How to solve this problem using laplace transform?

[haha1234]

Homework Statement

The differential equation given:
y''-y'-2y=4t²

Homework Equations

The Attempt at a Solution

I used the laplace transform table to construct this equation,and then I did partial fraction for finding the inverse laplace transform.But I'm now stuck at finding the inverse laplace transform of 1/s^3 and 1/s^2...

And the attached photo is the attempted solution.
https://www.physicsforums.com/attachments/92000

 

[Mark44]

Homework Statement

The differential equation given:
y''-y'-2y=4t²

Homework Equations

The Attempt at a Solution

I used the laplace transform table to construct this equation,and then I did partial fraction for finding the inverse laplace transform.But I'm now stuck at finding the inverse laplace transform of 1/s^3 and 1/s^2...

And the attached photo is the attempted solution.
https://www.physicsforums.com/attachments/92000

I didn't verify your work, but here is a table of Laplace transforms - http://web.stanford.edu/~boyd/ee102/laplace-table.pdf

 

[haha1234]

I didn't verify your work, but here is a table of Laplace transforms - http://web.stanford.edu/~boyd/ee102/laplace-table.pdf

But there is no conversion of 1/s^3 provided!

 

[Mark44]

But there is no conversion of 1/s^3 provided!

Look just below the one for 1/s². It's a more general formula.

 

[Ray Vickson]

Homework Statement

The differential equation given:
y''-y'-2y=4t²

Homework Equations

The Attempt at a Solution

I used the laplace transform table to construct this equation,and then I did partial fraction for finding the inverse laplace transform.But I'm now stuck at finding the inverse laplace transform of 1/s^3 and 1/s^2...

And the attached photo is the attempted solution.
https://www.physicsforums.com/attachments/92000

If you know the inverse transform of 1/s, then you can get the inverse transform of 1/s^2 by integration, and of 1/s^3 by integration again. Remember: there are some standard general transform results that are helpful. Below, let $f(t) \leftrightarrow g(s) = {\cal L}(f)(s)$. Then:
$$\begin{array}{l} \displaystyle \frac{df(t)}{dt} \leftrightarrow s g(s) - f(0+)\\ \int_0^t f(\tau) \, d \tau \leftrightarrow \displaystyle \frac{1}{s} g(s) \end{array}$$
These were given specifically in the table suggested by Mark44; did you miss them?