Solution to an ODE $f'(t) = S(0) + \mu f(t)$

Question

I came across this problem In a finance module of mine and I don't quite understand the solution. The actual equation is as follows:

$E[S(t)] = S(0) + \mu \int^{t}_{0} E[S(u)]du$

Where $S(t)$ is an asset price at time $t$ And $E$ is the expectation. The solution to the question is given as:

$E[S(t)] = S(0)e^{ut}$

and is stated that this is obtained from solving an ODE. I am probably missing something really obvious here, but if someone could fill in the steps then that would be great!

Answer 1

You differentiate both sides with respect to $t$ leading you to $$\frac d{dt} E[S(t)] = \mu E[S(t)].$$ This is a simple ODE which is known to have the solution $$E[S(t)] = E[S(0)] \exp(\mu t),$$ where $E[S(0)] = S(0)$, since you already know the price at the beginning.

Answer 2

The initial value problem $$ x'(t)=f(t,x(t)), \;x(0)=x_0 $$ is equivalent to the integral equation $$ x(t)=x_0+\int_0^t f(s,x(s))\,ds. $$ This is the basis for the existence theorems for ODE and the Picard iteration.

In your case you go backwards from the integral equation to the ODE. You could also use the usual integrating factor to find $$ \frac{d}{dt}\left(e^{-μt}\int_0^t\Bbb E[S(u)]du\right)=e^{-μt}\Bbb E[S(0)] $$ and so on without assuming the differentiability of $\Bbb E[S(t)]$.