Question

Working on Laplace transformation
y'''+2y''-y'-2y=sin(3t) y(0)=0 y'(0)=0 y''(0)=1

Checking my step with the solution manual but apparently I did something wrong.
They got Laplace(y)= (s^2-12)/(s-1)(s+1)(s+2)(s^2+9)
vs what I got
(3+(s^2+9))/((s+1)(s-1)(s^2+9))
could some tell me what I did wrong?

Answer 1

If you have the work that you did, it would be easier to identify the source of error. Otherwise, I will work through the problem myself.

Answer 2

give me a sec got to retype it

Answer 3

s^3Y(s)-y''(0)+2s^2Y(s)-1sY(s)-2Y(s)
Y(s)(s^3+2s^2-s-2)-y''(0)=3/(s^2+9)

Y(s)(s^2-1)(s+2)=3/(s^2+9) +y''(0)

Answer 4

clearly y''(0)=1 making it (3+(s^2+9))/((s+1)(s-1)(s^2+9))

how did they get (s^2-12)/(.....)
focusing on the -12 instead

Answer 5

Oh, I am not seeing where the -12 is coming from. They may have mistakenly assumed sin(at) had Laplace transform a/(s^2 - a^2) and corrected mid problem or something of that nature.

I am getting (s^2 + 12)/[(s + 2)(s + 1)(s - 1)(s^2 + 9)]

Answer 6

Besides the (s^2 - 12) being a typo for (s^2 + 12), then the answer you wrote just needs the (s + 2) factor also accounted for, and then it should be good.

Answer 7

that's the thing... I solve this problem twice with the answer I got
and solve this problem using the other method and apparently (s^2-12)/(.....) is the right factor

Answer 8

What was the "other method"? As in, (s^2 - 12) IS correct? Since I am not seeing how it is possible, but I am working out the problem myself.

Answer 9

using the 3rd order differential equation, transforming y'''+2y''-y'-2y=sin(3t)
to a homogeneous algebra problem

http://www.wolframalpha.com/input/?i=y%27%27%27%2B2y%27%27-y%27-2y%3Dsin%283t%29
[the solution is said to have a sin and a cos]

my solution are compose of 2 Partial fraction
3/(s+1)(s-1)(s^=9) + 1/(s^-1)(s+2)
where only sin is present :/

Answer 10

I think there is suppose to be a cosine... ahhh

Answer 11

\( \displaystyle \frac{s^2 + 12}{(s^2 + 9)(s + 2)(s - 1)(s + 1)} = \frac{As + B}{s^2 + 9} + \frac{C}{s + 2} + \frac{D}{s - 1} + \frac{E}{s + 1} \)
Broken up like this, that (As + B)/(s^2 + 9) would break up into a sine inverse and a cosine inverse, right?

Answer 12

I just used
Y(s)= 3/(s+1)(s-1)(s^2+9) +1/(s^2-1)(s+2) since it's a lot easier

http://www.wolframalpha.com/input/?i=partial+fraction+3%2F%28%28s%2B1%29%28s-1%29%28s^2%2B9%29%29

http://www.wolframalpha.com/input/?i=partial+fraction+1%2F%28%28s^2-1%29%28s%2B2%29%29

Answer 13

Y(s) -.5/(s+1)+1/6(s-1) +1/3(s+2) -3/20(s+1)+2/20(s-1)-3/10(s^2+9)

-3/10(s^2+9) produce the sin

Answer 14

Oh, but the first partial fraction is missing that (s+2) in the denominator.
The (s+2) has to be distributed to both terms as well.
http://www.wolframalpha.com/input/?i=partial+fraction+3%2F%28%28s%2B1%29%28s-1%29%28s%2B2%29%28s%5E2%2B9%29%29

Answer 15

\( \dfrac{3 + (s^2 + 9)}{\color{green}{(s + 2)}(s^2 - 1)(s^2 + 9)} = \dfrac{3}{\color{green}{(s + 2)}(s^2 - 1)(s^2 + 9)} + \dfrac{1}{\color{green}{(s + 2})(s^2 - 1)} \)

Answer 16

s+2... wow thanks!

Answer 17

Yup, glad to help! :)

Answer 18

where did i miss the s+2... lol
this problem is ridiculous

Answer 19

yup...just wow. I found it.

Answer 20

Yeah, the huge degree denominator to decompose is not very helpful here. lol
That was between your original answer (3 + (s^2 + 9))/... and where you broke it into two fractions. Although in the first post, I didn't notice the (s + 2) in the denominator either.