# Laplace and Series Question

• Nov 24th 2009, 08:03 PM
electricalphysics
Laplace and Series Question
I need help on two webwork questions.

The first is a Laplace
http://img16.imageshack.us/img16/670/webwork1.th.jpg

For this is transformed the equation and ended up with a linear differential eqn of the form:
Y'(s) +Y(s) (sē-6)/-2s= (8+7sē)/-2sē

Im not sure if this is the correct method. I'm also stuck after this step as I cannot find a suitable integrating factor.

Secondly,
I completed most of this question
I am just unsure as to what
" The closed form of solution a" is referring too.

http://img20.imageshack.us/img20/7262/45883123.jpg

Thanks !
• Nov 25th 2009, 12:39 AM
chisigma
First question...

Remembering that if $Y(s)= \mathcal{L} \{y(t)\}$ the following basic properties hold...

$\mathcal{L} \{y^{'}(t)\} = s\cdot Y(s) - y(0)$

$\mathcal{L} \{y^{''}(t)\} = s^{2}\cdot Y(s) - s\cdot y(0) - y^{'}(0)$

$\mathcal{L} \{t\cdot y(t)\} = - Y^{'} (s)$ (1)

... the DE...

$y^{''} + 2t\cdot y^{'} - 4\cdot y = 8$ , $y(0)=7$, $y^{'}(0)=0$ (2)

... in terms of LT is written as...

$s^{2}\cdot Y(s) -7s - 2\cdot s\cdot Y^{'}(s) - 2\cdot Y(s) -4\cdot Y(s) = \frac{8}{s}$ (3)

... and from (3) ...

$Y^{'}(s)= \frac{s^{2}-6}{2s}\cdot Y(s) - \frac{8 + 7\cdot s^{2}}{2\cdot s^{2} }$ (4)

So your solution is correct but the problem now is to solve the first order DE (4) in terms of s...

Kind regards

$\chi$ $\sigma$
• Nov 25th 2009, 09:36 AM
chisigma
In the last post we have shown that the linear second order DE in t...

$y^{''} + 2t\cdot y^{'} - 4\cdot y = 8$ , $y(0)=7$, $y^{'}(0)=0$ (1)

... can be written in terms of LT as the first order DE in s...

$Y^{'}(s)= \frac{s^{2}-6}{2s}\cdot Y(s) - \frac{8 + 7\cdot s^{2}}{2\cdot s^{2} }$ (2)

The (2) can be solved in 'standard fashion' and its general solution is...

$Y(s) = c \cdot \frac{s^{\frac{x^{2}}{4}}}{s^{3}} + \frac{36}{s^{3}} + \frac{7}{s}$ (3)

For the 'initial value theorem' from the condition $y(0)=7$ we detrive that it must be...

$\lim_{t \rightarrow 0} y(t) = \lim_{s \rightarrow \infty} s\cdot Y(s) = 7$ (4)

... and that means that in (3) must be $c=0$, so that the solution of (1) is...

$y(t)= \mathcal{L}^{-1} \{\frac{36}{s^{3}} + \frac{7}{s}\} = 18 t^{2} + 7$ (5)

Kind regards

$\chi$ $\sigma$