11-21-25 Second Order Linear Differential Equations

$P(x) y^{\prime\prime} + Q(x) y^\prime + R(x) y - G(x)$

When $G(x) = 0$, we have a homogenous linear equation

When $G(x) \ne 0$, we have a non-homogenous equation

When P, Q, and R are constant functions, we have:

• $ay^{\prime\prime} + by^\prime + cy = 0$

A typical solution of this equation is:

• $y=e^{rx} \Rightarrow y^\prime = re^{rx} \Rightarrow y^{\prime\prime} = r²e^{rx}$ 

This means:

• $ar²e^{rx}+bre^{rx}+ce^{rx}$

• $\Rightarrow e^{rx}(ar²+br+c) = 0$

• $\Rightarrow ar²+br+c=0$

Solving for the value of r:

• Option 1:

  • $r=\frac{-b\pm\sqrt{b²-4ac}}{2a}$

• Option 2:

  • Factor

The cases when solving for r:

• $b²-4ac > 0$

  • You have two real unequal roots/solution

  • $y=C_1e^{r_1x} + C_2 e^{r_2x}$ (General Solution where $C_1$ and $C_2$ are still in the solution

•  $b²-4ac = 0$

  • You have one real root

  • $y=C_1e^{rx} + C_2xe^{rx}$

• $b²-4ac < 0$

  • Imaginary numbers and have to use the quadratic formula$</p></li><li><p>$r_1 = \alpha + \beta i$</p></li><li><p>$r_2 = \alpha - \beta i$</p></li><li><p>$y=e^{\alpha x} (C_1 \cos (\beta x) + C_2 \sin (\beta x))$</p></li></ul></li></ul><div data-type="horizontalRule"><hr></div><p>Example:</p><p></p><p>$y^{\prime\prime} - 5y^\prime + 6y = 0$</p><p></p><p>$ar² + br + c = 0$</p><p></p><ul><li><p>$a = 1$</p></li><li><p>$b = -5$</p></li><li><p>$c = 6$</p></li></ul><p></p><p>$1r²-5r+6=0$</p><p></p><p>$(-5)² - 4 (1)(6) = 25 - 24 = 1 > 0$</p><ul><li><p>There are two real solutions</p></li></ul><p></p><p>$\frac {-b\pm\sqrt {b²-4ac}}{2a}$</p><ul><li><p>$\frac {- (-5)\pm \sqrt {1}}{2(1)}$</p></li><li><p>$\frac {5 + 1}{2}$and$\frac {5 - 1}{2}$</p><ul><li><p>$r_1 = 3$,$r_2 = 2$</p></li></ul></li></ul><p></p><p>$y=C_1e^{r_1x}+ C_2e^{r_2x}$</p><p>$y=C_1e^{3x} + C_2e^{2x}$</p><div data-type="horizontalRule"><hr></div><p>Example:</p><p></p><p>$y^{\prime\prime} - 6y^\prime + 9 = 0$</p><p></p><p>$1r² - 6r + 9$</p><p></p><p>$(-6)²-4(1)(9) = 36 - 36 = 0$</p><p></p><p>$\frac {-(-6)\pm\sqrt {36}}{2(1)} \Rightarrow \frac {6 \pm 6}{2}\Rightarrow $$\frac {12}{2}$and$\frac 02$$