Proving existence and uniqueness of solution to differential equation I have to prove existence and uniqueness of the solution to the following equations:
$f'_0(t)=-\lambda_0f_0(t)$
$f'_n(t)=-\lambda_n f_n(t)+\lambda_{n-1}f_{n-1}(t), n=1,2,...$
where $\lambda_n\geq0$ and $f_n(0)\geq0 $ for all n.
I prove by induction that  $f_n(t)\geq0 $ for all $t\geq0$.
I have no idea where to start. I appreciate any help.
 A: The conditions you have given are insufficient to guarantee that their solution will be unique. Suppose $\ \lambda_0\ne0\ $, but $\ \lambda_n=0\ $ for all $\ n\ge1\ $ for instance.   Then the conditions have the trivial solution
$$
f_n(t)=0
$$
for all $\ n\ $, but also the non-trivial solution
\begin{align}
f_0(t)&=e^{-\lambda_0t}\\
f_1(t)&=1-e^{-\lambda_0t}\\
f_n(t)&=0
\end{align}
for all $\ n\ge2\ $ (or any positive multiple of it).
In the general case, if you set
\begin{align}
\mathbf{g}_n(t)&=\pmatrix{f_0(t)\\f_1(t)\\\vdots\\f_n(t)}\ \text{, and}\\
\Lambda_n&=\pmatrix{-\lambda_0&0&0&\dots&0&0\\
\lambda_0&-\lambda_1&0&\dots&0&0\\
0&\lambda_1&-\lambda_2&\dots&0&0\\
\vdots&\vdots&\ddots&\ddots&\vdots&\vdots\\
0&0&0&\dots&\lambda_{n-1}&-\lambda_n}\ ,
\end{align}
then you can write the differential equations in vector form as
$$
\mathbf{g}_n'(t)=\Lambda_n\mathbf{g}_n(t)\ .
$$
This has the solution
\begin{align}
\mathbf{g}_n(t)&=e^{\Lambda_n t}\mathbf{g}_n(0)\\
&=\left(\sum_{k=0}^\infty\frac{t^k\Lambda_n^k}{k!}\right)\mathbf{g}_n(0)\ .\ 
\end{align}
which is unique if $\ \mathbf{g}_n(0)\ $ is uniquely specified, but not otherwise.
Perhaps what you are required to do is to show that the solution is unique for any given $\ f_1(0),$$ f_2(0), \dots,$$ f_n(0), \dots\ $, but if that is the case, the question should have been more clearly worded.
