Piecewise initial value Laplace

**Phyxius117** · November 22nd 2009, 01:06 PM

Here is the question:

y'' + 64y = {P

P =
2t, 0<=t<2
0, t>=2

y'(0) = 0
y(0) = 0

Find the equation you get by taking the Laplace transform of the differential equation and solve for

Y(s) = ?

Please help!!

I tried and got this as answer but it's wrong...

2/(s^2(s^2+64))

**chisigma** · November 22nd 2009, 08:20 PM

An alternative expression for p(t) is...

$p(t)= 2t\cdot \{1-\mathcal{U} (t-2)\}$ (1)

... and the DE in terms of LT is...

$s^{2}\cdot Y(s) + 64\cdot Y(s) = P(s)$ (2)

... where $Y(s) = \mathcal{L}\{y(t)\}$ and $P(s) = \mathcal{L}\{p(t)\}$ , so that is...

$Y(s)= \frac{P(s)}{s^{2} +64}$ (3)

What is $P(s)$ , i.e.the LT of (1)?...

Kind regards

$\chi$ $\sigma$

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