# What happens if you only get 1 solution to a 2nd Order Diff EQ?

1. Feb 17, 2006

### mr_coffee

Hello everyone, i'm slightly confused on this problem, when i factored it and solved for r, i came out with only 1 answer, r = -13/72
Here is my problem and work:
http://img213.imageshack.us/img213/685/lastscan15uk.jpg [Broken]

Last edited by a moderator: May 2, 2017
2. Feb 17, 2006

### benorin

If you get repeated roots in your auxilary equation, e.g.

$$y^{\prime\prime}-2y^{\prime}+1=0$$

has the auxilary equation

$$r^{2}-2r+1=0$$ so $$(r-1)^2=0$$

and hence r=1 (with multiplicity 2) so the general solution is of the form

$$y=Ae^t+Bte^t$$

The roots of the equation

$$5184r^2+1827r+169=0$$

are $$r=-\frac{13}{72}$$ (with multiplicity 2) so your general solution is of the form...

Last edited: Feb 17, 2006
3. Feb 17, 2006

### Tide

One way to see why the method proposed by Benorin is valid would be to write your solution as

$$y = A e^{r_1 t} + B e^{r_2 t}$$

and apply the initial conditions to determine the coefficients. Then let $r_2$ approach $r_1 = r$ (the root of your actual equation). Usually, l'Hopital's Rule can be applied to yield Benorin's solution.

4. Feb 18, 2006

### mr_coffee

Thanks guys, i don't know if its right or not, but i think my method is right:
http://img211.imageshack.us/img211/6219/lastscan0ku.jpg [Broken]

Last edited by a moderator: May 2, 2017