Question #3122
Given y(0) = 2 and y'(0) = 3.
Solve for the Laplace transform Y(s) using the differential equation below.
y'' - 2y' + y = 0
Solve for the Laplace transform Y(s) using the differential equation below.
y'' - 2y' + y = 0
Expert's answer
Let's solve the equation:
(s2-2s+1) Y(s) = (s-2) y(0) + y'(0)
(s-1)2 Y(s) =2(s-2) + 3 = 2s - 1
Y(s) = (2s-1) / (s-1)2 = (2(s-1)+1) / (s-1)2 = 2 / (s-1) + 1 / (s-1)2
We can find y(x) using inverse Laplace transform for each part and then use
the property of linearity:
L-1[1/(s-1)] = exp(x)
L-1[1/(s-1)2] = x exp(x)
Thus,
y(x) = L-1[Y(s)] = 2 exp(x) + x exp(x) = (x+2) exp(x)
(s2-2s+1) Y(s) = (s-2) y(0) + y'(0)
(s-1)2 Y(s) =2(s-2) + 3 = 2s - 1
Y(s) = (2s-1) / (s-1)2 = (2(s-1)+1) / (s-1)2 = 2 / (s-1) + 1 / (s-1)2
We can find y(x) using inverse Laplace transform for each part and then use
the property of linearity:
L-1[1/(s-1)] = exp(x)
L-1[1/(s-1)2] = x exp(x)
Thus,
y(x) = L-1[Y(s)] = 2 exp(x) + x exp(x) = (x+2) exp(x)
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