# Solve y'' + 4y = g(t); y(0) = 2, y'(0) = 2 using Laplace transforms, g(t) is given

1. Mar 22, 2012

### s3a

1. The problem statement, all variables and given/known data
Solve the differential equation y'' + 4y = g(t); y(0) = 2, y'(0) = 2 using Laplace transforms.

The g(t) function is a piecewise function and is attached as g.jpg.

2. Relevant equations
Laplace transforms of regular and unit step/Heaviside functions.

3. The attempt at a solution
My work is attached as MyWork.jpg. I suspect the part with the e to be the culprit but I'm not specifically sure as to what I did wrong and would appreciate it if someone could point it out to me.

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• ###### MyWork.jpg
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2. Mar 23, 2012

### LCKurtz

Re: Solve y'' + 4y = g(t); y(0) = 2, y'(0) = 2 using Laplace transforms, g(t) is give

I didn't check all your steps, but it looks to me you haven't taken into account the f(t) = t part. Your formula for that is$$f(t) = t(1-u(8\pi))+8\pi u(t-8\pi)$$

3. Mar 23, 2012

### s3a

Re: Solve y'' + 4y = g(t); y(0) = 2, y'(0) = 2 using Laplace transforms, g(t) is give

What does u(8π) mean?

4. Mar 23, 2012

### Ray Vickson

Re: Solve y'' + 4y = g(t); y(0) = 2, y'(0) = 2 using Laplace transforms, g(t) is give

Why don't you just use the definition, rather than applying formulas that can be misused (as you did)? For $g(t) = \min(t,8 \pi)$ we have
$$L[g](s) = \int_0^\infty e^{-st} g(t) \, dt = \int_0^{8 \pi} e^{-st} t \, dt + \int_{8 \pi}^\infty e^{-st} 8 \pi \, dt,$$ and just do both integrals.

RGV

Last edited: Mar 23, 2012
5. Mar 23, 2012

### LCKurtz

Re: Solve y'' + 4y = g(t); y(0) = 2, y'(0) = 2 using Laplace transforms, g(t) is give

Sorry, that was a typo; that expression should be $u(t-8\pi)$ but the time for me to edit and correct it has expired.

6. Mar 23, 2012

### euquila

Re: Solve y'' + 4y = g(t); y(0) = 2, y'(0) = 2 using Laplace transforms, g(t) is give

u(t-8∏) is a unit step function. You can think of it as energy going into your system after 8∏ time has elapsed. The function is 0 <= 8∏ and 1 after 8∏.