# Second Order Initial Value problem Query

Hey guys just wondering for Second Order IVP's how to solve the equation when the root is complex for example:

Solve the IVP: $$y'' +9y=0, \space y(0)=1, \space y'(0)=1$$ $$r^2 +9=0$$ $$r^2 =-9$$ $$r = \pm3i$$

I'm unsure what to do now I've found the roots.

I'd appreciate any guidance here.

• These make the solution $y_g=C_1e^{+3ix}+C_2e^{-3ix}$ and after simplifying give $y_g=C_1\sin3x+C_2\cos3x$ – Nosrati Apr 6 '17 at 13:57

As said the user in the comment the solution is $$y_g(x)=C_1e^{+3ix}+C_2e^{-3ix}$$(see the form of general solution in your book about this kind of ordinary differential equation). Then using $e^{iax}=\cos(ax)+i\sin(ax)$ and that the cosine is even and the sine function an odd function one has $$y_g(x)=C_1(\cos(3x)+i\sin(3x))+C_2(\cos(3x)-i\sin(3x))$$ that is
$$y_g(x)=(C_1+C_2)\cos(3x)+(C_1-C_2)i\sin(3x)$$ thus $$y'_g(x)=-3(C_1+C_2)\sin(3x)+3(C_1-C_2)i\cos(3x).$$
From your conditions you get (if there are no mistakes in my calculations) $$1=y_g(0)=(C_1+C_2)\cdot 1+0=C_1+C_2$$ and $$1=y'_g(0)=0+3(C_1-C_2)\cdot 1=3C_1-3C_2,$$ thus $C_1=2/3$ and $C_2=1/3$.