second order differential equation using laplace/inverse lap

[trickslapper]

my original is: x'' +(w^2)*x=Fsin(w*t); with initial conditions x(0)=x'(0)=0.

I applied laplace transform to all three pieces and i got:

X*s^2 + w^2/s^2 = F*w/s^2+w^2.

The answer i'm supposed to get is F*w/(s^2 + w^2)^2 but i can't get to it.. also i've been given a hint that says to express the final answer as (F/2w^2) * 2w^3/(s^2+w^2)^2

is there some algebra trick i need to apply somewhere, because when i solve the algebraic equation for X i don't get the answer i'm supposed to get. Any ideas to get the correct Laplace transform? If i can get there then i'll try the inverse laplace on my own (i know what that is supposed to be as well)

 

[galactus]

Re: second order differential equation using laplace/inverse

\(\displaystyle x''+w^{2}x=Fsin(wt), \;\ x'(0)=0, \;\ x(0)=0\)

Using \(\displaystyle L(x'')=s^{2}X-sx(0)-x'(0)\)

Since x(0)=0 and x'(0)=0, we get

\(\displaystyle L(x'')=s^{2}X\)

So, making the sub into the DE, we get:

\(\displaystyle s^{2}X+w^{2}X=\frac{Fw}{s^{2}+w^{2}}\)

\(\displaystyle X(s^{2}+w^{2})=\frac{Fw}{s^{2}+w^{2}}\)

\(\displaystyle X=\frac{Fw}{(s^{2}+w^{2})^{2}}\)

Now, do your inverse Laplace thing.

 

[trickslapper]

Re: second order differential equation using laplace/inverse

Ok, the hints were actually for the second question which is to find the inverse laplace. The hint is to rewrite X(s) as F/2w[sup:1xcqvl0x]2[/sup:1xcqvl0x] * 2w[sup:1xcqvl0x]3[/sup:1xcqvl0x]/(s[sup:1xcqvl0x]2[/sup:1xcqvl0x] + w[sup:1xcqvl0x]2[/sup:1xcqvl0x])[sup:1xcqvl0x]2[/sup:1xcqvl0x]

The answer is supposed to be : F/2w[sup:1xcqvl0x]2[/sup:1xcqvl0x] (sinwt-wtcoswt), do you guys see how to get to this solution?

 

[galactus]

Re: second order differential equation using laplace/inverse

Yes, do a little rewriting and you can look it up in a Laplace table.

\(\displaystyle tcos(wt)=\frac{s^{2}-w^{2}}{(s^{2}+w^{2})^{2}}\)

\(\displaystyle sin(wt)=\frac{w}{s^{2}+w^{2}}\)

\(\displaystyle \frac{F}{w^{2}}\left(\frac{w}{s^{2}+w^{2}}-\frac{w(s^{2}-w^{2})}{(s^{2}+w^{2})^{2}}\right)=\frac{Fw}{(s^{2}+w^{2})^{2}}\)

 

[trickslapper]

Re: second order differential equation using laplace/inverse

ok cool, at first i was wondering how you got t*cos(wt) to be that but then i saw in my notes one of the properties that says L[t*f(t)] = -F'(s).

thanks man !

Edit: If i didn't already know the answer, how would we solve this problem? Or...how did the first person to solve this problem solve it?

 

[galactus]

Re: second order differential equation using laplace/inverse

Get a Laplace table and do some algebraic fanaggling.

As I showed in my last post. \(\displaystyle \frac{Fw}{(s^{2}+w^{2})^{2}}\) can be rewritten so that resembles a familiar Laplace form(s) .

Looking at it, as is, tells us there will be sin and cos involved.