- #1

- 76

- 1

## Homework Statement

Use the Laplace Transform to solve: y"+2y'+2y=t y(0)=y'(0)=1

## Homework Equations

L{y(t)} = Y(s)

L{y'(t)} = sY(s)-y(s)

L{y"(t)} = s

^{2}Y(s)-sy(0)-y'(0)

using the laplace transform table: t

^{n}= n!/(s

^{n+1}) where n=1

## The Attempt at a Solution

Take laplace on both sides:

L{y"(t)} + 2L{y'(t)} + 2L{y(t)} = L{t}

s

^{2}Y(s) - sy(0) - y'(0) + 2[sY(s)-y(0)] + 2Y(s) = 1/s

^{2}

after plugging in the initial conditions I get:

s

^{2}Y(s) - s - 3 + 2sY(s) + 2Y(s) = 1/s

^{2}

isolate Y(s):

Y(s)[s

^{2}+2s+2] = 1/s

^{2}+ s + 3

Y(s) = (1/s

^{2}+ s + 3) / (s

^{2}+2s+2)

I completed the square in the denominator to get: (m+1)

^{2}+1

Y(s) = s / [(s

^{2}+1)(s+3)

^{2}-8]}

Take the Laplace inverse

L

^{-1}{ s / [(s

^{2}+1)(s+3)

^{2}-8]}

I add and subtract 3 in the numerator to get:

L

^{-1}{ (s+3)-3 / [(s

^{2}+1)(s+3)

^{2}-8]}

Use linearity property of inverse transform to get from L

^{-1}{Y(s)} to y(t):

L

^{-1}{ (s+3) / [(s

^{2}+1)(s+3)

^{2}-8]} - 3L

^{-1}{ 1 / [(s

^{2}+1)(s+3)

^{2}-8}

How do I apply partial fraction decomposition to get y(t)?

Last edited: