Y'' - 9y = 4t - δ_2(t); y() = -2, y'(0) = -1

[s3a]

Homework Statement

Solve the differential equation y'' - 9y = 4t - δ_2(t); y() = -2, y'(0) = -1 for y(t) when 0 <= t < 2 as well as when t > 2.

Homework Equations

Dirac delta function. Heaviside function. Laplace and Inverse Laplace Transforms.

The Attempt at a Solution

I'll type the relevant information here since my handwriting for my work is ugly (but I am uploading it anyways just in case you need it). I get to Y(s) = 4/[s^2 * (s^2 - 9)] - e^(-2s)/(s^2 - 9) - (2s+1)/(s^2 - 9) and I'm stuck at getting the inverse Laplace Transform of the bolded/italicized term. Could someone please show me how to do that part? I know it involves the unit step/Heaviside function but I don't know what to do specifically.

From Wolfram Alpha,
The inverse Laplace Transform of that bolded/italicized term is:
http://www.wolframalpha.com/input/?i=inverse+laplace+transform+of+{-exp(-2s)/(s^2+-+9)}
The inverse Laplace Transform of the whole (s) = 4/[s^2 * (s^2 - 9)] - e^(-2s)/(s^2 - 9) - (2s+1)/(s^2 - 9) equality is:
http://www.wolframalpha.com/input/?i=inverse+laplace+transform+of+{4/[s^2+*+(s^2+-+9)]+-+exp(-2s)/(s^2+-+9)+-+(2s%2B1)/(s^2+-+9)}

(Just to note, the Wolfram Alpha answers would be wrong if my work leading up to the equation I retyped in this forum post from my work is incorrect.)

Thanks in advance!

 

[Mark44]

Homework Statement

Solve the differential equation y'' - 9y = 4t - δ_2(t); y() = -2, y'(0) = -1 for y(t) when 0 <= t < 2 as well as when t > 2.

Homework Equations

Dirac delta function. Heaviside function. Laplace and Inverse Laplace Transforms.

The Attempt at a Solution

I'll type the relevant information here since my handwriting for my work is ugly (but I am uploading it anyways just in case you need it). I get to Y(s) = 4/[s^2 * (s^2 - 9)] - e^(-2s)/(s^2 - 9) - (2s+1)/(s^2 - 9) and I'm stuck at getting the inverse Laplace Transform of the bolded/italicized term. Could someone please show me how to do that part? I know it involves the unit step/Heaviside function but I don't know what to do specifically.

From Wolfram Alpha,
The inverse Laplace Transform of that bolded/italicized term is:
http://www.wolframalpha.com/input/?i=inverse+laplace+transform+of+{-exp(-2s)/(s^2+-+9)}
The inverse Laplace Transform of the whole (s) = 4/[s^2 * (s^2 - 9)] - e^(-2s)/(s^2 - 9) - (2s+1)/(s^2 - 9) equality is:
http://www.wolframalpha.com/input/?i=inverse+laplace+transform+of+{4/[s^2+*+(s^2+-+9)]+-+exp(-2s)/(s^2+-+9)+-+(2s%2B1)/(s^2+-+9)}

(Just to note, the Wolfram Alpha answers would be wrong if my work leading up to the equation I retyped in this forum post from my work is incorrect.)

Thanks in advance!

Have you tried splitting up 1/(s² - 9) into two terms using partial fraction decomposition?

 

[s3a]

I haven't because I didn't foresee any upcoming successful step.

Here is my work for what you mentioned.

 

[Mark44]

That work looks OK.

What you need to deal with the expressions you're asking about is this formula:

$$L^{-1}(e^{-as}~G(s)) = u_a g(t - a)$$

Here G(s) = $L {g(t)}$

Edit: For some reason, the LaTeX isn't rendering correctly. Also, I neglected to indicate that I was talking about the inverse Laplace transform in the first equation above.

 

[s3a]

Did you mean:

L{u_c(t) f(t-c)} = e^(-cs) L{f(t)} ?

 

[Mark44]

Did you mean:

L{u_c(t) f(t-c)} = e^(-cs) L{f(t)} ?

The first formula should have been for the inverse Laplace transform. It's now fixed.

 

[s3a]

Wrong thread sorry.