# Math Help - A quick question on complimentary functions

1. ## 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

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

3. 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 $y = (Ax + B) e^{\lambda x}$.

4. 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 $\lambda = \alpha \pm i \beta$.

5. 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.

6. 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 $2$.

7. 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.