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I have the differential equation $x^2y''+xy'+m^2x^{2m}y=0$ where $x>0$ and $m\neq 0$. I know a general method to solve this kind of equation with Bessel functions: $$x^2y''+(2p+1)xy'+(a^2x^{2r}+b^2)y=0 \Rightarrow y=x^{-p}\Big( c_1J_{q/r}(\frac{a}{r}x^r) + c_2Y_{q/r}(\frac{a}{r}x^r ) \Big)$$ where $q=\sqrt{p^2-b^2}$ so my equation is solved for $y=c_1J_0(x^m)+c_2Y_0(x^m)$.

Is it possible to get the same solution without the previous formula? I've tried some substitution (for example $t=x^m$) with the goal of rewrite the equation in a more "Bessel-like form" ($x^2y''+xy'+(x^2-p^2)y=0$) but without any kind success.

For the specific equation the approach is possible or it's complicated to the point that is just more pratical to use the formula?

injo
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    Try $y(x)=x^az(t)$, where $t=bx^c$. (See https://math.stackexchange.com/questions/1804280/verify-y-xaz-p-leftbxc-right-is-a-solution-to-y-left-frac1-2ax-ri.) – Gonçalo Aug 16 '24 at 11:51
  • @Gonçalo Thanks, my equation is a special case with $a=p=0$, $b=1$, $c=m$ so with you suggestion the equation becomes $(mx^m)^2z''+m(mx^m)z'+(mx^m)^2z=0$. I'm much closer even if for some reason I have an extra $m$ in the middle term. – injo Aug 16 '24 at 12:32
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    It's a small point but writing the equation in Sturm-Liouville form helps: $$x \frac{d}{dx}\left( x \frac{dy}{dx}\right)+m^2 x^{2m}y=0$$ Then the change of variables $t=x^m$ amounts to the operator substitution $x\dfrac{d}{dx}=mt\dfrac{d}{dt}$. – Semiclassical Aug 16 '24 at 14:32
  • @Semiclassical Really helpful, thanks. – injo Aug 16 '24 at 15:52
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    If you think you've worked out the answer, I suggest writing it out below and submitting it yourself so that this question can be concluded. – Semiclassical Aug 16 '24 at 15:55
  • @Semiclassical Actually using you suggestion I get $mt\frac{d}{dt}\Big(mt\frac{dy}{dt}\Big)+m^2t^2y=0 \Rightarrow mt (my'+mty'')+m^2t^2y \Rightarrow (mt)^2y'' +m(mt)y' + m^2t^2y=0$ so the extra $m$ factor in the second term is still here because I'm expecting something like $(mt)^2y'' +(mt)y' + (mt)^2y=0$. I'm doing something wrong? – injo Aug 16 '24 at 16:02
  • Your expectation isn't correct. If you clear $m^2$, you get $t^2 y''+ty'+t^2 y=0$ which is already a case of Bessel's equation. – Semiclassical Aug 16 '24 at 16:10

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Following suggestions in comment section, the problem can be solved writing the equation in Sturm-Liouville form with $x^m=t$. In this case the differential operator becomes $x\frac{d}{dx}=x\frac{dt}{dx}\frac{d}{dt}=x\cdot mx^{m-1}\frac{d}{dt}=mt\frac{d}{dt}$. \begin{equation} x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} + m^2x^{2m}y=0 \\ x\frac{d}{dx}\Big(x\frac{dy}{dx}\Big) + m^2x^{2m}y=0 \\ mt\frac{d}{dt}\Big(mt\frac{dy}{dt}\Big) + m^2t^2y=0 \\ mt\Big(m\frac{dy}{dt} + mt\frac{d^2y}{dt^2}\Big) + m^2t^2y=0 \\ m^2t^2y''+m^2ty'+m^2t^2y=0 \\ t^2y''+ty'+t^2y=0 \end{equation}

This is a Bessel differential equation of order $p=0$ so its general solution for $x>0$ is $$y=c_1J_0(t)+c_2Y_0(t) \Rightarrow y=c_1J_0(x^m)+c_2Y_0(x^m)$$

injo
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    Well done. A good follow-up exercise is to use these techniques ($y\to t^n z, x(d/dx)\to m t(d/dt)$) to verify the general formula given initially. – Semiclassical Aug 16 '24 at 19:21