Solve {(y''-3y'+2y=g),(y(0)=y'(0)=0):} whereas g(t)={(0,t>pi),(sin t,0<=t<=pi):}

Beckett Henry 2022-09-11 Answered
Solve
{ y 3 y + 2 y = g y ( 0 ) = y ( 0 ) = 0
whereas
g ( t ) = { 0 , t > π sin t , 0 t π .
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Answers (1)

Mateo Tate
Answered 2022-09-12 Author has 18 answers
Should be:
y 3 y + 2 y = g s 2 L y 3 s L y + 2 L y = 0 π sin t e t s d t s 2 L y 3 s L y + 2 L y = 1 + e π s 1 + s 2 L y ( s 2 3 s + 2 ) = 1 + e π s 1 + s 2 L y = 1 + e π s ( 1 + s 2 ) ( s 1 ) ( s 2 ) = ( 1 + e π s ) ( 3 10 s 1 + s 2 + 1 10 1 1 + s 2 + 1 5 1 s 1 1 2 1 s 2 )
Let L z = 3 10 s 1 + s 2 + 1 10 1 1 + s 2 + 1 5 1 s 1 1 2 1 s 2
Then z = ( 3 10 cos t + 1 10 sin t + 1 5 e t 1 2 e 2 t ) u ( t )
So we have L y = ( 1 + e π s ) L z = L z + e π s L z
Therefore y ( t ) = z ( t ) + z ( t π )
Substitute to find:
y ( t ) = { 0 if  t < 0 1 10 [ 3 cos ( t ) + sin ( t ) + 2 e t 5 e 2 t ] if  0 t π 1 10 [ 2 ( e t + e t π ) 5 ( e 2 t + e 2 ( t π ) ) ] if  t > π
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