Would this be the correct approach for solving y'' - 4y = 4H(t-2)

Question

If not, what would be a better way?

Please see attached imageenter image description here

Answer 1

You get from the general setup of such tasks $y(t)=0$ for $t<0$, which includes $y'(0)=0$. In consequence also $y(t)=0$, $y'(t)=0$ for $t<2$. For $t>2$ the equation reads $$ y''-4y=4,~~y(2)=0,~~y'(2)=0 $$ with a solution $$ y(t)+1=A\sinh(2(t-2))+B\cosh(2(t-2)), $$ so that $B=1$ and $A=0$. In total $$ y(t)=H(t-2)(\cosh(2(t-2))-1)=2H(t-2)\sinh^2(t-2). $$

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Please explain, in typed text, not another image, where the contemplation of the Laplace transform of $tf(t)$ for the treatment of the right side comes from.