A tutorial on how to solve second order differential equations with auxiliary equation having 2 distinct complex solutions. Examples with detailed solutions are included.
The auxiliary equation of a second order differential equation
\[ \dfrac{d^2y}{dx^2} + b \dfrac{dy}{dx} + c y = 0 \]
is given by
\[ k^2 + b k + c = 0 \]
If \( b^2 - 4c \) is less than zero, the above quadratic equation has two complex conjugate solution of the form
\[ k_1 = r + t i \] and \[ k_2 = r - t i \], where \( i \) is the imaginary unit.
In such case, it can be shown that the general solution to the second order differential equation may be written as follows
\[ y = e^{r x} \left( A \cos x t + B \sin x t \right) \] where \( A \) and \( B \) are constants.
Solution to Example 1
The auxiliary equation is given by
\( k^2 + k + 2 = 0 \)
Solve for \( k \) to obtain 2 complex conjugate solutions
\[ k_1 = -\dfrac{1}{2} - i \dfrac{\sqrt{7}}{2} \]
and \[ k_2 = -\dfrac{1}{2} + i \dfrac{\sqrt{7}}{2} \],
\( r = -\dfrac{1}{2} \) (real part)
and \( t = \dfrac{\sqrt{7}}{2} \) (imaginary part)
The general solution to the given differential equation is given by
\[ y = e^{- x / 2} \left( A \cos ((\sqrt{7} /2) x) + B \sin ((\dfrac{\sqrt{7}}{2}) x) \right) \]
where \( A \) and \( B \) are constants.
Solution to Example 2
The auxiliary equation is given by
\[ k^2 + \sqrt{3} k + 3 = 0 \]
Solve the quadratic equation to obtain
\[ k_1 = - \dfrac{\sqrt{3}}{2} + \dfrac{3}{2} i \] and \[ k_2 = - \dfrac{\sqrt{3}}{2} - \dfrac{3}{2} i \]
The general solution to the given differential equation is given by
\[ y = e^{-\dfrac{\sqrt{3}}{2} x} \left( A \sin (\dfrac{3}{2})x + B \cos (\dfrac{3}{2})x \right) \]
The initial condition \( y(0) = 1 \) gives
\[ y(0) = e^0 \left( A \sin 0 + B \cos (0) \right) = 1 \] which gives \( B = 1 \)
\( y'(0) = 0 \) gives
\[ y'(0) = -(\dfrac{\sqrt{3}}{2})e^0 \left( A \sin 0 + B \cos 0 \right) + e^0 \left( (\dfrac{3}{2}) A \cos 0 - (\dfrac{3}{2}) B \sin 0 \right) \]
Solve the system of equations \( B = 1 \) and \(-(\dfrac{\sqrt{3}}{2}) B + (\dfrac{3}{2}) A = 0 \) to obtain
\( A = \dfrac{\sqrt{3}}{3} \) and \( B = 1 \)
The solution may be written as
\[ y = e^{- \dfrac{\sqrt{3}}{2} x} \left( (\dfrac{\sqrt{3}}{3} \sin (\dfrac{3}{2}) x + \cos (\dfrac{3}{2}) x \right) \]