How to solve this ODE with the Laplace transform?

Question

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 !

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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?

Answer 1

$$y'''-y''-y'+y= -10 \cos (2t-1)+5 \sin(2t-1)$$ 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.

Answer 2

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.