$f(t) = te^t + \int_0^t \tau f(t-\tau)d\tau$

Question

How can I solve for $f(t)$ given $f(t) = te^t + \int_0^t \tau f(t-\tau)d\tau$

I began by taking the laplace transform of all the terms to get $$F(s) = \frac{1}{(s-1)^2} + \frac{F(s)}{s^2}$$ I am stuck on what to do when I solve for $$F(s) = \frac{1}{(s-1)^2}\cdot \frac{1}{1-\frac{1}{s^2}}$$ Any help would be appreciated.

Answer 1

I don't know whether you are still looking for an answer, but here is a perhaps longer but easier alternative approach, which makes use of series expansion instead of Laplace transform.

Firstly, let's recall that the solution can be Taylor expanded (around zero) as $f(t) = \displaystyle \sum_{n\ge0} f^{(n)}(0)\frac{t^n}{n!}$, where $f^{(n)}$ denotes the $n^\mathrm{th}$ derivative. It is easy to check by differentiating the original equation that the latter satisfies $$ f^{(n)}(t) = te^t + \int_0^t \tau f^{(n)}(t-\tau) \,\mathrm{d}\tau + ne^t + tf^{(n-1)}(0) + f^{(n-2)}(0), $$ hence the simple inhomogeneous first-order linear recurrence relation $f^{(n)}(0) = f^{(n-2)}(0) + n$, which is solved by $f^{(n)}(0) = A + B(-1)^n + \frac{1}{4}n^2 + \frac{1}{2}n - \frac{3}{8}$, where $A$ and $B$ are constants to be determined thanks to the the initial conditions $f(0) = 0$ and $f'(0) = 1$.

Finally, plugging this expression into the Taylor series will lead to a solution made of exponentials with polynomial prefactors, such as gpmath's inverse Laplace transform in the comment section.

Answer 2

I believe the use of the Laplace transform is quite unnecessary given that the process of finding the inverse image of it is most of the time quite annoying and time-consuming and you can easily avoid it. In particular, if you differentiate both sides of the given equation we achieve the following: $$f^{'}(x)=(x+1)e^x+\int_{0}^{x}f(t)dt.$$ Repeating the same process we finally get $$f^{''}(x)=(x+2)e^x+f(x)$$ or $$f^{''}(x)-f(x)=(x+2)e^x.$$ Thus, by setting $y=f(x)$ we obtain a 2nd-order non-homogeneous linear differential equation, in particular we have that $$y^{''}-y=(x+2)e^x,$$ with the initial conditions being $y(0)=0$ and $y^{'}(0)=1$. Thus, using variation of parameters we see that $$f(x)=\frac{1}{8}e^x(e^{-2x}(2x^2+6x+1)-1).$$