Solving Laplace Transform: y"+2y'+2y=t, y(0)=y'(0)=1

[sami23]

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)} = s²Y(s)-sy(0)-y'(0)
using the laplace transform table: tⁿ = n!/(sⁿ⁺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}
s²Y(s) - sy(0) - y'(0) + 2[sY(s)-y(0)] + 2Y(s) = 1/s²

after plugging in the initial conditions I get:
s²Y(s) - s - 3 + 2sY(s) + 2Y(s) = 1/s²

isolate Y(s):
Y(s)[s²+2s+2] = 1/s² + s + 3
Y(s) = (1/s² + s + 3) / (s²+2s+2)
I completed the square in the denominator to get: (m+1)²+1

Y(s) = s / [(s²+1)(s+3)²-8]}

Take the Laplace inverse
L^-1{ s / [(s²+1)(s+3)²-8]}

I add and subtract 3 in the numerator to get:
L^-1{ (s+3)-3 / [(s²+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) / [(s²+1)(s+3)²-8]} - 3L^-1{ 1 / [(s²+1)(s+3)²-8}

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

 

[mighty2000]

To apply partial fraction decomposition, you need to factor the denominator of the expression you have obtained for Y(s). In this case, the denominator is (s^2+1)(s+3)^2-8. This can be factored further as (s^2+1)((s+3)^2-4), using the difference of squares formula. Then, you can write this as (s^2+1)(s^2+6s+9-4), which simplifies to (s^2+1)(s^2+6s+5).

Now, you can set up the partial fraction decomposition by writing Y(s) as a sum of simpler fractions with unknown coefficients A, B, C, and D:
Y(s) = A/(s^2+1) + B/(s+3) + C/(s+3)^2 + D/(s^2+6s+5)

To solve for the unknown coefficients, you can multiply both sides by the denominator of Y(s) and then equate the coefficients of the terms on both sides. This should give you a system of equations that you can solve to find the values of A, B, C, and D.

Once you have the values of the coefficients, you can substitute them back into the expression for Y(s) and then take the inverse Laplace transform to obtain y(t).