2nd order Diff EQ with x^2*y'', what shall i do i missed it

[mr_coffee]

Hello everyone!
i'm confused on how to approach this problem, my professor did an example and he used m^2-m-4m+6 = 0, if u have t^2*y''-4ty' +6y = 0;

So i tried to do the following, but the answer is wrong. Anyone see?
http://img88.imageshack.us/img88/3603/lastscan7jj.jpg

THanks!

 

[georgeh]

Maybe I am wrong but this is what I did..
t^2y"-4ty'+6y=0
y"-4t^-1*y+6t^-2=0
X^r(r^2-4t^-1+6t^-2)=0
quadratic..
y=4t^-1(+-)sqrt(16/t^2+24/t^2)/2

y= 4/t (+-) sqrt( 16+24/t^2)/2
y=2/t + sqrt(10)/t, as long as t > 0 or it can be when t < 0, depending on initial conditions.

 

[qbert]

first i see a differential equation in y.
then some mixed equation in r,x.
then finally an expression in m.
and i don't have any idea how you've gone
from one to the other.


the idea for this is not to try solutions
that look like y = exp(kt) but to instead
look for solutions like y = x^m.

[the reason is that x^2 y'' will have the same
power of x as xy' and as y].

 

[HallsofIvy]

What you are doing is a method that works for equations with constant coefficients when your equation has variable coefficents.

e^rx works for constant coefficients because, since you don't have any functions in the coefficients, the derivatives have to be the "same type" in order to cancel.

For problems like these, "Euler type" or "equi-potential" equations, see what happens if you try y= x^r rather than y= e^rx.

Another way to handle these equations is to make the substitution
t= ln x. If you do use the chain rule correctly, that will give you a differential equation for y as a function of t that has constant coefficients.

 

[mr_coffee]

Thanks again Ivey! our professor never showed us, that, its weird he said i was doing this problem the right way! but i got the rigth answer with your technique i got:
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/5f/83961a711684a50c9332dfd970dc391.png