# solving an initial value ODE with a twist

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

y' = -f(x)y; y(0) = 1;

where f(x) = (1 if x<=2, 3/2 if x>2):

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

I just dont understand what part (c) is asking, if anyone could help me with the concepts I would really appreciate it.

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## 1 Answer

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.

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Thank you! To be perfectly honest, I had integrated with respect to t instead of x and had absolutely no idea where to go from there. My bad :) – user6895 Feb 11 '11 at 5:01