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Mathematics LibreTexts

3.2: Complex Roots of the Characteristic Equation

  • Page ID
    399
  • [ "article:topic", "complex roots", "characteristic equation", "authorname:green" ]

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    We have already addressed how to solve a second order linear homogeneous differential equation with constant coefficients where the roots of the characteristic equation are real and distinct. We will now explain how to handle these differential equations when the roots are complex. The example below demonstrates the method.

    Example \(\PageIndex{1}\)

    Solve 

    \[ y'' - 4y' + 13y = 0. \nonumber \]

    Solution

    As before we assume that \(y = e^{rt}\) is a solution. We have

    \[ y' = re^{rt} \;\;\; y'' = r^2e^{rt}. \nonumber\]

    Substituting back into the original differential equation gives

    \[ r^2e^{rt} - 4re^{rt} + 13e^{rt} = 0 \nonumber\]

    \[ r^2 - 4r + 13 = 0 \;\;\; \text{ dividing by $e^{rt}$. }\nonumber\]

    This quadratic does not factor, so we use the quadratic formula and get the roots

     \[ r = 2 + 3i \;\;\; \text{ and } \;\;\; r = 2 - 3i \nonumber\]

    We can conclude that the general solution to the differential equation is

    \[ \begin{align*} y &= a_1e^{(2 + 3i)t} + a_2e^{(2-3i)t}\\ &= a_1e^{2t}e^{3it} & \text{ Using the rule of exponents } \\ &= e^{2t} (a_1e^{3it} + a_2e^{-3it}) & \text{ Factoring out the $e^{2t}$.} \end{align*}\]

    Although this gives the general solution, it it not satisfactory since the solution involves complex exponents. To deal with this we use Euler's formula

    \[ e^{iq} = \cos \, q + i \sin \, q. \nonumber\]

    This gives 

    \[ y = e^{2t} [a_1( \cos (3t) + i \sin (3t)) + a_2 ( \cos (-3t) + i \sin (-3t)). ] \nonumber\]

    Since the \( \cos x \) is an even function and \( \sin x \) sin x is an odd function, we get

    \[ y = e^{2t} [ a_1( \cos (3t) + i \sin (3t)) + a_2( \cos (3t) - i \sin (3t))] \nonumber\]

    or

    \[ y = e^{2t} [(a_1 + a_2) \cos (3t) + (a_1 - a_2)i \sin (3t) ]. \nonumber\]

    Finally let 

    \[ c_1 = a_1 + a_2 \;\;\; \text{and} \;\;\; c_2 = i(a_1 - a_2) \nonumber\]

    and we get 

    \[ y = e^{2t} [ c_1 \cos (3t) + c_2 \sin (3t) ] .\nonumber\]

    General Solution

    In general if 

    \[ ay'' + by' + cy = 0 \]

    is a second order linear differential equation with constant coefficients such that the characteristic equation has complex roots

    \[ r = l + mi \;\;\; \text{and} \;\;\; r = l - mi \]

    Then the general solution to the differential equation is given by 

    \[ y = e^{lt}\left[ c_1 \cos(mt) + c_2 \sin(mt) \right ] \label{gen1}\]

    Example \(\PageIndex{2}\): Graphical Solutions

    Solve

    \[ y'' - 10y' + 29 = 0 \;\;\; \text{given } y(0) = 1, \;\; y'(0) = 3 \nonumber\]

    Solution

    The characteristic equation is

    \[r^2 - 10r + 29 = 0\nonumber \]

    which has roots 

    \[ r = 5 + 2i \;\;\; \text{and} \;\;\; r = 5 - 2i. \nonumber\]

    The general solution is 

    \[ y = e^{5t}[ c_1 \cos (2t) + c_2 \sin (2t) ]. \nonumber\]

    We use the initial values to find the constants. Plug in \( y(0) = 1 \) 

    \[ 1 = 1 [ c_1 (1) + c_2 (0) ] \nonumber\]

    so that \( c_1 = 1\). We have

    \[ y' = 5e^{5t}[ \cos (2t) + c_2 \sin (2t) ] + e^{5t}[ -2 \sin (2t) + 2c_2 \cos (2t)].\nonumber \]

    Plugging in \(y' (0) = 3 \)

    \[ \begin{align*} 3 &= 5 [ 1 + 0] + 1[0 + 2c_2] \\[5pt] &= 5 + 2c_2 .\end{align*}\]

     Hence \(c_2 = -1 \).

    The final solution is 

    \[ y = e^{5t} [ \cos (2t) - \sin (2t) ] .\nonumber\]

    Let's investigate the graphs of solutions for the general solutions in Equation \ref{gen1}

    • For \(l = 0\), the graph is periodic.
    • For \(l > 0\), the amplitude increases exponentially
    • For \(l < 0\), the amplitude approaches 0 exponentially.

    The three types are pictured below.

    Contributors