MHB William's question at Yahoo Answers regarding a 3rd order Cauchy-Euler equation

[MarkFL]

Here is the question:

There are three linearly independent solutions of the differential equation?


There are three linearly independent solutions of the differential equation t^3y'''+3t^2y''-6ty'+6y=0 of the form t^p. Find the possible values of p.

a. 1,2,3
b. 1,2,-3
c. 1,-2,3
d.1,-2,-3
e. -1,2,3
f. -1,2,-3
g. -1,-2,3
h. -1,-2,-3

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello William,

We are given the linear ODE:

$$t^3y'''+3t^2y''-6ty'+6y=0$$

We are then told to assume that there are 3 linearly independent solutions of the form:

$$y(t)=t^p$$

Hence:

$$y'(t)=pt^{p-1}$$

$$y''(t)=p(p-1)t^{p-2}$$

$$y'''(t)=p(p-1)(p-2)t^{p-3}$$

Substituting into the ODE, we obtain:

$$t^3p(p-1)(p-2)t^{p-3}+3t^2p(p-1)t^{p-2}-6tpt^{p-1}+6t^p=0$$

Dividing through by $$t^p\ne0$$ we obtain the characteristic equation:

$$p(p-1)(p-2)+3p(p-1)-6(p-1)=0$$

$$(p-1)(p(p-2)+3(p-2))=0$$

$$(p-1)(p-2)(p+3)=0$$

Hence, the possible values for $p$ are:

$$p=-3,1,2$$

This is choice b.).