# Solving a Differential Eq using Laplace and Unit-Step function (1 Viewer)

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!!!Solving a Differential Eq using Laplace and Unit-Step function

1. The problem statement, all variables and given/known data
I am having a terrible time just starting this. I need some explanation here. I have given up on the text I am using...

I need to solve the following piecewise function using Laplace and Unit-Step function:

y"+4y=f(t) y(0)=-1, y'(0)=0

where f(t)= t, t<2
5, t>2

3. The attempt at a solution

Now I condensed it to one line using the Unit Step function giving:

$y''+4y=t-t*U(t-2)+5*U(t-2)$

Applying Laplace to the LHS is easy enough, but what is the Laplace of the unit step function? I am having a hard tome extrapolating it from the text.

So far I have

$s^2Y(s)-sy(0)-y'(0)+4Y(s)=\frac{1}{s^2}-???$

Could somebody help me out with the "U" terms here?

Thank you

SKIP AHEAD TO POST #10 That is where I need help

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$$L(-t*U(t-2))$$

I cannot figure this out!!!!

$$L(-t*U(t-2))$$
Is this even "Laplacable"? I guess I do not understand the rules

nobody has any advice?

#### HallsofIvy

Since the U(t-2) just changes the limits of integration to 2 to infinity, rather than 0 to infinity, you need to change the variable from x to u= x- 2 to get back the 0 limit.

Doing that gives a general rule that the Laplace transform of f(x)U(x-a) is e-asF(s) where F(s) is the Laplace transform of f(x).

So for $-t*U(t-2)$ I should get $e^{-2s}*-\frac{1}{s^2}$?

I don't think this is correct...

Since the U(t-2) just changes the limits of integration to 2 to infinity, rather than 0 to infinity, you need to change the variable from x to u= x- 2 to get back the 0 limit.

Doing that gives a general rule that the Laplace transform of f(x)U(x-a) is e-asF(s) where F(s) is the Laplace transform of f(x).
Halls, I just found this in my text as a side note. It would appear that the Laplace of f(t)*U(t-a)=e^{-as}*L{g(t+a)} ........ not just +L{g(t)}

Okay so I have got it down to this point:

$$Y(s)[s^2+4]=\frac{1}{s^2}-e^{-2s}*\frac{1}{(s+2)^2}+5e^{-2s}*\frac{1}{s+2}-s$$

I was going to write the RHS over the LCD and then divide out by [s^2+4] .....sound good? .... there must be a neater way to so this... isn't there? This looks like it will get very sloppy very quick!

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This equation is B.S. F#\$k this.

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