# (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:

$\mathcal{L}\{\sigma\}(s) = E_1\cdot\mathcal{L}\{\epsilon\}+\eta\cdot\bigl[\mathcal{L}\{\epsilon\}(s)\cdot s - 0\bigr]$

$\mathcal{L}\{\epsilon\} = \frac{1}{E_1+\eta\cdot s}\cdot\mathcal{L}\{\sigma\}(s)$

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!

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 $\epsilon$

• thank you! but ... how can I use this in the Laplace transformated equation? My formulary says: $\mathcal{L}\{f'(t)\} = sF(s)-f(0+)$ ... so where to put the $\dot\epsilon_0$? – eniem Dec 2 '14 at 16:16
• @eniem I have worked out the problem in full. – user_of_math Dec 2 '14 at 17:14
• Thanks a lot! That helped me really! – eniem Dec 2 '14 at 18:46