A quick question on complimentary functions

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June 8th 2009, 02:06 AM
craig

A quick question on complimentary functions

Hi.

If the auxiliary equation of a second order differential equation has 2 complex repeated roots, would I still use:

$e^{\alpha x}(A \cos{\beta x} + B\sin{\beta x})$

Just in the form:

$A \cos{\beta x} + B\sin{\beta x}$ ?

Thanks
June 8th 2009, 02:16 AM
craig

Doesn't matter, found out that you use $e^{\alpha x}(A \cos{\beta x} + B\sin{\beta x})$
June 8th 2009, 02:53 AM
mr fantastic

Quote:

Originally Posted by craig

Hi.

If the auxiliary equation of a second order differential equation has 2 repeated roots, would I still use:

$e^{\alpha x}(A \cos{\beta x} + B\sin{\beta x})$

Just in the form:

$A \cos{\beta x} + B\sin{\beta x}$ ?

Thanks

If the root is repeated then the general solution is $y = (Ax + B) e^{\lambda x}$ .
June 8th 2009, 02:54 AM
mr fantastic

Quote:

Originally Posted by craig

Doesn't matter, found out that you use $e^{\alpha x}(A \cos{\beta x} + B\sin{\beta x})$

This is the general solution for when the roots are not real, that is, the roots have the form $\lambda = \alpha \pm i \beta$ .
June 8th 2009, 05:41 AM
craig

Hi thanks for the reply, there was actually a typo in my original post, it was for roots such as $\pm 2i$ .

Have fixed the typo now.
June 9th 2009, 08:38 PM
Sampras

Quote:

Originally Posted by craig

Hi thanks for the reply, there was actually a typo in my original post, it was for roots such as $\pm 2i$ .

Have fixed the typo now.

Assume the coefficients are real. Then the roots come in pairs. I would think the general solution would be $y = c_{1}e^{ax} \cos bx+c_{2}xe^{ax} \cos bx+c_{3}e^{ax} \sin bx + c_{4}xe^{ax} \sin bx$ for $\lambda = a \pm bi$ if degree of multiplicity is $2$ .
June 14th 2009, 02:46 AM
HallsofIvy

Quote:

Originally Posted by Sampras

Assume the coefficients are real. Then the roots come in pairs. I would think the general solution would be $y = c_{1}e^{ax} \cos bx+c_{2}xe^{ax} \cos bx+c_{3}e^{ax} \sin bx + c_{4}xe^{ax} \sin bx$ for $\lambda = a \pm bi$ if degree of multiplicity is $2$ .

No, that the same as $(c_1+ c_3)e^{ax}cos bx+ (c_2+ c_4)e^{ax}sin bx$ so you have nothing new. For a+ bi a double root of the characteristic equation, you want $y= e^{ax}(c_1cos(bx)+ c_2sin(bx))+ xe^{ax}(c_3cos(bx)+ c_4sin(bx))$ .

That is, multiply the "base" solution by x as mr. fantastic said.