solve 3x^3y'''+10x^2y''-8xy'-4y=0 using cauchy-euler

Question

​Looking to solve the 3rd order differential equation. Need the auxiliary equations also.

Answer 1

Assume y=x^z, where z is a constant, so y'=zx^(z-1), y''=z(z-1)x^(z-2), y'''=z(z-1)(z-2)x^(z-3).

Substitute for y and its derivatives:

3zx^z(z-1)(z-2)+10zx^z(z-1)-8zx^z-4x^z=0;

x^z(3z(z-1)(z-2)+10z(z-1)-8z-4)=0. Trivial case is x=0.

Auxiliary equation is: 3z(z^2-3z+2)+10z^2-10z-8z-4=0;

3z^3-9z^2+6z+10z^2-10z-8z-4=0=3z^3+z^2-12z-4=3z(z^2-4)+z^2-4=(3z+1)(z-2)(z+2).

So we have z=-1/3, 2 or -2. Therefore y=x^(-1/3), y=x^2, y=x^-2.

Let's check the simplest of these y=x^2: y'=2x, y''=2, y'''=0:

0+20x^2-16x^2-4x^2=0 OK.

Now y=x^-2: y'=-2x^-3, y''=6x^-4, y'''=-24x^-5:

-72x^-2+60x^-2+16x^-2-4x^2=0 OK. Looking good!