Why is the substitution x=e^z valid for solving a cauchy euler equation? Doesn't it imply that the domain of the solution y(x) is restricted to x>0? What about solutions with domain x<0
Justification for the substitution used for Cauchy Euler equation
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ordinary-differential-equations
1 Answers
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For any non-zero constant $c$, the substitution $ce^z$ would work just as well, because after substitution, you can divide the $c$ out. So by taking $c<0$, you recover your negative solutions. (But in the usual method, we just let the constants take care of this at the end.)
Note that when you divide the original equation by $x^2$, the functions usually called $p(x)$ and $q(x)$ have a discontinuity at $x=0$. So we have to pick one side of $0$ or the other before we even start.