1. ## Is this true?

If y1 = e^(ax)*Cos(Bx) is a solution of a certain homogenrous second order linear equation with constant coefficents then y2 = e^(ax)*Sin(Bx) is also a solution of this equation?

2. Originally Posted by DCU
If y1 = e^(ax)*Cos(Bx) is a solution of a certain homogeneous second order linear equation with constant coefficents then y2 = e^(ax)*Sin(Bx) is also a solution of this equation?
Saying that $e^{ax}Cos(bx)$ is a solution of a certain homogenreous second order linear equation with constant coefficents implies that a+ Bi is a solution to the characteristic equation.

IF the differential equation (and therefore characteristic equation) has real coefficients, then a- Bi must also be a solution to the characteristic equation and so $e^{ax}Cos(bx)$ is an independent solution to the differential equation. However, if any of the coefficients is non-real complex, that is not necessarily true.

3. So it's true if the coefficients are real numbers but not if the coefficents aren't?