Solving ODE Step function with Laplace

[xVladx]

Homework Statement

Hi all came across this problem whilst doing some revision and i can't work out the answer

Solve the following equation with laplace transformation

Homework Equations

y''+16y = f(t) = { 1 t < pi ] with y(0) = 0 and y'(0) = 0
_____________{ 0 t >= pi ]

The Attempt at a Solution

I find the laplace transform and get:

Y(s) = (1/16s)-(1/(16(s^2+4^2)))-(e^(-pi*s)/(s^2+4^2))

I am kind of stuck when doing the final step since never delt with this type of problem and the sine/cosine transforms.

I come up with y(t)=1/16 - sin(4t)/64 - (sin(4t)* u(t - pi)) /16 but not sure if it is right or how to get it back to the final form.

Any help is appreciated

 

[vela]

I find the laplace transform and get:

Y(s) = (1/16s)-(1/(16(s^2+4^2)))-(e^(-pi*s)/(s^2+4^2))

Show us your work to get to this point. It doesn't look right to me.

 

[xVladx]

Take Laplace of both sides:
s^2Y(s) -sy(0)-y'(0) +16Y= 1/s - e^-pi*s
y(0) = y'(0) = 0
Y(s^2+16)=1/s - e^-pi*s
Y = [1/s(s^2+16)] - e^-pi*s/(s^2+16)
Using partial fractions [1/s(s^2+16)] = 1/16s - 1/16(s^2+4^2)
Y(s) = (1/16s)-(1/(16(s^2+4^2)))-(e^(-pi*s)/(s^2+4^2))

 

[xVladx]

Also I've seen in other posts that some people can post the equation writing similar to the way it is done in MS word, how do you do that? would be much neater then

 

[vela]

You took the Laplace transform of the righthand side incorrectly. In the time domain, you can write it as 1-u(t-pi). You didn't calculate the transform of the second term correctly.

Regarding posting equations, start here:

https://www.physicsforums.com/showthread.php?t=386951

 

[xVladx]

Also I've seen in other posts that some people can post the equation writing similar to the way it is done in MS word, how do you do that? would be much neater then

Nevermind this I found out about the Latex thing. Just going to restate the problem to give it a try

\ddot{y} + 16y = f(t) = \left\{\begin{array}{cc}1&amp;t&lt;{\pi}\\0 &amp; t\geq{\pi}\end{array}\right] with y(0) = 0 and \dot{y}(0) = 0

 

[xVladx]

Should I have gotten Laplace (1-u(t-\pi)) = \LARGE\frac{1}{s}-\frac{e^{-\pi s}}{s}

 

[vela]

Yes.

 

[xVladx]

Would my answer then be
y(t) = \left\{\begin{array}{cc}\frac{1}{16}-\frac{1}{64\sin(4t)}&amp;t&lt;{\pi}\\0 &amp; t\geq{\pi}\end{array}\right]

 

[vela]

No, that's not correct.

 

[xVladx]

OK guess I am stuck then.

 

[vela]

Recheck your partial fractions decomposition of

\frac{1}{s(s^2+16)}

And how did you end up with the sine in the denominator?

 

[xVladx]

Hmm mistyped my Latex, I ment:

y(t) = \left\{\begin{array}{cc}\frac{1}{16}-\frac{\cos(4t)}{16}&amp;t&lt;{\pi}\\0 &amp; t\geq{\pi}\end{array}\right][/QUOTE]

 

[vela]

You got it.

 

[xVladx]

Thanks very much for you help! :)