​solve using Laplace transformation in ODE

Mathematics
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solve using Laplace transformation 

x''−2x' + x = 0, x(0) = 1, x'(0) = 0 

May 19th, 2015

According to given question

x''−2x' + x = 0

laplace (x) = X 
laplace (x') = pX -x(0) 
laplace (x'') = (p^2)X - px(0) - x'(0) 
so replace them in eq: 
(p^2)X -px(0) -x'(0) -2(pX +x(0)) +X =0   where P=2i/3
<=> (P^2)X-pX(0)-X(0)-2p(X)-2x(0)+X=0 
<=> X(p^2 -p -2) =2 
<=> X = 2/(p^2 -p -2) 
<=> X = 2/(p^2 -p +1/4 -9/4) 
<=> X = 2/((p-1/2)^2 -9/4) 
<=> X = (-4i/3)(3i/2)/((p-1/2)^2 +(3i/2)^2) 
<=> x = (-4i/3)(e^(t/2))(sin(3it/2))

Hope This Will help you 

May 19th, 2015

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