Laplace inverse of $(s+2)U(s)=0$ and $(s+1)U(s)=0.$

Question

I asked this question here and I was given an answer but with some steps unfolded Solve the following problem, $u'(t)+p(t)u(t)=0,\;\;u(0)=0,$ $p(t)=\begin{cases}2& 0\leq t< 1,\\1 &t\geq 1\end{cases}.$.

My question now is, how do I take the La

Taking the Laplace transform of both sides $$L(u'(t))+L(p(t)u(t))=0,$$ $$sU(s)-u(0)+L(p(t)u(t))=0,$$ For $0\leq t<1$ $$sU(s)+2U(s)=0,$$ $$(s+2)U(s)=0.$$

For $t\geq 1$ $$sU(s)+U(s)=0,$$ $$(s+1)U(s)=0.$$

Can someone show me how to take the Laplace inverse of this to get $u(t)=e^{-2t}$ and $u(t)=e^{-t}.$

Answer 1

Hint: Correct your formulas with step function: $$(s+2)U(s)=\dfrac1s$$ and $$(s+1)U(s)=\dfrac1s$$

Answer 2

First for $0 \le t \le 1$

$$ sU(s)+2U(s)=u(0)\Rightarrow u(t) = u(0) e^{-2t}(\phi(t)-\phi(t-1)) $$

and with $u(1) = u(0)e^{-2}$ we proceed solving for $1 \le t $

$$ sU(s)+U(s) = u(1)\Rightarrow u(t) = u(1)e^{-(t-1)}\phi(t-1) $$

finally

$$ u(t) = u(0) e^{-2t}(\phi(t)-\phi(t-1))+u(0)e^{-2}e^{-(t-1)}\phi(t-1) $$

or

$$ u(t) = u(0)\left(e^{-2t}(\phi(t)-\phi(t-1))+e^{-t-1}\phi(t-1)\right) $$

Here $\phi(t)$ is the Heavside unit step function.