I'm looking for a substitution that would convert:
$$y''+(ax-b^2)y=0$$
To the more familiar format (Airy DV):
$$y''(u)+k^2uy(u)=0$$
Would something like $k^2u=ax-b^2$ do that trick?
Airy style differential equation
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$\begingroup$
ordinary-differential-equations
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0You need to be more specific. What does (u) mean? As a function of u or multiplied by? Priming derivatives risk becoming ambiguous whenever changes of variable occur. I would recommend to rewrite specifying exactly which partial derivatives you take. – 2017-02-20
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1Rather, try $$u = \frac{ax-b^2}{(ak)^{2/3}}$$ – 2017-02-20