11-21-25 Second Order Linear Differential Equations
When G(x) = 0, we have a homogenous linear equation
When , we have a non-homogenous equation
When P, Q, and R are constant functions, we have:
A typical solution of this equation is:
This means:
Solving for the value of r:
Option 1:
Option 2:
Factor
The cases when solving for r:
b²-4ac > 0
You have two real unequal roots/solution
(General Solution where and are still in the solution
You have one real root
b²-4ac < 0
Imaginary numbers and have to use the quadratic formular_1 = \alpha + \beta ir_2 = \alpha - \beta iy=e^{\alpha x} (C_1 \cos (\beta x) + C_2 \sin (\beta x))y^{\prime\prime} - 5y^\prime + 6y = 0ar² + br + c = 0a = 1b = -5c = 61r²-5r+6=0(-5)² - 4 (1)(6) = 25 - 24 = 1 > 0\frac {-b\pm\sqrt {b²-4ac}}{2a}\frac {- (-5)\pm \sqrt {1}}{2(1)}\frac {5 + 1}{2} and \frac {5 - 1}{2}r_1 = 3, r_2 = 2y=C_1e^{r_1x}+ C_2e^{r_2x}y=C_1e^{3x} + C_2e^{2x}y^{\prime\prime} - 6y^\prime + 9 = 01r² - 6r + 9(-6)²-4(1)(9) = 36 - 36 = 0\frac {-(-6)\pm\sqrt {36}}{2(1)} \Rightarrow \frac {6 \pm 6}{2}\Rightarrow \frac {12}{2} and \frac 02$$