Question

y'' -5y' + 6y = 0, y(0) = 1, y'(0) =2
[s^2 F(s) - s f(0) - f'(0)] -5 [F(s) - f(0)] + 6[F(s)] = 0
(s^2 +1)F(s) - (s -5)f(0) - f'(0) = 0 (s^2 + 1)F(s) - (s-5)(1) - 2 = 0
(s^2 + 1)F(s) = s -3 F(s) = (s-3)/(s^2 + 1)

Here's where I'm stuck. I can't factor the denominator to get partial fractions, so I feel like I should just divide up the fraction like this: s/(s^2 + 1) -3/(s^2 + 1)
And then use cost - 3sint, but that's not the answer in my book.

Answer 1

wut? It looks like you should be able to find the solutions readily from the char.polynomial. :)

(r-2)(r-3) ==> \(y=c_1e^{2x}+c_2e^{3x}\)

Now use the power of your initial conditions to solve for c1 and c2 ! :D

Answer 2

I know, but we have to use Laplace. :(

Answer 3

You're sure about what you've done so far?

Answer 4

I'm playing with it a little now. . . trying completing the square and stuff, but I'm pretty confused at this point.

Answer 5

I can say that you've reached to the point where \(F(s)=\frac{s}{s^2+1}-\frac{3}{s^2+1}\), and now you're trying to find the inverse laplace transform?

Answer 6

indeed

Answer 7

Latex is not working here, is it working for you?

Answer 8

nope

Answer 9

i can read not latex notation though

Answer 10

OK. The inverse laplace transform of both terms are very common and can be easily proved using the definition of laplace transform and integration by parts. For the first term, you should use \(L\{ \cos kt\}=\frac{s}{s^2+k^2}\), with k=1. For the second term you can use \(L\{\sin kt\}=\frac{k}{s^2+k^2}\) with \(k=1\) as well, and take \(3\) as a constant.

Answer 11

Note: "L{f(t)} denotes the laplace transform of f(t)"

Answer 12

Right, that's what I did! But I'll tell you what my book says:

Answer 13

y = e^2t

Answer 14

Using the previous two formulas we can easily find that: \(f(t)=\cos t-3\sin t\). But that does not look like the right solution to the differential equation!!

Answer 15

Give me a minute to check what you did :D

Answer 16

ok take your time

Answer 17

Well, here's the full answer:
Taking the laplce transform of the given differential equation we have \(s^2Y(s)-sy(0)-y'(0)-5(sY(s)-y(0))+6Y(s)=0\). Substituting the given initial conditions, we can find that \(Y(s)(s^2-5s+6)=s-3\). Divide both sides by \(s^2-5s+6\), which can be factored to \((s-3)(s-2)\), we end up with \(Y(s)=\frac{s-3}{(s-3)(s-2)}=\frac{1}{s-2}\). So, the solution of the differential equation is the laplace transform of thar last expression which is \(y(t)=e^{2t}\).

Answer 18

If there is a step that seems unclear to you, just let me know!

Answer 19

wait one sec....where did the 5s come from in the s2−5s+6?

Answer 20

oh wait

Answer 21

I know what I've been doing wrong

Answer 22

Good :)

Answer 23

f'(t) = sF(s) - f())

Answer 24

and I did F(s) -f(0)

Answer 25

THANK YOU THANK YOU THANK YOU THANK YOUTHANK YOU THANK YOU

Answer 26

I see!! So you know where it came from?

Answer 27

Haha, You're welcome!!

Answer 28

know where what came from?

Answer 29

the 5s.

Answer 30

yes, i do, thanks so much for your time

Answer 31

No problem!!

Answer 32

not anymore there isn't ;)

Answer 33

Ok, time to go try these problems again. Thanks again.