How to solve this ODE with the Laplace transform? I want to solve this ODE
$$ y'''-y''-y'+y= -10 \cos (2t-1)+5 \sin(2t-1) $$
with
$y( \frac12)= 1 $ , $ y'( \frac12 )=2 $, $y''( \frac12 )=1 $
$ t \in [ \frac12 , + \infty [ $
using the Laplace-Transformation.
usually I use the differential-approach:
$$ ( L( f^{(k)}))(s)=s^k(Lf)(s)- \sum_{j=0}^{k-1} s^j f^{(k-1-j)} (0) $$
so setting $Y(s)= L \{ y(t) \} $
I get $s^3 Y(s)-s^0y''(0)-s^1y'(0)-s^2y(0)-s^2Y(s)+s^0 y'(0)+s^1y(0)-sy(s)+s^0y(0)+Y(s)... $
I can't continue here, because of the initital values.
How can I solve this ODE? Do I need to shift it somehow?
Or do $ y''(0) etc..$ vanish, because $ t \in [ \frac12, \infty[ $?
Appreciate any help !

EDIT
Thank you for the help so far !
after applying the shift theorem I solve
$$z'''-z''-z'+z=-10 \cos (2t) +5 \sin(2t) $$
with $ z(0)=1,z'(0)=2,z''(0)=1 $
Using the differential approach it comes to
$$s^3 Z(s)-1-2s-s^2-s^2Z(s)+2+s-sZ(s)+1+Z(s)= \frac{-10s+10}{(s^2+4} $$
$$ \Leftrightarrow Z(s)(s^3-s^2-s+1)= \frac{-10s+10}{(s^2+4)} -2+s^2+s $$
$$ \Leftrightarrow Z(s)= \frac{(-10s+10)-(s^2+4)(2+s^2+s)}{(s^3-s^2-s+1)(s^2+4)} $$ $$ \Leftrightarrow Z(s)= - \frac32 \frac{1}{s-1}- 2 \frac{1}{(s-1)^2} + \frac12 \frac{1}{s+1} + \frac{2}{s^2+4} $$
looking up the inverse it comes to
$z(t)= - \frac32 e^t - 2t e^t+ \frac12 e^{-t}+ \sin(2t) $
but $ z'(t)= \cos(t)-2te^t- \frac{7 e^t}{2}- \frac{e^{-t}}{2} $
and so $ z'(0)=-2 \neq 2 $
where is my Mistake?
 A: $$y'''-y''-y'+y= -10 \cos (2t-1)+5 \sin(2t-1)$$
Substitute $u=2t-1$ the equation becomes:
$$8y'''-4y''-2y'+y= -10 \cos (u)+5 \sin(u)$$
Where you have the initial conditions:
$$u(0)=1,u'(0)=2,u''(0)=1$$
$$u \in [0 , + \infty [$$

$$\Leftrightarrow Z(s)(s^3-s^2-s+1)= \frac{-10s+10}{(s^2+4)} -2+s^2+s$$
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.
A: 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.
