IVP problem using Laplace transform and partial fractions

• member 731016
member 731016
Homework Statement
Relevant Equations
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!

ChiralSuperfields said:
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.
ChiralSuperfields said:
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!

member 731016

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