Solve the initial value problem: ODE with discontinuous coefficients?

Question

Solve the initial value problem $$ y'-y = \left\lbrace \begin{aligned} &1 & & \text{when}\quad 0<t<1 \\ &0 & & \text{when}\quad t>1 \end{aligned} \right. ,\qquad y(0)=0 \, . $$

So I understand that $p(t)=-1$ and $g(t)$ is $1$ and $0$ (depending on the $t$ value), and that I'm supposed to solve it as two separate DEs. I got for the first case that $y=-1$ and for the second case that $y=0$.

• Is this correct?
• Is this the solution? Or do I have to do something with the initial value $y(0)=0$?

Any help is appreciated !!

Answer 1

For the first case $(t<1)$, you should have: $$y'-y=1$$ $$\int \dfrac {dy}{1+y}=t+K$$ $$\ln(y+1)=t+K$$ $$(y+1)=e^{t+K}=Ke^t$$ $$y=Ke^t-1$$ Since $y(0)=0 \implies K=1$ $$\boxed{y=e^t-1}$$

Answer 2

Define \begin{equation} f(t) \;=\; \begin{cases} 1 & 0 < t < 1\\ 0 & t > 1 \end{cases}\, . \end{equation} Then, we have $$ \frac{dy}{dt} - y \;=\; f(t)\, . $$ Multiply both sides of this by $e^{-t}$: \begin{equation} e^{-t}\left(\frac{dy}{dt} - y\right)\;=\; \frac{d}{dt}\left(e^{-t} y\right) \;=\; e^{-t} f(t) \end{equation} Now integrate both sides of this from $0$ to $t$ (using $s$ as a dummy variable of integration): \begin{equation} \int_0^t \frac{d}{ds}\Bigl(e^{-s} y(s)\Bigr) \, ds \;=\; e^{-s} y(s) \Biggr\vert_{s = 0}^{s = t} \;=\; e^{-t} y(t) \;-\; y(0) \;=\; \int_0^t e^{-s} f(s)\, ds \end{equation} Using $y(0) = 0$, this can be rearranged into \begin{align} y(t) &\;=\; \int_0^t e^{t-s} f(s)\, ds\\ &\;=\; \int_0^{\min(t,1)} e^{t-s}\, ds \qquad\qquad \text{(Using the definition of $f$)}\\ &\;=\; e^t \;-\; e^{t - \min(t,1)}\\ &\;=\; \begin{cases} e^t - 1 & t < 1\\ e^t\left(1 - 1/e\right) & t > 1 \end{cases} \end{align}

Note that this solution is continuous at $t = 1$, as it should be.

Answer 3

I would do this as two separate problems. First solve y'- y= 1, y(0)= 0. That is a non-homogeneous differential equation. The solution to the associated homogeneous equation, y'- y= 0, is $y(t)= Ce^t$ for C an arbitrary constaant. If we try y= A, a constant for the entire equation, we get -A= 1 so A= -1. The general solution to the first equation if $y(t)= Ce^t- 1$. With initial condition $y(0)= C- 1= 0$, C= 1. The solution to this initial value problem is $y(t)= e^t- 1$.

Now look at the second equation, y'- y= 0. The general solution to that is $y(t)= Be^t$. To determine B we need an initial value, a value at t= 1. Since we must have the derivatives existing, y must be continuous so $y(1)= Be= e- 1$ from the previous solution. We must have $B= (e- 1)/e= 1- e^{-1}$. The solution to this initial value problem is $y(t)= (1- e^{-1})e^t= e^{t}- e^{t-1}$.

The solution to the original problem is

$y(t)= e^t- 1$ for $t\le 1$

$y(t)= e^{t}- e^{t- 1}$ for $t> 1$.