## Homework 6 Solutions

p. 226: 1.

$\mathcal{L}[t^2-2t]= \mathcal{L}[t^2]-2\mathcal{L}[t]$

$= \frac{2}{s^3}-2\frac{1}{s^2}$

Odd solutions in the back of the book.

p. 226: #2:

$\mathcal(t^2 - 3)^2$

$= (t^2 - 3)(t^2 - 3) = t^4 - 6t^2 + 9$

So

$\mathcal{L}[(t^2-3)^2] = \mathcal{L}[t^4]-6\mathcal{L}[t^2]+9\mathcal{L}[1]=$

$= \frac{4!}{s^5} - 6\frac{2!}{s^3} + \frac{9}{s}$

p. 226: #4:

$\ sin^2 4 t = \frac {1 - cos2(4t)}{2}$

So the Laplace Transform can be gotten from the table.

p. 226: #23.

$\mathcal{L}[f(t)]=\int_0^\infty e^{-st}f(t)\ dt=F(s).$

So

$\mathcal{L}[f(ct)]=\int_0^\infty e^{-st}f(ct)\ dt.$

Make the change of variables

$\tau=ct$

to get

$\mathcal{L}[f(ct)]=\int_{\tau=0}^\infty e^{-(s/c)\tau}f(\tau)\ (1/c) d\tau=$

$\frac{1}{c}F(s/c).$

p. 226: 10.

$\mathcal{L}[-8sin(0.2t)]=\frac{-1.6}{s^2+0.04}$

p. 226: 12.

$\mathcal{L}[(t+1)^3]=\frac{6}{s^4}+\frac{6}{s^3}+\frac{3}{s^2}+\frac{1}{s}$

p. 226: 30.

$\mathcal{L}^{-1}[\frac{2s+16}{s^2-16}]=2cosh(4t)+4sinh(4t)$

(AJ) I have the same solutions for p 226 #10 and #12, but on #30, I factored the denominator and used partial fraction decomposition to get

$\mathcal{L}^{-1}=-e^{-4t}+3e^{4t}$


AJ, I was able to get your answer and verified that it is correct. However, I am unable to see why my initial answer was wrong. I seperated as follows:

$\mathcal{L}^{-1}[2*\frac{s}{s^2-4^2}+4*\frac{4}{s^2-4^2}]$

then used (8) and (9) from Table 6.1. Thoughts?

They are actually the exact same thing, so both answers should be correct. This can be proven using:

cosh(bx) = (1/2)*(e^(bx) + e^(-bx))

sinh(bx) = (1/2)*(e^(bx) - e^(-bx))

--Idougla 23:40, 7 October 2010 (UTC)

Thank you!

P. 226: 39. Can somebody post a solution for 39? I must be missing something on this one.

Re-write as:

$\mathcal{L}^{-1}[\frac{1}{s^2+5}-\frac{1}{s+5}]=\mathcal{L}^{-1}[\frac{1}{\sqrt{5}}*\frac{\sqrt{5}}{s^2+\sqrt{5}^2}-\frac{1}{s-(-5)}]$

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Dhruv Lamba, BSEE2010