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

Expert's answer

Let's solve the equation:

(s^{2}-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)

(s

(s-1)

Y(s) = (2s-1) / (s-1)

We can find y(x) using inverse Laplace transform for each part and then use

the property of linearity:

L

L

Thus,

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