Solving ODE Roots: y'' + 2y' + 5y = 0

[kasse]

y'' + 2y' + 5y = 0 (*)

OK, what I have done is computing the two roots y1 = exp(-x)*cos2x and y2 = exp(-x)*sin2x.

However, when I compute the derivatives of these two, and substitute into (*), the eq. doesn't equate 0.

Are my roots wrong?

 

[Dick]

In that case you haven't taken the derivatives correctly, because the roots are correct.

 

[wbclark]

Let us first look at the characteristic equation of the ODE.

$$P(\lambda) = \lambda^2 + 2\lamda + 5 = 0$$
$$(\lambda + 1)^2 = -4$$
$$\lambda + 1 = \pm2i$$
$$\lambda = -1 \pm2i$$

Your roots appear to be correct.

 

[kasse]

Let us first look at the characteristic equation of the ODE.

$$P(\lambda) = \lambda^2 + 2\lamda + 5 = 0$$
$$(\lambda + 1)^2 = -4$$
$$\lambda + 1 = \pm2i$$
$$\lambda = -1 \pm2i$$

Your roots appear to be correct.

Yes...

The derivative of cos2x is -2sin2x, isn't it? I'm going to check this tomorrow. Now I need some sleep.