# Solve the differential equation y′′+2y′+5y=sin 2t using Laplace transform when y=0 and y′=0 when t=0

Jonah Jacobson 2022-09-14 Answered
Solve the differential equation ${y}^{″}+2{y}^{\prime }+5y=\mathrm{sin}2t$ using Laplace transform when y=0 and y′=0 when t=0
I've done the transform part and got a function -
$y\left(S\right)=\frac{2}{\left({s}^{2}+4\right)\left({s}^{2}+2s+5\right)}=\frac{A}{{s}^{2}+4}+\frac{Bs+C}{{s}^{2}+2s+5}$
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nizkem0c
You will need

Multiply out:

equate coefficients,

solve, which I leave up to you. When the denominators are quadratic there are typically no really good short cuts. You could try in (∗) substituting s=2i to give
$2=\left(2Ai+B\right)\left(1+4i\right)$
then dividing by 1+4i and equating real and imaginary parts, but this may well be more trouble than it's worth.

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Darius Nash
Try this:
$y\left(S\right)=\frac{2}{\left({s}^{2}+4\right)\left({s}^{2}+2s+5\right)}=\frac{As+B}{{s}^{2}+4}+\frac{Cs+D}{{s}^{2}+2s+5}$
Then i got:
$A=\frac{-4}{17},\phantom{\rule{thinmathspace}{0ex}}B=\frac{2}{17},\phantom{\rule{thinmathspace}{0ex}}C=\frac{4}{17},\phantom{\rule{thinmathspace}{0ex}}D=\frac{6}{17}.$

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