# Initial Value Problem. Laplace Transform

## Homework Statement

The question is solve using laplace transforms: y" - 4y = 1-2t where y(0) =0 and y'(0) = 0

Ans is: y = t/2 - 1/4 + 1/4*e^(-2t)

## Homework Equations

http://www.stanford.edu/~boyd/ee102/laplace-table.pdf" [Broken]
partial fractions

## The Attempt at a Solution

(s^2)Y - sy(0) - y'(0) - 4Y = L{1-2t}
(s^2)Y - 4Y = L{1-2t}
Y(s-4) = (1/s) - (2/s^2)
Y = [ (1/(s(s-4)) ] - [ 2/(s^2(s-4)) ]
L^-1 { [ s/(s^2(s-4)) ] - [ 2/(s^2(s-4)) ] = { (s - 2) / (s^2(s-4)) }

partial fractions to get
A/s + B/(s^2) + C/(s-4) = -1/(8s) + 1/(2s^2) + 1/8((s-4))

inverse laplace transform to get
y = -1/8 + t/2 + 1/8e^(4t)

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## Homework Statement

The question is solve using laplace transforms: y" - 4y = 1-2t where y(0) =0 and y'(0) = 0

Ans is: y = t/2 - 1/4 + 1/4*e^(-2t)

## Homework Equations

http://www.stanford.edu/~boyd/ee102/laplace-table.pdf" [Broken]
partial fractions

## The Attempt at a Solution

(s^2)Y - sy(0) - y'(0) - 4Y = L{1-2t}
(s^2)Y - 4Y = L{1-2t}
Y(s-4) = (1/s) - (2/s^2)
The above should be Y(s2 - 4) = 1/s - 2/s2
Y = [ (1/(s(s-4)) ] - [ 2/(s^2(s-4)) ]
L^-1 { [ s/(s^2(s-4)) ] - [ 2/(s^2(s-4)) ] = { (s - 2) / (s^2(s-4)) }

partial fractions to get
A/s + B/(s^2) + C/(s-4) = -1/(8s) + 1/(2s^2) + 1/8((s-4))

inverse laplace transform to get
y = -1/8 + t/2 + 1/8e^(4t)

Last edited by a moderator: