Solve the following problem, $u'(t)+p(t)u(t)=0,\;\;u(0)=0,$ $p(t)=\begin{cases}2& 0\leq t< 1,\\1 &t\geq 1\end{cases}.$

Question

Using Laplace transform, solve the following problem.

$$u'(t)+p(t)u(t)=0,\;\;u(0)=0,$$ $$p(t)=\begin{cases}2& 0\leq t< 1,\\1 &t\geq 1\end{cases}.$$

Here's what I've done:

Taking the Laplace transform of both sides $$L(u'(t))+L(p(t)u(t))=0,$$ $$sU(s)-u(0)+L(p(t)u(t))=0,$$ $$sU(s)+L(p(t)u(t))=0.$$

I'm stuck at this point. Please, how do I proceed?

Answer 1

For $0< t< 1$, we have $u'(t)+2u(t)=0$, $u(0)=0$ for which there is trivial solution $u(t)=0$.

For $t\ge 1$, we have $u'(t)+u(t)=0$ for which the solution is $u(t)=Ae^{-t}$ where $A$ is arbitrary constant.

Added: As per David's comments, using the continuity of $u(t)$ at $t=1$ we get $A=0$. Hence the solution will be $u(t)=0$ for $t\ge 0$

Answer 2

$$\int_0\dfrac{d u(t)}{u(t)}=\int_0-p(t)\ dt=\begin{cases}2t& 0\leq t\leq 1,\\t &t\geq 1\end{cases}.$$ then $$u(t)=\begin{cases}0& t\leq 0,\\e^{2t}& 0\leq t\leq 1,\\e^t &t\geq 1\end{cases}.$$