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Solving "Impossible" Differential Equations with Mathematica

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Say we wanted to find the value of the initial value problem's solution at a particular value of x. How about at x = 2 for now? Then all we need to do is evaluate sol1 at x = 2 as follows:

sol1/.x->2

Go ahead and try it. Come back when you have.

Let's see what you should have gotten...


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ODE Laboratories: A Sabbatical Project by Christopher A. Barker

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