# Quick question (Euler)

1. Jan 2, 2010

### manenbu

Given the homogeneous equation:
x3y''' + 15x2y'' + 61xy' + 64y = 0

I get 3 solutions which are all -4.

Does this mean that the solution for y should be:

y = c1x-4 + c2x-4lnx + c3x-4ln2x
?

2. Jan 2, 2010

### matematikawan

I think you get it right.

3. Jan 2, 2010

### pbandjay

If they are all solutions to the differential equation, then, by the superposition principle, the sum is a solution.

4. Jan 3, 2010

### HallsofIvy

Staff Emeritus
Notice, by the way, that the substitution t= ln(x) changes the differential equation into the "constant coefficients" equation $d^3y/dt^3+ 12d^2y/dt^2+ 48dy/dt+ 64y= 0$ which has characteristice equation $r^3+ 12r^2+ 48r+ 64= (r+ 4)^2= 0$ so the general solution to the constant coefficients equation is $y(t)= C_1e^{-4t}+ C_2te^{-4t}+ C_3t^2 e^{-4t}$ and the general solution to the original equation is $y(x)= C_1e^{-4ln(x)}+ C_2 ln(x) e^{-4ln(x)}$$+ C_3 (ln(x))^2 e^{-4ln(x)}$$= C_1x^{-4}$$+ C_2 ln(x) x^{-4}+ C_3 (ln(x))^2 x^{-4}$, just as you say.