Is There a Theorem Relating Exponential and Trigonometric Functions?

[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.

 

[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.

 

[daster]

Thanks Daniel.

 

[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)].

 

[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.

 

[mathwonk]

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