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Bonsoir je cherche les solutions de l'équation differentielle de type

$$x^3y''(x)+(ax^3+bx^2+cx+d)y(x) =0$$

Merci d'avance

Good evening, I'm searching solutions of a differential equation of the type:

$x^3y''(x) + (ax^3+bx^2+cx+d)y(x) = 0.$

Thanks is advance.

Ivo Terek
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Said LAACHIR
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  • I do not think you can have a closed form solution. – Mhenni Benghorbal Sep 22 '13 at 14:42
  • @Maesumi: wouldn't this work only if the constant of integration were zero and if the ratio of the coefficients were constant? – automaton 3 Sep 22 '13 at 14:54
  • @automaton3 you are right. back to series! – Maesumi Sep 22 '13 at 15:09
  • Mathematica gives answers for the linear case, when $a=b=0$. – Maesumi Sep 22 '13 at 19:56
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    \begin{equation} x^3y^{''}(x) + ( ax^3 + b x^2 + c x + d + e/x) y(x)=0 \end{equation} is mapped onto the Heun equation as outlined in here https://math.stackexchange.com/questions/2934638/hunt-for-exact-solutions-of-second-order-ordinary-differential-equations-with-va . – Przemo Oct 26 '18 at 11:08

1 Answers1

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Hint:

$x^3y''(x)+(ax^3+bx^2+cx+d)y(x)=0$

$\dfrac{d^2y}{dx^2}+\left(a+\dfrac{b}{x}+\dfrac{c}{x^2}+\dfrac{d}{x^3}\right)y=0$

Let $r=\dfrac{1}{x}$ ,

Then $\dfrac{dy}{dx}=\dfrac{dy}{dr}\dfrac{dr}{dx}=-\dfrac{1}{x^2}\dfrac{dy}{dr}=-r^2\dfrac{dy}{dr}$

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(-r^2\dfrac{dy}{dr}\right)=\dfrac{d}{dr}\left(-r^2\dfrac{dy}{dr}\right)\dfrac{dr}{dx}=\left(-r^2\dfrac{d^2y}{dr^2}-2r\dfrac{dy}{dr}\right)\left(-\dfrac{1}{x^2}\right)=\left(-r^2\dfrac{d^2y}{dr^2}-2r\dfrac{dy}{dr}\right)(-r^2)=r^4\dfrac{d^2y}{dr^2}+2r^3\dfrac{dy}{dr}$

$\therefore r^4\dfrac{d^2y}{dr^2}+2r^3\dfrac{dy}{dr}+(dr^3+cr^2+br+a)y=0$

doraemonpaul
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