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So I'm trying to solve this differential equation:

$y'' - 4y' + 3y = f(t)\:\:\:$ with initial conditions $y(0) = y'(0) = 0$

using laplace transform.

After taking the Laplace transform and doing some algebra and then taking the inverse I reach:

$$y(t) = \mathscr{L^{-1}} \left[\frac{F(s)}{(s-3)(s-1)}\right]$$

I'm thinking of solving this using convolution theorem but I'm unsure if it works for the product of 3 functions.

I took the convolution of $\frac{1}{(s-3)}$ and $\frac{1}{(s-1)}$ and got:

$$\frac{e^{3t}}{2} - \frac{e^{t}}{2}$$

Now if I were to take convolution again and get:

$$y(t) = \frac{1}{2}\int_0^t (e^{3u} - e^{u})\: F(t-u)\: du$$

Is this the correct answer to my initial problem?

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1 Answer 1

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I'm thinking of solving this using convolution theorem but I'm unsure if it works for the product of 3 functions.

You are looking at this being $F(s) p(s) q(s)$. But the product of two functions is also a function, so $F(s) z(s)$, it being two functions, is completely equivalent.

As such, your method will work. Note that you could also take the function $\frac{1}{s-3} \frac{1}{s-1}$ and write it as a sum $\frac{A}{s-3} + \frac{B}{s-1}$:

$\frac{1}{s-3} \frac{1}{s-1} = \frac{A}{s-3} + \frac{B}{s-1} = \left( As - A + Bs - 3B\right)\frac{1}{s-3} \frac{1}{s-1}$

So that it follows that: $A + B = 0$, $-A - 3B = 1 \to A = \frac{1}{2}$ and $B = - \frac{1}{2}$.

This means that $H(s) = \frac{1}{2} \frac{1}{s-3} - \frac{1}{2} \frac{1}{s-1}$ and $h(t) = \frac{1}{2} \left[ e^{3t} - e^{t} \right]$, as you had found.

It follows that you have indeed found the correct result, the convolution of $f(t)$ with $h(t)$, assuming that you accidentally wrote a capital $F$ instead of the lowercase $f$ in your solution.

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  • $\begingroup$ Oh yes now I just feel silly thanks! :P About the capital and small F, people interchange them sometimes so I always get confused which one to use. $\endgroup$
    – Silver
    Commented Oct 5, 2014 at 11:51
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    $\begingroup$ Usually the capital is used on the laplace-domain, and the lowercase on the time-domain. It's a good convention :) $\endgroup$
    – Daimonie
    Commented Oct 5, 2014 at 11:59

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