Solving y''-6y+9y=2 with Laplace Transforms

f00lishroy
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Homework Statement


y''-6y+9y=2 y(0)=y'(0)=0

*Note* Professor will NOT allow use of partial fractions, so please don't use it.

Homework Equations


Laplace transform table
Y=[y'(0)+sy(0)+ay(0)+R]/[(s^2)+as+b]

The Attempt at a Solution


Y=L(2)/(s-3)^2
L(2)=2/s
Y=(2/s)[1/(s-3)^2]
Y=2*1/[s^3-6s^2+9s]
From here I cannot figure out how to continue without using partial fractions since I can't get the roots, from which I would be able to invert and use the Laplace Transform table.
 
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You could also invert Y(s) using the Bromwich integral
y(t)=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty} Y(s)e^{st}\,ds.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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