Answer to Question #3122 in Differential Equations for John

Question

Given y(0) = 2 and y'(0) = 3.
Solve for the Laplace transform Y(s) using the differential equation below.

y'' - 2y' + y = 0

Answer 1

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)