Solving a Differential Equation with Unit Step Function

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Homework Help Overview

The discussion revolves around solving a differential equation involving a unit step function. The equation presented is i'' + 2i = f(t), with initial conditions i'(0)=i(0)=0, where f(t) is defined as the difference of two unit step functions, u(t-10) - u(t-20).

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the Laplace transform of the function and the process of inverting it. There are questions regarding the application of the shifting theorem and the correct form of the inverse transform, particularly whether it should involve sine or cosine functions. Some participants express uncertainty about handling the exponential shifts due to the presence of multiple terms.

Discussion Status

The discussion is ongoing, with participants offering insights into the inversion of the Laplace transform and questioning the correctness of the forms derived. There is no explicit consensus on the method to proceed, but various approaches are being explored.

Contextual Notes

Participants note potential confusion regarding the application of the exponential shift theorem and the need to break down the fractions for clarity. There is also mention of correcting terms in the equations presented.

icystrike
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Homework Statement


Its not homework anyway:
I'm asked to find the solution to the differential equation:

i'' + 2i = f(t)
i'(0)=i(0)=0

f(t) = u(t-10) - u(t-20) Unit step function (I've found in part a of the question)

Then I've gotten:

[tex]\mathscr{L}(i) = \frac{e^{-10s}-e^{-20s}}{s(s^{2}+2)}[/tex]

Homework Equations


The Attempt at a Solution

 
Last edited:
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icystrike said:

Homework Statement


Its not homework anyway:
I'm asked to find the solution to the differential equation:

i'' + 2i = f(t)
i'(0)=i(0)=0

f(t) = u(t-10) - u(t-20) Unit step function (I've found in part a of the question)

Then I've gotten:

[tex]\mathscr{L}(i) = \frac{e^{-10s}-e^{-20s}}{s(s^{2}+2)}[/tex]

Do you know how to invert

[tex]\frac 1 {s(s^2+2)}[/tex]

by itself? That is what you do then use the shifting theorem with regard to the exponentials.
 
[tex]\frac 1 {s(s^2+2)}[/tex] ~~> [tex]\frac{1}{2} + \frac{1}{2} sin (\sqrt{2}t)[/tex]

Uhh.. I am not too sure abt the exponential shift as its sum of two exponents instead of just one of them ...
 
icystrike said:
[tex]\frac 1 {s(s^2+2)}[/tex] ~~> [tex]\frac{1}{2} + \frac{1}{2} sin (\sqrt{2}t)[/tex]

Close but not quite. Are you sure it isn't a cosine form?

Uhh.. I am not too sure abt the exponential shift as its sum of two exponents instead of just one of them ...

Break it into two fractions with the exponentials separated into different numerators and work those.
 
attachment.php?attachmentid=33726&stc=1&d=1301509859.jpg

The last equation should be f(t) instead of F(s)
 

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