Use Laplace transforms to find the solutions of $y''(t) +9y(t) = 3\delta_{2\pi/3} +9H_{\pi}(t)$

Question

Use Laplace transforms to find the solutions of the following IVP

$y''(t) +9y(t) = 3\delta_{2\pi/3} +9H_{\pi}(t), $

(where $H_{\pi}(t)$ is the Heaviside function of $\pi$)

with initial values: $y(0) = 0 $ and $y'(0) = 1$

Answer 1

Since it's not clear what you've already tried I'd give sketch of useful properties that are essential to solving this IVP. Note there are a lot of great tables for $\mathcal{L}$ and $\mathcal{L}^-1$, but I'd recommend proving each identity at least once so you'd be comfortable with these type of calculations. (Which generally aren't that hard and prove to be a great exercises too!)

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HINT:

(1) Relation between LT and the derivative: $$\mathcal{L}(f')=s\mathcal{L}(f)-f(0)$$ In your case, note you'll need to apply this property twice.

(2) The LT of Dirac's delta 'function' is given by: $$\mathcal{L}(\delta(t-a))=e^{-as}$$

(3) The LT of the Heaviside function is given by: $$\mathcal{L}(H(t-a))=\frac{e^{-as}}{s}$$

The last two identities could be calculated directly from the definition of LT: $$\mathcal{L}(f)=\int_{0}^{\infty}f(t)e^{-st}dt$$

Answer 2

To compare with an alternative solution method:

• The usual Laplace-method assumption is that $y(t)=0$ for $t<0$, which implies also $y=0$ for $t<\frac{2\pi}3$.
• At $t=\frac{2\pi}3$ the derivative has a jump of $3$, which means that on the interval $[\frac{2\pi}3,\pi]$ we have $y$ equal to the solution of the IVP with $y(\frac{2\pi}3)=0$, $y'(\frac{2\pi}3)=3$, giving $$y(t)=\sin(3(t-\frac{2\pi}3))=\sin(3t).$$
• Continuing from $t=\pi$, we have $y$ equal to the solution of $(y-1)''+9(y-1)=0$, $y(\pi)=\sin(3\pi)=0$, $y'(\pi)=3\cos(\pi)=-3$, so that $$ y(t)=1-\cos(3(t-\pi))-\sin(3(t-\pi))=1+\cos(3t)+\sin(3t) $$

Combined into one formula, this gives $$ y(t)=H_{\frac{2\pi}3}(t)\sin(3t)+H_\pi(t)(1+\cos(3t)) $$