This tutorial explains how to solve second order differential equations whose auxiliary equation has two distinct complex conjugate roots. Step-by-step examples and exercises are included for practice.
Consider a second order differential equation:
\[ \frac{d^2y}{dx^2} + b \frac{dy}{dx} + c y = 0 \]
Its auxiliary equation is:
\[ k^2 + b k + c = 0 \]
If the discriminant \( b^2 - 4c < 0 \), the quadratic has two complex conjugate roots:
\[ k_1 = r + t i, \quad k_2 = r - t i \]
The general solution of the differential equation is:
\[ y(x) = e^{r x} \left( A \cos(t x) + B \sin(t x) \right) \] where \( A \) and \( B \) are arbitrary constants.
Solve the differential equation:
\[ \frac{d^2y}{dx^2} + \frac{dy}{dx} + 2y = 0 \]
Solution:
The auxiliary equation is:
\[ k^2 + k + 2 = 0 \]
Solving gives two complex conjugate roots:
\[ k_1 = -\frac{1}{2} + \frac{\sqrt{7}}{2} i, \quad k_2 = -\frac{1}{2} - \frac{\sqrt{7}}{2} i \]
Thus, \( r = -\frac{1}{2} \) and \( t = \frac{\sqrt{7}}{2} \). The general solution is:
\[ y(x) = e^{-x/2} \left( A \cos\left(\frac{\sqrt{7}}{2} x\right) + B \sin\left(\frac{\sqrt{7}}{2} x\right) \right) \]
Solve the differential equation with initial conditions \( y(0)=1, y'(0)=0 \):
\[ \frac{d^2y}{dx^2} + \sqrt{3} \frac{dy}{dx} + 3y = 0 \]
Solution:
Auxiliary equation:
\[ k^2 + \sqrt{3} k + 3 = 0 \]
Complex roots:
\[ k_1 = -\frac{\sqrt{3}}{2} + \frac{3}{2} i, \quad k_2 = -\frac{\sqrt{3}}{2} - \frac{3}{2} i \]
General solution:
\[ y(x) = e^{- \frac{\sqrt{3}}{2} x} \left( A \cos\left(\frac{3}{2} x\right) + B \sin\left(\frac{3}{2} x\right) \right) \]
Apply initial conditions:
\( y(0) = 1 \Rightarrow B = 1 \)
\( y'(0) = 0 \Rightarrow -\frac{\sqrt{3}}{2} B + \frac{3}{2} A = 0 \Rightarrow A = \frac{\sqrt{3}}{3} \)
Final solution:
\[ y(x) = e^{- \frac{\sqrt{3}}{2} x} \left( \frac{\sqrt{3}}{3} \cos\left(\frac{3}{2} x\right) + \sin\left(\frac{3}{2} x\right) \right) \]
Solve the following differential equations:
Answers: