}{s+2} $$ and finally $$ f(t) = \mathcal{L}^{-1}\left[\frac{1}{s+2}\right](t) = e^{2t}u(t), $$ with $u(t)$ the unit step function, which can be dropped in the case where you consider $t \ge 0$ only.

Answer 3

Set $g(t)=e^{-t}f(t)$ to get $$ g(t)=1+\int_0^t g(\tau)\,d\tau. $$ This means first that $g$ is infinitely often continuously differentiable. Then the integral equation is equivalent to the IVP $$ g'(t)=g(t),~~~g(0)=1. $$ This has a well-known solution.