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: `Lap^{:-1:} {(n!)/((s-a)^[n+1])}=e^[at]t^n`, with a = 1 and n = 2.

So

`Lap^{:-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

`Lap^{:-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`

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