Solving a Differential Equation with Unit Step Function

[icystrike]

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

 

[LCKurtz]

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.

 

[icystrike]

[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 ...

 

[LCKurtz]

[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.

 

[icystrike]

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