Laplace Transform Solution to Second Order ODE IVP

1. The problem statement, all variables and given/known data

y''+6y=f(t), y(0)=0, y'(0)=-2



f(t)= t for 0≤t<1 and 0 for t≥1



2. Relevant equations





3. The attempt at a solution



L{y''}+6L{y}=L{t}-L{tμ(t-1)} where μ(t-1) is Unit Step



Y(s)=L{y}

sY(s)-y(0)=L{y'} and y(0)=0

s²Y(s)-sy(0)-y'(0)+6Y(s) where y(0)=0 and y'(0)=-2



LHS:

s²Y(s)+2+6Y(s)

Y(s)(s²+6)-2



RHS:

L{t}-L{tμ(t-1)}

L{t}=1/s²

L{tμ(t-1)}=L{(t-1)μ(t-1)+L{μ(t-1)}=e^-s/s²+e^-s/s

L{t}-L{tμ(t-1)}=1/s²-e^-s/s²-e^-s/s



Y(s)= -1/(s²+6) -e^-s/(s²(s²+6))-e^-s/s(s²+6)



taking the inverse laplace of the first term

-L-1{1/(s²+6)}=(-1/6)L-1{1/s}+(1/6)L-1{1/(s+6)}=(-1/6)+1/6e^-6t)



how could I take the laplace of the other 2 terms?

I was thinking L-1{e^-s/(s²(s²+6))}=y(t-1)μ(t-1)?



or I could just break it up into (1/6)L-1{1/s²}-(1/6)L-1{1/(s²+6)} which I know the inverse laplace transforms for, however when I do so I get

y(t)= (1/6)t-(1/6)+(1/6)e^-6t

which seems a bit bizarre to me, I would have assumed that the answer would have a unit step function



Thank you for your time

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