Phantom L.

Solve the following initial-value problem using Laplace transforms.

y''- y' - 2y = 5sin(x) ; y(0) = 1 , y'(0) = -1

By: Tutor
4.4 (32)

Mathematics/Statistics Tutor

Phantom L.

Don't I have to derive Y(x) twice before plugging it in? And this method of solving seems to be the method of undetermined coefficient, is this how I should solve the problem even if the problem states solve using Laplace transform?
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04/28/18 Bobosharif S.

Laplace Transform (LP)
You have to use properties of LP:
L{y'}=sL{y(s)}-y(0)
L{y''}=s2L{y(s)}-sy(0)-y'(0)
L{5sin(x)}=5/(1+s2)
Take LP from both sides of
y''- y' - 2y = 5sin(x)
L{y''}-L{y'}-2L{y}=5L{sin(x)}
s2L{y}-sy(0)-y'(0)-sL{y}+sy(0)-2L{y}=5/(1+s2)
L{y}(s2-s-2)=5/(1+s2)-1
L{y}=(4-s2)/(1+s2) (1/(s2-s-2))
L{y}=-(2+s)/(1+s+s2+s3)
So far what we did is LP from the solution of  DE. To find solution intself, we haveto take inverse LP:
y(x)=L-1{-(2+s)/(1+s+s2+s3)}
y(x)=-e-x/2+(1/2)(cos(x)-3sin(x))

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04/29/18

Phantom L.

Thanks Mr. Sharif
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04/29/18

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