# ODE - question to the solution of $y''+4y=0$

The solution of the ODE (using the constant coefficients method) is of the form $$y(x)=e^{\alpha x}(c_1 cos(\beta x)+c_2 sin(\beta x)) \quad \quad (1)$$

which in this case is $$y(x)=c_1 cos(2x)+c_2 sin(2x) \quad \quad (2)$$ https://www.wolframalpha.com/input/?i=y%27%27(t)%2B4y(t)%3D0

the solution of the the quadratic equation resulting from the ODE $r^2+4=0$ is a complex number $r=\frac{\sqrt{-8}}{2}= \sqrt{-2} = \sqrt{2} i$

so $\alpha=0$
complex number has a form $\alpha + \beta i$ why then $\beta=2$ and not $\beta=\sqrt{2}$ in (2)?

• $r^2+4=0 \not \implies r=\frac{\sqrt{-8}}{2}= \sqrt{-2} = \sqrt{2} i$ – Zack Ni Aug 2 '16 at 2:11
• I don't get what you want to say – Michal Aug 2 '16 at 2:16
• See my answer, and I also suggest another look at the quadratic formula. – Sean Roberson Aug 2 '16 at 2:17
• I see, $r=\frac{\sqrt{-16}}{2}= 2 i$, dyscalculia... – Michal Aug 2 '16 at 2:24
• The equation is easier to solve in this case by just subtracting 4 and taking the square root. I think one is less likely to make a mistake using that method as opposed to using the quadratic formula which seems to be the case here. – jdods Aug 2 '16 at 2:36

Careful: with the quadratic formula, $a=1, b=0, c=4$. The discriminant is $-16,$ not $-8$ ($b^2-4ac$).
Anytime we have complex roots $u+iv$ from the characteristic equation, we end up having $e^{ux} \left(\cos vx + \sin vx \right)$ as a solution.