IVP problem using Laplace transform and partial fractions

[member 731016]

Homework Statement

please see below

Relevant Equations

Please see below

For this problem,

The solution is,

However, I'm confused by the partial fraction decomposition of $\frac{2}{s^4(s^2 + 1)}$

I never done that sort of thing before. However, I think it would be done like this (Please correct me if I am wrong, the algebra is crazy here).

$\frac{2}{s^4(s^2 + 1)} = \frac{A}{s} + \frac{Bs + C}{s^2} + \frac{Ds^2 + Es + F}{s^3} + \frac{Gs^3 + Hs^2 + Js + K}{s^4} + \frac{Ls + M}{s^2 + 1}$

It seems rather tedious because of the $\frac{1}{s^4}$, but is that correct (There may be a simpler method but I can't see it).

Thanks!

 

[Mark44]

Homework Statement: please see below
Relevant Equations: Please see below

For this problem,
View attachment 346713
The solution is,
View attachment 346714
However, I'm confused by the partial fraction decomposition of $\frac{2}{s^4(s^2 + 1)}$

I never done that sort of thing before. However, I think it would be done like this (Please correct me if I am wrong, the algebra is crazy here).

$\frac{2}{s^4(s^2 + 1)} = \frac{A}{s} + \frac{Bs + C}{s^2} + \frac{Ds^2 + Es + F}{s^3} + \frac{Gs^3 + Hs^2 + Js + K}{s^4} + \frac{Ls + M}{s^2 + 1}$

No, that isn't right. For repeated linear factors and their powers (i.e., $s, s^2, s^3, s^4$), the decomposition would look like this:
$\frac{2}{s^4(s^2 + 1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s^3} + \frac{D}{s^4} + \frac{Es + F}{s^2 + 1}$.
The $As + B$ numerators would go with irreducible quadratic factors, such as the $s^2 + 1$ factor in the denominator.

It seems rather tedious because of the $\frac{1}{s^4}$, but is that correct (There may be a simpler method but I can't see it).

Thanks!