Solve Piecewise Equation with Laplace

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Homework Help Overview

The discussion revolves around solving a piecewise initial value problem involving a second-order differential equation with Laplace transforms. The equation is defined differently over the intervals [0, pi/2] and [pi/2, infinity), with specific initial conditions provided.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss rewriting the piecewise equation in a linear format and applying Laplace transforms. There is confusion regarding the notation and the correct interpretation of the right-hand side (RHS) of the equation. Some participants suggest using Heaviside functions to facilitate the transformation.

Discussion Status

The discussion is active, with participants clarifying notation and exploring the implications of using Heaviside functions. There is acknowledgment of previous misunderstandings, and some guidance has been offered regarding the application of Laplace transforms to the piecewise function.

Contextual Notes

Participants note potential confusion arising from the use of specific values like pi/2 in the context of the problem, as well as the need for clarity in notation when discussing transforms. There is also mention of the basic definition of the Laplace transform and its application to piecewise functions.

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Solve Piecewise Equation with Laplace (Problem Solved ty)

Homework Statement



Solve the initial value problem:

v''=(1/3)v-| sin t for t in [0,pi/2]
| 0 for t in [pi/2,infinity)

v(0)=v'(0)=0

Homework Equations



N/A

The Attempt at a Solution



First I rewrite the piecewise part of the equation in linear format:

v''=(1/3)v-sin(t)-sin(t)*u_(pi/2)

Next I need to apply transforms to both sides of the equations, but I'm stuck on:
sin(t)*u_(pi/2)

I don't see any table entry that would transform this...
 
Last edited:
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I am confused as pi/2>1, then you would have a double RHS. Did you mean this?
 
hunt_mat said:
I am confused as pi/2>1, then you would have a double RHS. Did you mean this?

I corrected the typo. Thanks.
 
You could write the RHS as a Heviside function times sin t I suppose, that would convert it to an expression which can be laplace transformed.
 
hunt_mat said:
You could write the RHS as a Heviside function times sin t I suppose, that would convert it to an expression which can be laplace transformed.

Does RHS mean right hand side?
 
Yes, it's common shorthand.
 
hunt_mat said:
Yes, it's common shorthand.

I thought I already attempted to write it as a heviside function in my attempt...
 
Okay, I didn't understand your notation sorry, you're basically asking what is:

<br /> \int_{0}^{\infty}e^{-st}H(t-\pi /2)\sin tdt<br />

Right? All the rest you can do? Well the above integral is just:

<br /> \int_{0}^{\infty}e^{-st}H(t-\pi /2)\sin tdt=\int_{0}^{\frac{\pi}{2}}e^{-st}\sin tdt<br />

Can you do it now?
 
Last edited:
CarbonWater said:

Homework Statement



Solve the initial value problem:

v''=(1/3)v-| sin t for t in [0,pi/2]
| 0 for t in [pi/2,infinity)

v(0)=v'(0)=0

Homework Equations



N/A

The Attempt at a Solution





First I rewrite the piecewise part of the equation in linear format:

v''=(1/3)v-sin(t)-sin(t)*u_(pi/2)

Next I need to apply transforms to both sides of the equations, but I'm stuck on:
sin(t)*u_(pi/2)

I don't see any table entry that would transform this...

If g(t) = sin(t) for 0<= t <= pi/2 and g(t) = 0 for t > pi/2, then its Laplace transform L(g)(s) Iis integral{ sin(t)*exp(-s*t) dt, t = 0 .. pi/2}. You can do this integral easily enough. It is the type of thing you got for homework in calculus 101; or, you can consult a table of integrals (for the indefinite integral) or submit it to an on-line integrator, such as
http://integrals.wolfram.com/index.jsp .

RGV
 
  • #10
Thanks hunt and ray, I forgot about about the basic definition of the laplace transform. With that said, I don't remember doing any examples using heaviside functions with the basic definition. Is there a generalized form of hunt's second equation?
 
  • #11
Yes, it is given by:

<br /> \int_{0}^{\infty}e^{-st}H(t-\alpha )f(t)dt=\int_{0}^{\alpha}e^{-st}f(t)dt<br />

But I am sure that you couls have figured this out for yourself.
 
  • #12
hunt_mat said:
Yes, it is given by:

<br /> \int_{0}^{\infty}e^{-st}H(t-\alpha )f(t)dt=\int_{0}^{\alpha}e^{-st}f(t)dt<br />

But I am sure that you couls have figured this out for yourself.

I think you had a typo in your previous post which is what was confusing me. (2pi instead of pi/2) Thanks for the help.
 
  • #13
Thanks for pointing this out.
 

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