(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

[itex] tx'' + (4t-2)x' + (13t-4)x = 0 [/itex]

Use laplace transform to solve.

3. The attempt at a solution

I've split up the X'(s) and X(s) and integrated to get [itex] X(s) = \frac{c}{(s^2 + 4s + 13)^2} = \frac{c}{9} \left[ \frac{3}{(s+2)^2 + 9} \right]^2 [/itex]

From this I'm guessing convolution product, using [itex] F(s) = G(s) = \frac{3}{(s+2)^2 + 9} [/itex]

This gives [itex] f(t) = g(t) = e^{-2t}sin(3t) [/itex]

Now I'm trying to evaluate

[itex]x(t) = \frac{ce^{-2t}}{9} \int_0^t sin(3x)sin(3(t-x)) dx [/itex]

which I get from the convolution formula (the e^-2x cancel out nicely).

Here's the problem:

I use integration by parts twice, but each time I do, no matter what I choose my u and v to be, I always end up with the integral being equal to 0... Where is my mistake???

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# Homework Help: Solving differential equations using convolution, the dreaded 0=0.

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