General Solution to Second Order Differntial Eq

oxx

Sep 2009
5
0
This is on a review sheet for the exam, and I have the solution, but I can't figure out how the professor got to the first step.

The question reads: Find the general solution to:

[(x^2)/2]y'' + 2xy' - (7/8)y = (x^4)/2 x>0


The first line on the solution sheet reads:

Characteristic eq: (1/2)r(r-1) + 2r - (7/8) = 0

I understand the substituting r for y's, but I don't see how he was able to get rid of the x's and get the r(r-1) part.

It would help a ton if someone could explain this to me.
Thanks a lot in advance.
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
This is on a review sheet for the exam, and I have the solution, but I can't figure out how the professor got to the first step.

The question reads: Find the general solution to:

[(x^2)/2]y'' + 2xy' - (7/8)y = (x^4)/2 x>0


The first line on the solution sheet reads:

Characteristic eq: (1/2)r(r-1) + 2r - (7/8) = 0

I understand the substituting r for y's, but I don't see how he was able to get rid of the x's and get the r(r-1) part.

It would help a ton if someone could explain this to me.
Thanks a lot in advance.
Is there a certain method you should be using to solve this DE?
 

oxx

Sep 2009
5
0
It doesn't specify. The professor went on to get r= 1/2 and -7/2

y = Ax^(1/2) + Bx^(-7/2)

Then using variation of parameters his final solution was
y=(4/105)x^4 + Ax^(1/2) + Bx^(-7/2)
 

chisigma

MHF Hall of Honor
Mar 2009
2,162
994
near Piacenza (Italy)
May be that the best approach is to 'attack' first the 'incomplete equation'...

\(\displaystyle \frac{x^{2}}{2} y^{''} + 2 x y^{'} - \frac{7}{8} y = 0\) (1)

The 'coefficients' of this linear DE aren't constants , so that the 'standard approach' doesn't work. It is easy to verify that solutions of (1) are of the type \(\displaystyle y(x)= x^{r}\)... then try to find the constants \(\displaystyle r \) able to solve (1)...

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
Last edited:

oxx

Sep 2009
5
0
I can't believe I couldn't figure that out. That's a lot for your help!