# Answer to Question #3122 in Differential Equations for John

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:

(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,

(s

^{2}-2s+1) Y(s) = (s-2) y(0) + y'(0)(s-1)

^{2}Y(s) =2(s-2) + 3 = 2s - 1Y(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)Need a fast expert's response?

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