Question

Laplace transform of y"+9y=|sin(3t)| y(0)=0, y'(0)=0

Answer 1

I am assuming the crux of the problem is that absolute value of sine since there are formulas to integrate a derivative... Do you know of a formula to take the Laplace transform of a periodic function?

Answer 2

The first step of taking the Laplace transform:
\( \displaystyle \mathcal{L} \left[ y' ' + 9y \right] = \mathcal{L} \left[ \left| \sin (3t) \right| \right] \)
The left side, linearity applies:
\( \displaystyle \mathcal{L} \left[ y' ' \right] + 9 \mathcal{L} \left[ y \right] = \mathcal{L} \left[ \left| \sin (3t) \right| \right] \)

The left side has the Laplace transform of a second derivative, which we can reduce using the Laplace transform of a derivative formula.

The right side has the Laplace transform of a periodic function with half the normal period of sin(3t). This is because the first half of sine is positive, then the second half is (normally) negative. But absolute values pull back the negative part to make it the same as the first half.
This is where this formula would come in handy, for a function f(t) with period T:
\( \displaystyle \mathcal{L} \left[ f(t) \right] = \dfrac{1}{1 - e^{-sT}} \int_{0}^{T} e^{-st} f(t) \ dt \)

Answer 3

Once you have an appropriate period set up, sin(3t) is always positive over the integral region, so the absolute values may be omitted and the function integrated normally!