Eulers Equation confusion Are these the right equations?

[mr_coffee]

Hello everyone, I just went thorugh the pratice exam in about 15 minutes and got them all right but right when we were about to leave he also said you will need to do the following and he wrote the following:

Euler:
for 1 real solution:
USE:
y = c1*t^a + c2*t^a*ln(t);



2 complex:
USE:
y = t^a(c1*sin(b*ln(t)) + c2*cos(b*ln(t));
a is the real part
b is the imageary;

BUt he left out, what if you get repeated roots, also what if you get 2 real roots?

Also in his first equation:
for 1 real solution:
USE:
y = c1*t^a + c2*t^a*ln(t);

What is a? is that like r?
y''+18y'+85 = 0;
r^2+18r+85 = 0;
r1 = -9+2i
r2 = -9-2i

like how those are r's, is that what he ment by a?


So i asked him, when would we use those forms? And he wrote, when you have a differential in the form of:
a*x^2y'' +bxy' + cy = 0;


So i wanted to test it out to see if it works, so i tried:
http://suprfile.com/src/1/2hef20/lastscan.jpg
And using it the old way obviouslly doesn't work, can someone tell me the other forms? like for 2 year solutions like this problem requires and for repeated roots? Thank you! Once i get these other equations i think i'll be able to figure it out.


Ivey I looked at what you posted in the last 2 threads i posted but the professor said its memorizing a few simple formulas, he doesn't want us to work it out or put it in a different form.

An "Euler type equation" has x to a power equal to the order of the derivative (that's why it's also called "equipotential"). Your example,
x2y"+ 2xy'- 30y= x3 is exactly of that kind. As for the substitution, u= ln x, I've also told you about that before.

If u= ln x, by the chain rule, dy/dx= (du/dx)(dy/du)= (1/x)dy/du and
d2y/dx2= (-1/x2)dy/du+ (1/x2)d2y/du2. Also x= eu so x3= e3u Your equation becomes
d2y/du2+ dy/du- 30u= e3u.
That should be easy.

THanks!

 

[mr_coffee]

I think i got it,
m^2-m-6m-8 = 0;
m = 8,-1

y = Ax^r1+Bx^r2;