How to solve this ODE with the Laplace transform?

$$y'''-y''-y'+y= -10 \cos (2t-1)+5 \sin(2t-1)$$ Substitute $u=2t-1$ the equation becomes: $$8y'''-4y''-2y'+y= -10 \cos (u)+5 \sin(u)$$ Where you have the initial conditions: $$u(0)=1,u'(0)=2,u''(0)=1$$ $$u \in [0 , + \infty [$$

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$$\Leftrightarrow Z(s)(s^3-s^2-s+1)= \frac{-10s+10}{(s^2+4)} -2+s^2+s$$ Divide both sides by $s-1$: $$ Z(s)(s^2-1)= \frac{-10}{(s^2+4)} +s+2$$ $$ Z(s)= \frac{2}{(s^2+4)} - \frac{2}{(s^2-1)}+\dfrac 1{s-1}+\dfrac 1{s^2-1}$$ $$ Z(s)= \frac{2}{(s^2+4)} - \frac{1}{(s^2-1)}+\dfrac 1{s-1}$$ $$ Z(s)= \frac{2}{(s^2+4)} + \frac{1}{2(s+1)}+\dfrac 1{2(s-1)}$$ $$z(t)=\sin(2t)+\dfrac 12e^{-t}+\dfrac 12 e^t$$ $$\implies z(t)=\sin(2t)+\cosh (t)$$

And you have that $$z(0)=1,z'(0)=2,z''(0)=1$$ As expected.

Hint: The shift theorem for Laplace transforms states that $\mathcal L[z(t)]=e^{-as}\mathcal L[y(t)]$ where $z(t)=y(t-a)$.

This transforms the initial conditions into $z(0),z'(0),z''(0)$ from which you can substitute into the expressions $\mathcal L[z],\mathcal L[z'],\mathcal L[z'']$ and $\mathcal L[z''']$.

Then the right-hand side of your transformed equation will be the quotient of polynomials involving $s$ which calls for partial fraction decomposition to be used.