solve with Laplace $y''-2y'+2y=0$ detect my mistakes

Question

$y''-2y'+2y=0$ and $y(0)=2,y'(0)=0$. My try: Ly will be the Laplace function of y.

$\begin{array}{c} y''-2y'+2y=0\\ y(0)=2,y'(0)=0\\ Ly''=sLy'\\ Ly'=sLy-2\\ Ly''=s(sLy-2)=s^{2}Ly-2s\\ s^{2}Ly-2s-2(sLy-2)+2Ly=0\\ s^{2}Ly-2(sLy)+2Ly=2s-4\\ Ly(s^{2}-2s+2)=2s-4\\ Ly=\frac{2s-4}{(s^{2}-2s+2)}\\ Ly=\frac{2s-4}{(s-1)^{2}+1^{2}}\\ Ly=2\frac{1}{(s-1)^{2}+1^{2}}s-4\frac{1}{(s-1)^{2}+1^{2}}\\ y=-2(e^{t}\sin t)'-4(e^{t}sint)=-4 e^t sin(t) + 2 e^t (cos(t) + sin(t)) \end{array}$

which of course is incorrect.

Answer 1

$$y=-2(e^{t}\sin t)'$$ $$y=-2(e^t\sin t +e^t \cos t)$$ Take the Laplace transform : $$Ly=-2\left(\frac{1}{(s-1)^{2}+1^{2}}+\frac{s-1}{(s-1)^{2}+1^{2}}\right)$$ $$Ly=-\frac{2s}{(s-1)^{2}+1^{2}}$$ There is a sign mistake. You had this : $$Ly= \color {red}{2\frac{1}{(s-1)^{2}+1^{2}}s}-4\frac{1}{(s-1)^{2}+1^{2}}$$ So that you should have: $$y=\color {red}{+2}(e^{t}\sin t)'-4(e^{t}\sin t)$$ This agrees with @Lutz Lehmann 's answer. $$y=2e^t\cos t-2e^t\sin t$$

Answer 2

You need to apply the shift to all occurrences of $s$ in $$ L\{y\}=\frac{2s-4}{(s-1)^{2}+1^{2}} =\frac{2(s-1)-2}{(s-1)^{2}+1^{2}} =2L\{\cos t\}(s-1)-2L\{\sin t\}(s-1) $$ so that $$ y=2e^t\cos t-2e^t\sin t $$

Your approach is also correct, if done correctly to the end, \begin{align} L\{y\}(s)&=2sL\{\sin t\}(s-1)-4L\{\sin t\}(s-1)\\ &=2L\{(e^t\sin t)'\}-4L\{e^t\sin t\} \\[1em]\hline y(t)&=2(e^t\sin t)'-4e^t\sin t\\ &=2e^t\cos t-2e^t\sin t \end{align}

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Testing without Laplace $$ (e^{-t}y)''+e^{-t}y=0, ~~ (e^{-t}y(t))'|_{t=0}=y'(0)-y(0)=-2 \\ \implies e^{-t}y=2\cos t - 2\sin t $$ You had the signs right, my memory of the Laplace relations is incomplete. Corrected thanks @Aryadeva