# solving second order differential equation

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:

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

• I do not think you can have a closed form solution. Commented Sep 22, 2013 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? Commented Sep 22, 2013 at 14:54
• @automaton3 you are right. back to series! Commented Sep 22, 2013 at 15:09
• Mathematica gives answers for the linear case, when $a=b=0$. Commented Sep 22, 2013 at 19:56
• $$x^3y^{''}(x) + ( ax^3 + b x^2 + c x + d + e/x) y(x)=0$$ is mapped onto the Heun equation as outlined in here math.stackexchange.com/questions/2934638/… . Commented Oct 26, 2018 at 11:08

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

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

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