How to solve this equation y'''-8y"+4y'+48= 4x^2+1 using Laplace Transformation.

CMIIh 2020-11-23 Answered
How to solve this equation y8y"+4y+48=4x2+1 using Laplace Transformation.
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Expert Answer

Macsen Nixon
Answered 2020-11-24 Author has 117 answers
Step 1
Converting the equation in Laplace transform:
s3Y8s2Y+4sY+48Y=4d2ds21s+1s
s3Y8s2Y+4sY+48Y=8s3+1s
s3Y8s2Y+4sY+48Y=(8+s2)(s3)
(s38s2+4s+48)Y=(8+s2)(s3)
Y=(8+s2)s3(s38s2+4s+48)
Step 2
Converting into partial fraction and solving:
Y=(8+s2)s3(s38s2+4s+48)
=(8+s2)s3(s4)(s+2)(s6)
=1(6s3)1(72s2)132(s4)+43864s132(x+2)+11864(s6)
=112d2ds21s+172dds1s132(s4)+43864s132(s+2)+11864(s6)
Step 3
Thus finding inverse Laplace transform find y(t):
y(t)=(x2)12+x72e4x32+43864e2x32+11e6x864
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