Solving separable ODEsdxdt=f(x)g(t) with x(0)=x0

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Cason Singleton · Accepted Answer

Step 1The point is that if x≡x⋅ then x′=0 from direct differentiation and also f(x)g(t)=0, so a constant function equal to a zero of f is a solution to the ODE. For it to be a solution to the IVP then you need x0=x⋅.Step 2This needs a hypothesis that allows us to ensure that the solution is unique, so that one cannot have a constant solution and a non-constant solution to the same IVP (which would happen if any non-constant solution ever hit a zero of f, since then you could consider the IVP initialized when the non-constant solution hit the zero of f and get two solutions). Here that hypothesis is that f&#039; is continuous. (As an aside, this particular hypothesis is not necessary for uniqueness, but merely f and g being continuous is not sufficient for uniqueness. So some additional hypothesis is required, this is just what they chose to use to write this problem.)Step 3Basically this is asking you to find a separable equation with non-unique solution to an IVP with x(0)=x0 and f(x0)=0. Hopefully you have already seen some examples of this otherwise it will be a bit challenging to make one up from scratch.

Solving separable ODEs \(\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={f{{\left({x}\right)}}}{g{{\left({t}\right)}}}\) with \(\displaystyle{x}{\left({0}\right)}={x}_{{0}}\)

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