# Laplace transform of $y'' + 3y' + 2y = f(t), \; y(0) = y'(0) = 0$ [closed]

Can you do this? This is part of my final year EE work. I need to solve this in order to figure out how my sensor is behaving.

$$y'' + 3y' + 2y = f(t), \; y(0) = y'(0) = 0$$ where $f(t)$ is a square wave.

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## closed as off-topic by dustin, graydad, Rory Daulton, voldemort, Sujaan KunalanJan 18 '15 at 19:56

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You should probably mention things you've tried, or else people here may think you're just looking for easy answers for homework problems. – james Jan 20 '13 at 22:45
The things I've tried are far too long to put on this page. Literally pages and pages of workings. But I keep on getting to the wrong answer... This is part of my final year project in EE. I need to solve this in order to use a sensor – user59055 Jan 20 '13 at 23:18

Since this is a EE project, I will not solve everything for you, but this should help you move forward.

We have:

$$\tag 1 y'' + 3 y' + 2y = f(t), ~y(0) = y'(0) = 0, ~ \text{where f(t) is a square wave}$$

Taking the Laplace Transform of $(1)$, yields:

$\tag 2 \mathcal L(y'') +3 \mathcal L(y') + 2\mathcal L(y) = \mathcal L(f(t))$

with:

$\tag 3 \mathcal L(y'') = s^{2}y(s) - s(0) -(0) = s^{2}y(s)$

$\tag 4 3 \mathcal L(y') = 3(s y(s) -(0)) = 3 s y(s)$

$\tag 5 2 \mathcal L(y) = 2 y(s)$

For periodic functions, $f(x + \omega) = -f(x)$, the Laplace Transform is given by (you can also calculate it by steps - it is not too bad):

$$\mathcal L(f(x)) = \frac{\int_0^\omega e^{-s x} f(x) dx}{1 + e^{- \omega s}}$$ Now, we have no idea what sort of square wave you have, so I assume it is $f(x+1) = -f(x)$, but you will have to modify the result below if your square wave varies from that (in amplitude and frequency).

Hence, with $\omega = 1$, we obtain:

$\tag 6 \mathcal L(f(t)) = \Large \frac{\int_0^\omega e^{-s t} f(t) dt}{1 + e^{- \omega s}} = \frac{\int_0^1 e^{-s t} (1) dt}{1 + e^{- s}} = \frac{(\frac{1}{s})(1 - e^{-s})}{1+e^{-s}} = \frac{1}{s} \tanh \frac{s}{2}$

Now, substituting $(3)$ through $(6)$ into $(2)$, yields:

$$(s^{2} + 3s +2)y(s) = \frac{1}{s} \tanh \frac{s}{2}$$

Solving for $y(s)$, yields:

$$y(s) = \frac{\tanh \frac{s}{2}}{(s)(s^2+ 3s + 2)} = \frac{\tanh \frac{s}{2}}{(s)(s+1)(s+2)}$$

Since this is a project, I think this is where I will stop, since you should now be able to move it forward given that you are in this class.

Against my better judgement, I will also use WolframAlpha, to solve and plot this DEQ. This is a further hint to help you to resolve the point where I stopped above.

Regards

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All this work, and good work, at that, needs recognition! +1 – amWhy May 7 '13 at 2:01
Yes, I understand. I think, for some askers, they are so absorbed in the urgency of solving a problem, they forget to return and acknowledge help...especially when the asker's use of the site is minimal and/or sporadic. – amWhy May 7 '13 at 2:07