## 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). $

Even solutions (added by Adam M on Oct 5, please check results):

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)}] $