# LSO – Two complex roots

Up a level : Differential Equations
Previous page : LSO - Two equal real roots Say we got two complex roots p+qi and pqi. Using the solution for two different roots we get $\begin{gathered} y = C{e^{{r_1}x}} + D{e^{{r_2}x}} \hfill \\ \quad = C{e^{(p + qi)x}} + D{e^{(p - qi)x}} \hfill \\ \quad = C{e^{px}}{e^{qix}} + D{e^{px}}{e^{ - qix}} \hfill \\ \quad = {e^{px}}(C{e^{qix}} + D{e^{ - qix}}) \hfill \\ \end{gathered}$

Using the fact that ${e^{i\theta }} = \cos \theta + i\sin \theta$

we get that $\begin{gathered} y = {e^{px}}(C{e^{qix}} + D{e^{ - qix}}) \hfill \\ \quad = {e^{px}}(C\cos (qx) + Ci\sin (qx) + D\cos ( - qx) + Di\sin ( - qx)) \hfill \\ \quad = {e^{px}}(C\cos (qx) + Ci\sin (qx) + D\cos (qx) - Di\sin (qx)) \hfill \\ \quad = {e^{px}}((C + D)\cos (qx) + (C - D)i\sin (qx)) \hfill \\ \end{gathered}$

Now we want both C+D and (C–D)i to be real, We can choose C and D to be complex conjugates, say C=U+Vi this gives us $\begin{gathered} C + D = U + Vi + U - Vi = 2U, \hfill \\ (C - D)i = (U + Vi - U + Vi)i = - 2V \hfill \\ \end{gathered}$

In that way the constants may be any value. Let us write C instead of 2U and D instead of -2V, this will finally give us the solution $y = {e^{px}}(C\cos (qx) + D\sin (qx))$

An example

Let us check against a special case if this really works. Say the equation has pure complex roots then we must have started with an equation of the form $y'' + by = 0$

that will have the characteristic equation ${r^2} + b = 0$

If b is positive we get the solutions ${r^2} = \pm i\sqrt b$

This would give us $y = C\cos (\sqrt b x) + D\sin (\sqrt b x)$

as a solution to the differential equation. This is something we already found when we solved the equation $y'' = - \frac{k}{m}y$

where we found the solution to be $y = A\sin \left( {\sqrt {\frac{k}{m}} t - \theta } \right)$

that, with a bit of algebra, can be found to be equivalent to the solution we found now. We can use the formula $\sin (\alpha - \beta ) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$

and we let t be x and we can let the constant in front of t be ω.  This would give us $y = A\sin \left( {\omega x - \theta } \right) = A\sin (\omega x)\cos \theta - A\cos (\omega x)\sin \theta$

so q= ω, C=Acosθ and D=Asinθ. Up a level : Differential Equations
Previous page : LSO - Two equal real roots Last modified: Mar 26, 2019 @ 21:15