# Suppose a, b, c are three real numbers such that

1. Oct 18, 2016

### nysnacc

1. The problem statement, all variables and given/known data

2. Relevant equations
character equation

3. The attempt at a solution
Should I set a = ax2 b= bx c =c in the character equation?

2. Oct 18, 2016

### Ssnow

Hi, you must start to substitute $y_{1}(x)=x^{r_{1}}$ in the differential equation and find $r_{1}$ that satisfy the equation ...,
so start to calculate $y'_{1}(x)$ and $y_{1}''(x)$ ...

The same for $y_{2}(x)=x^{r_{2}}\ln{x}$ ...

3. Oct 18, 2016

### nysnacc

y1 = xr1
y1' = r1* xr1-1
y1" = r1 (r1-1)* xr1-2 = ( r12 - r1) *xr1-2

then plug in ODE?

4. Oct 18, 2016

### nysnacc

Then (I simply say r instead of r1)
a(r2-r)xr +b r xr + cxr

xr ( a(r2-r) + b r +c )
And solve for r ?

5. Oct 18, 2016

### Staff: Mentor

You can't solve the above, because they aren't equations!
The equation you need to solve is a(r2-r)xr +b r xr + cxr = 0,
which is equivalent to xr ( a(r2-r) + b r +c ) = 0
Now solve for r.

6. Oct 18, 2016

### nysnacc

Oh yeah!! but I don't know a, b and c... there are four unknowns then

7. Oct 18, 2016

### Staff: Mentor

But you have a relationship involving a, b, and c given in your problem statement.

8. Oct 18, 2016

### nysnacc

So I solve for r which is r = .... (in terms of a, b, c) probably be -b +/- sqrt (b^2 - 4 ac) / 2a

Then use the relationship involving a, b, and c given in problem statement?

But do I need one more equation?

9. Oct 18, 2016

### nysnacc

For r1, it is -B/2A which is -(b-a)/2a
Do I end up wth the letter as coefficient, or any further?

10. Oct 18, 2016

### Staff: Mentor

No it isn't.

Start from this:
xr ( a(r2-r) + b r +c ) = 0
The part on the left has to be identically equal to zero; i.e., for all values of x.

So either xr = 0 (can't be true)
or a(r2-r) + b r +c = 0
Solve this equation for r.

11. Oct 18, 2016

### Staff: Mentor

Yes, this is what I get: r = -(b - a)/(2a), which is the same as (a - b)/(2a)

12. Oct 18, 2016

### nysnacc

Great, so this is it, for r1 ?? :)

13. Oct 18, 2016

### Staff: Mentor

Yes.

14. Oct 18, 2016

### nysnacc

For r2,
I have y2 = xr2 ln (x)
y2' = xr2-1 +r2 xr2-1 ln (x)
y2" = (r2-1)xr2-2 +r2 xr2-2 + (r2^2-r2)xr2-2 ln(x)

Which then plug in to the ODE, i found r2 = -(b-a)/ 2a same as r1

and whole ODE can be 0 if x =1