The questions is to solve the differential equation:

$$\sqrt{x}\frac{d^2y}{dx^2}+\lambda y=0$$

I never took differential equations so this has me stumped. I have only solved differential equations using the Frobenius Method which I do not think will work here. Can anyone show me how to do this?

  • $\begingroup$ The WolframAlpha solution is hideous and contains the Bessel function of the first kind. I doubt it's one you can solve by hand unless there is some really clever simplification that WA missed (that has happened). $\endgroup$ – Arthur Aug 4 '16 at 20:46

$$ \sqrt{x}y''+\lambda y = 0 $$ let $x = t^a$ then we have $$ \dfrac{d}{dx} = \dfrac{dt}{dx}\dfrac{d}{dt}=\frac{1}{at^{a-1}}\dfrac{d}{dt}\\ \dfrac{d^2}{dx^2} = \frac{1}{at^{a-1}}\dfrac{d}{dt}\frac{1}{at^{a-1}}\dfrac{d}{dt}=\frac{1-a}{a^2t^{a-1}t^a}\dfrac{d}{dt} +\frac{1}{a^2t^{2(a-1)}}\dfrac{d^2}{dt^2} $$ so we have $$ t^{a/2}\frac{1-a}{a^2t^{a-1}t^a}\dfrac{dy}{dt} +t^{a/2}\frac{1}{a^2t^{2(a-1)}}\dfrac{d^2y}{dt^2} = \lambda y $$ then we have $$ t^2\frac{1}{a^2t^{2a-a/2}}\ddot{y} + \frac{1-a}{a^2t^{2a-a/2}}t\dot{y} =\lambda y $$ so re-arranging we have $$ t^2\ddot{y} +(1-a)t\dot{y} - a^2t^{2a-a/2}\lambda y = 0 $$ if we have $$ 2a-a/2 = \frac{3}{2}a = 2\implies a = \frac{4}{3} $$ we have $$ t^2\ddot{y}-\frac{1}{3}t\dot{y}-\frac{16}{9}t^2\lambda y = 0 $$ which is Bessel like.


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