LSO – Two complex roots -

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Say we got two complex roots p+qi and p–qi. 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

$y'' + by = 0$