Taking Laplace transform of both sides:

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

Applying the initial condition and simplifying gives:

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

`(s^2-2s+1)Y` `=(1)/(s-1)-2s+1 `

`(s-1)^2Y=(1)/(s-1)-2s+1 `

Solving for Y:

`Y=(1)/((s-1)^3)+(-2s+1)/((s-1)^2)` `=1/2(2)/((s-1)^3)+(-2s+1)/((s-1)^2) `

For the first term, we use: `mathcal{L^[-1]} {(n!)/((s-a)^[n+1])}=e^[at]t^n`, with a = 1 and n = 2.

So

`mathcal{L^[-1]}{1/2 (2)/((s-1)^3)}` `=1/2 e^t t^2 `

For the second term, we express in partial fractions:

`(-2s+1)/((s-1)^2)` `=(A)/(s-1)+(B)/((s-1)^2) `

`-2s+1=A(s-1)+B `

Comparing coefficients:

`-2s=As ` gives `A = -2`.

`1=-A+B ` gives `B = -1`.

So `(-2s+1)/((s-1)^2)` `=-(2)/(s-1)-(1)/((s-1)^2) `

And

`mathcal{L}^[-1] { - (2)/(s-1) - (1)/((s-1)^2)}` `=2e^t - te^t`

Putting our inverse Laplace transform expressions together, the solution for y is:

`y(t)=1/2 t^2 e^t - 2e^t - te^t`