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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