- #1
jhmz
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im trying to work out how to use partial fractions on a fraction with repeated roots. I am learning about laplace transforms at the moment, i don't remeber the lecturer specifically going through how to solve the transforms that have repeated roots and can't find it anywhere in the lecture material however he has organised the study questions into distinct/repeated/purely imaginary/complex roots.
Anywho I am stuck on the first question.
2y'' + 8y' + 8y = 0
y(0) = 1
y'(0) = 0
which transforms to:
[tex]Y_{(s)}=\frac{s+4}{(s+2)^2}[/tex]
Now the partial fraction I've got so far from looking at websites such as:
http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/PartialFraction/RootsRepeat.html
looks like:
[tex]Y_{(s)}=\frac{s+4}{(s+2)^2}=\frac{a}{s+2}+\frac{b}{(s+2)^2}[/tex]
Now from here it says to multiply both sides by the denominator of the coefficient you want and let s=-2 in the case of the denominator of a, s+2 as doing this cancels everything on the left hand side which looks like:
[tex]a=(s+2)\frac{s+4}{(s+2)^2}[/tex] where s = -2
but then it equals zero
and i don't get what i am doing wrong
Anywho I am stuck on the first question.
2y'' + 8y' + 8y = 0
y(0) = 1
y'(0) = 0
which transforms to:
[tex]Y_{(s)}=\frac{s+4}{(s+2)^2}[/tex]
Now the partial fraction I've got so far from looking at websites such as:
http://www.swarthmore.edu/NatSci/echeeve1/Ref/LPSA/PartialFraction/RootsRepeat.html
looks like:
[tex]Y_{(s)}=\frac{s+4}{(s+2)^2}=\frac{a}{s+2}+\frac{b}{(s+2)^2}[/tex]
Now from here it says to multiply both sides by the denominator of the coefficient you want and let s=-2 in the case of the denominator of a, s+2 as doing this cancels everything on the left hand side which looks like:
[tex]a=(s+2)\frac{s+4}{(s+2)^2}[/tex] where s = -2
but then it equals zero
and i don't get what i am doing wrong
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