Solving differential equations using convolution, the dreaded 0=0.

[Locoism]

Homework Statement

tx'' + (4t-2)x' + (13t-4)x = 0

Use laplace transform to solve.

The Attempt at a Solution

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

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

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

Now I'm trying to evaluate

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

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?

 

[Dick]

Homework Statement

tx'' + (4t-2)x' + (13t-4)x = 0

Use laplace transform to solve.

The Attempt at a Solution

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

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

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

Now I'm trying to evaluate

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

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?

Hard to say. But I wouldn't try to do the integral that way. sin(A)sin(B)=(cos(A-B)-cos(A+B))/2. Use that identity.