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 + 5 y = sin 2 t using Laplace transform when y=0 and y′=0 when t=0
I've done the transform part and got a function -
y ( S ) = 2 ( s 2 + 4 ) ( s 2 + 2 s + 5 ) = A s 2 + 4 + B s + C s 2 + 2 s + 5
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Answers (2)

nizkem0c
Answered 2022-09-15 Author has 13 answers
You will need
2 ( s 2 + 4 ) ( s 2 + 2 s + 5 ) = A s + B s 2 + 4 + C s + D s 2 + 2 s + 5   .
Multiply out:
( ) 2 = ( A s + B ) ( s 2 + 2 s + 5 ) + ( C s + D ) ( s 2 + 4 )   ;
equate coefficients,
A + C = 0   , 2 A + B + D = 0   , 5 A + 2 B + 4 C = 0   , 5 B + 4 D = 2   ;
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 = ( 2 A i + B ) ( 1 + 4 i )
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
Answered 2022-09-16 Author has 3 answers
Try this:
y ( S ) = 2 ( s 2 + 4 ) ( s 2 + 2 s + 5 ) = A s + B s 2 + 4 + C s + D s 2 + 2 s + 5
Then i got:
A = 4 17 , B = 2 17 , C = 4 17 , D = 6 17 .

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