# solving an initial value ODE with a twist

The goal of this problem is to solve the initial value problem

$$y' = -f(x)y;\quad y(0) = 1;$$

where $$f(x)= \begin{cases} 1,&\text{if }x\leq 2\\ \frac{3}{2},& \text {if } x>2. \end{cases}$$ Since $f$ is discontinuous, it is necessary to solve the above ODE separately in each of the intervals where $f$ is continuous.

(a) Determine the intervals where $f$ is continuous.

(b) Solve the equation in each of these intervals. Note that each of the solutions obtained will have a different constant of integration.

(c) Match the solutions at the points where $f$ is discontinuous, in order to make the solution $y$ continuous on $\Bbb R$. Note that in this case it is impossible to make $y'$ continuous at the points where $f$ is discontinuous.

I just don't understand what part (c) is asking, if anyone could help me with the concepts I would really appreciate it.

On $x\le 2$, the solution is $y=\exp(-x)$, on x>2 it is $y=C\exp(-3x/2)$. The reason there is no constant in the first is you are given $y(0)$, and can find the constant is 1. If you take the limit of $f(x)$ as $x \to 2^-$ and as $x \to 2^+$ the second will depend on $C$ and the first will not, so you can choose $C$ to make $f(2)$ consistent across $x=2$.The note says that if you take the derivative as a limit as $x \to 2^-$ you get one answer and if you take the limit as $x \to 2^+$ you get a different one. If you plot it there will be a kink in the curve.