Inverse Laplace (stuck @ Partial Fraction)

[Italo Campoli]

Homework Statement

Find the Inverse laplace transform of:

http://www4c.wolframalpha.com/Calculate/MSP/MSP14541hg721e74730d4fb00004644i96f59549h1d?MSPStoreType=image/gif&s=30&w=201.&h=40.

Result http://www4c.wolframalpha.com/Calculate/MSP/MSP14591hg721e74730d4fb000042gcebh89c38eib7?MSPStoreType=image/gif&s=30&w=505.&h=36.

Homework Equations

=

maybe e^(at) forms on the table too since there´s s-a form in a part of the bottom

The Attempt at a Solution

[/B]
when getting to the final step of the partial fraction i get:

http://www4b.wolframalpha.com/Calculate/MSP/MSP14771df8fbfi46b7gihh000059eh9936e24diiii?MSPStoreType=image/gif&s=24&w=267.&h=22. (sorry about minus, idk what happen to the program ...)

now i could asume that http://www5a.wolframalpha.com/Calculate/MSP/MSP172120i2249gd4beb17900003d9dfc9b6ef7i9e8?MSPStoreType=image/gif&s=54&w=63.&h=20. that being 2i but it gets more complicated than i would expect, see, teacher sayd it is only 1 page long exercise, i would imagine there must be a much simpler way than going by imaginary roots.

I would appreciate if possible a step by step explanation since as a fun fact we were given this exercise without any class about inverse transform and i myself did some research. Thanks in advance

 

[Mark44]

Homework Statement

Find the Inverse laplace transform of:

http://www4c.wolframalpha.com/Calculate/MSP/MSP14541hg721e74730d4fb00004644i96f59549h1d?MSPStoreType=image/gif&s=30&w=201.&h=40.

Result http://www4c.wolframalpha.com/Calculate/MSP/MSP14591hg721e74730d4fb000042gcebh89c38eib7?MSPStoreType=image/gif&s=30&w=505.&h=36.

Homework Equations

=

maybe e^(at) forms on the table too since there´s s-a form in a part of the bottom

The Attempt at a Solution

[/B]
when getting to the final step of the partial fraction i get:

http://www4b.wolframalpha.com/Calculate/MSP/MSP14771df8fbfi46b7gihh000059eh9936e24diiii?MSPStoreType=image/gif&s=24&w=267.&h=22. (sorry about minus, idk what happen to the program ...)

You've omitted a lot of work, but it looks like you started with this:
$$\frac{s}{(s^2 + 6s + 18)((s^2 + 4)} = \frac{As + B}{s^2 + 6s + 18} + \frac{Cs + D}{s^2 + 4}$$
If you multiply both sides by $(s^2 + 6s + 18)((s^2 + 4)$, you get something different from what you show above; namely,
$s = (As + B)(s^2 + 4) + (Cs + D)(s^2 + 6s + 18)$
Note the extra parentheses I show that you don't show. These parentheses make a significant difference.
Expand the right side, and group the terms by powers of s to solve for A, B, C, and D.

now i could asume that http://www5a.wolframalpha.com/Calculate/MSP/MSP172120i2249gd4beb17900003d9dfc9b6ef7i9e8?MSPStoreType=image/gif&s=54&w=63.&h=20. that being 2i but it gets more complicated than i would expect, see, teacher sayd it is only 1 page long exercise, i would imagine there must be a much simpler way than going by imaginary roots.

Yes, there is. Don't break up the quadratics.

The formulas you want are these:
$$\mathcal{L}^{-1}[\frac{s}{s^2 + a^2}] = \cos(at)$$
and
$$\mathcal{L}^{-1}[\frac{1}{s^2 + a^2}] = \sin(at)$$

You'll need to complete the square in the one with $s^2 + 6s + 18$ to make it fit one of these formulas.

I would appreciate if possible a step by step explanation since as a fun fact we were given this exercise without any class about inverse transform and i myself did some research. Thanks in advance

 

[axmls]

Would you mind writing out all of your steps? It would make it easier to check.

 

[Italo Campoli]

You've omitted a lot of work, but it looks like you started with this:
$$\frac{s}{(s^2 + 6s + 18)((s^2 + 4)} = \frac{As + B}{s^2 + 6s + 18} + \frac{Cs + D}{s^2 + 4}$$
If you multiply both sides by $(s^2 + 6s + 18)((s^2 + 4)$, you get something different from what you show above; namely,
$s = (As + B)(s^2 + 4) + (Cs + D)(s^2 + 6s + 18)$
Note the extra parentheses I show that you don't show. These parentheses make a significant difference.
Expand the right side, and group the terms by powers of s to solve for A, B, C, and D.
Yes, there is. Don't break up the quadratics.

The formulas you want are these:
$$\mathcal{L}^{-1}[\frac{s}{s^2 + a^2}] = \cos(at)$$
and
$$\mathcal{L}^{-1}[\frac{1}{s^2 + a^2}] = \sin(at)$$

You'll need to complete the square in the one with $s^2 + 6s + 18$ to make it fit one of these formulas.

yes you're quite right that's were i started, i see now, with those parentheses there are less things to group up, ill give it a try thanks for that observation

Would you mind writing out all of your steps? It would make it easier to check.

sorry about that, on mark post above there is the beginning of the exercise, altho ill redo it from his point of view an post any doubt i have later on :P thanks both of you