Solve Piecewise Equation with Laplace

[CarbonWater]

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...

 

[hunt_mat]

I am confused as pi/2>1, then you would have a double RHS. Did you mean this?

 

[CarbonWater]

I am confused as pi/2>1, then you would have a double RHS. Did you mean this?

I corrected the typo. Thanks.

 

[hunt_mat]

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.

 

[CarbonWater]

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?

 

[hunt_mat]

Yes, it's common shorthand.

 

[CarbonWater]

Yes, it's common shorthand.

I thought I already attempted to write it as a heviside function in my attempt...

 

[hunt_mat]

Okay, I didn't understand your notation sorry, you're basically asking what is:

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

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

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

Can you do it now?

 

[Ray Vickson]

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

 

[CarbonWater]

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?

 

[hunt_mat]

Yes, it is given by:

$$\int_{0}^{\infty}e^{-st}H(t-\alpha )f(t)dt=\int_{0}^{\alpha}e^{-st}f(t)dt$$

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

 

[CarbonWater]

Yes, it is given by:

$$\int_{0}^{\infty}e^{-st}H(t-\alpha )f(t)dt=\int_{0}^{\alpha}e^{-st}f(t)dt$$

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.

 

[hunt_mat]

Thanks for pointing this out.