# Quick question

1. Jan 29, 2005

### daster

I was just wondering if there's a known result or theorem that states if:

$$y=e^{px}(A\cos (nx)+B\sin (mx))$$

Then:

$$ay'' + by' + cy = 0$$

I don't see why this should be the case, but I read this somewhere and was intrigued.

2. Jan 29, 2005

### dextercioby

No,in general,that is not true...
Try to do it viceversa...Solve the ODE...And then find the conditions the coefficients need to fulfil,as the solution would have the form u've given.

Daniel.

3. Jan 29, 2005

### daster

Thanks Daniel.

4. Jan 29, 2005

### mathwonk

perhaps you read that e^(px)[Acos(qx) + Bsin(qx)] solves an equation of form

ay'' + by' + cy = 0, where p+qi and p-qi are imaginary roots of the equation

aX^2 + bX + c = 0.

the reason for this is the solutions of this equation are all linear combinations of

e^(p+qi)x and e^(p-qi)x, i.e. of e^(px)[cos(qx) +isin(qx)]

and e^(px)[cos(qx) -isin(qx)], equivalently, of

e^(px)[cos(qx)] and e^(px)[sin(qx)].

I.e. of form Ae^(px)[cos(qx) + B e^(px)[sin(qx)] =

e^(px)[Acos(qx) + Bsin(qx)].

5. Jan 30, 2005

### daster

Yeah, I know that the form I posted was the complementary function of the ODE if the auxilary equation produced complex roots, but what I read was stating something generally, which I wasn't too sure about.

6. Feb 5, 2005

### mathwonk

i'm just saying you were very close, except you need to take n =m.