Laplace Transforms - Convolution Theorem

[JJBladester]

Homework Statement

Use the Laplace transform to solve the given initial-value problem.

$$y'' + 9y = cos(3t), y(0)=2, y'(0)=5$$

Homework Equations

$$L\left\{y''\right\} = s^{2}Y(s)-sy(0)-y'(0)$$

$$L\left\{y\right\} = Y(s)$$

$$L\left\{cos(kt)\right\} = \frac{s^{2}}{s^{2} + k^{2}}, s>0$$

$$L\left\{sin(kt)\right\} = \frac{k}{s^{2} + k^{2}}, s>0$$

$$L^{-1}\left\{F(s-a)\right\} = e^{at}f(t)$$ (shifting property)

$$L\left\{F(s)G(s)\right\} = f \ast g$$ (convolution of f and g)

The Attempt at a Solution

$$y'' + 9y = cos(3t), y(0)=2, y'(0)=5$$

$$L\left\{y''\right\} + 9L\left\{y\right\} = L\left\{cos(3t)\right\}$$

$$\left[s^{2}Y(s)-sy(0)-y'(0)\right] + 9Y(s) = \frac{s}{s^{2}+9}$$

$$Y(s)\left[s^{2} - 2s + 9\right] = 5 + \frac{s}{s^{2} + 9}$$

$$Y(s) = \frac{5}{s^{2} - 2s + 9} + \frac{s}{(s^{2} - 2s + 9)(s^{2} + 9)}$$

$$\frac{5}{(s-1)^{2} + 8} + \frac{s}{\left[(s-1)^{2} + 8)\right](s^{2} + 9)}$$

$$y(t) = \frac{5}{2\sqrt{2}}L^{-1}\left\{\frac{2\sqrt{2}}{s^{2} + (2\sqrt{2})^{2}}_{s\rightarrow(s-1)}\right\} + L\left\{F(s)G(s)\right\}$$ where $$F(s) = \frac{s}{(s-1)^{2} + 8}$$ and $$G(s) = \frac{1}{s^{2} + 9}$$

$$f(t) = L^{-1}\left\{\frac{(s-1) + 1}{(s-1)^{2} + (2\sqrt{2})^{2}}\right\} = L^{-1}\left\{\frac{s}{s^{2} + (2\sqrt{2})^{2}}_{s\rightarrow(s-1)}\right\} + \frac{1}{2\sqrt{2}}L^{-1}\left\{\frac{2\sqrt{2}}{s^{2} + (2\sqrt{2})^{2}}_{s\rightarrow(s-1)}\right\} = e^{t}cos(2\sqrt{2}t) + \frac{1}{2\sqrt{2}}\left[e^{t}sin(2\sqrt{2}t)\right]$$

$$g(t) = \frac{1}{3}L^{-1}\left\{\frac{3}{s^{2} + 3^{2}}\right\} = \frac{1}{3}sin(3t)$$

$$L^{-1}\left\{F(s)G(s)\right\} = f \ast g$$

And I get stuck here. The book's answer is:

$$y = 2cos(3t) + \frac{5}{3}sin(3t) + \frac{1}{6}tsin(3t)$$

 

[JJBladester]

Nevermind... Got it.

 

[JJBladester]

It was a good exercise in Latex, though. Count your positives! Happy Easter everyone.