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

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

2,968
2
!!!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:

[itex]y''+4y=t-t*U(t-2)+5*U(t-2)[/itex]

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

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

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

Thank you
 
2,968
2
SKIP AHEAD TO POST #10 That is where I need help
 
Last edited:
2,968
2
Can somebody please help me with the Laplace Transform

[tex]L(-t*U(t-2))[/tex]

I cannot figure this out!!!!
 
2,968
2
Can somebody please help me with the Laplace Transform

[tex]L(-t*U(t-2))[/tex]
Is this even "Laplacable"? I guess I do not understand the rules
 
2,968
2
nobody has any advice?
 

HallsofIvy

Science Advisor
41,626
821
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).
 
2,968
2
So for [itex]-t*U(t-2)[/itex] I should get [itex]e^{-2s}*-\frac{1}{s^2}[/itex]?
 
2,968
2
I don't think this is correct...
 
2,968
2
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)}
 
2,968
2
Okay so I have got it down to this point:

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

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!
 
Last edited:
2,968
2
This equation is B.S. F#$k this.
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top