(Another) linear ODE first order - solving by Laplace transformation? I don't know what's up, but I got stuck another time in a first order linear ODE. - The problem can be seen on p. 26 of a Lecture-Script. There's also the solution, which is comprehensible.
For my own interest, I've tried to solve this equation by Laplace-transformation but I don't get the same solution and don't know why.
Given is the following equation: $\sigma = E_1\cdot\epsilon + \eta\cdot\dot\epsilon$


*

*as well as it's initial conditions: $\epsilon(t = 0) = 0$ and $\sigma(t = 0) = \sigma_0$

*The disturbing function $\sigma(t)$ is given by: $\sigma(t) = \sigma_0 = \text{const.}$
By using Laplace, I get:
$\mathcal{L}\{\sigma\}(s) = E_1\cdot\mathcal{L}\{\epsilon\}+\eta\cdot\bigl[\mathcal{L}\{\epsilon\}(s)\cdot s - 0\bigr]$
This leads me to:
$\mathcal{L}\{\epsilon\} = \frac{1}{E_1+\eta\cdot s}\cdot\mathcal{L}\{\sigma\}(s)$
and with $\sigma(t) = \sigma_0$ finally to:
$\epsilon(t) = \frac{\sigma_0}{\eta}\cdot e^{-\frac{E_1}{\eta}t}$
... but that's actually not the same thing from this Script (p. 27), which says:
$\epsilon(t) = \frac{\sigma_0}{E_1}\bigl(1-e^{-\frac{E_1}{\eta}t}\bigr)$
Does anyone know what's wrong or where I make a mistake?! ...
Thank you so much in advance!
 A: Well, at $t=0$, $\epsilon=0$ so 
$\sigma=E_1 \epsilon + \eta \dot{\epsilon}$ 
gives you $\dot{\epsilon}_0 = \sigma_0 / \eta$, not $\dot{\epsilon}_0=0$.
ETA: Consider the problem. You have to understand that the unknown function is $\epsilon(t)$.
You are given $\sigma(t)=\sigma_0$, so your differential equation is
$$
\eta \dot{\epsilon} + E_1 \epsilon  = \sigma_0
$$
Notice that this is a first-order ODE, so you need only one initial condition - which is given to you as $\epsilon(0)=0$. Now take Laplace transforms of both sides, we'll call $L(\epsilon(t))$ as $F(s)$. 
Recall the formula for Laplace transform of a derivative
$$
L\left(\frac{d\epsilon}{dt}\right) = s F(s) - \epsilon(0^-)
$$
The Laplace transform of the RHS, a constant is just $\dfrac{\sigma_0}{s}$.
So you have
$$
\eta s F(s) -\eta \epsilon(0^-) + E_1 F(s) = \frac{\sigma_0}{s}
$$
Since the initial $\epsilon$ is 0, on rearranging you'll have
$$
F(s)= \frac{\sigma_0}{s(E_1 + \eta s)}
$$
Take partial fractions and you'll immediately see that the first time gives you a constant, and the second term the negative exponential.
$$
\epsilon(t) = \frac{\sigma_0 }{E_1} \left(1 - \exp \left( \frac{-E_1 \; t}{\eta} \right) \right)
$$
Now you can go back and check the initial value of $\dot{\epsilon}$. 
$$
\left. \dot{\epsilon} \right |_{t=0} = \sigma_0 / \eta
$$
But you did not really require it to solve the problem - because it is a first order ODE in $\epsilon$
