Homogeneous SDEs.md

tags
Math, Math/Pure

%% links: [[Differential Equations]] %%

Homogeneous SDEs

• A homogeneous SDE is an [[./Second-order Differential Equations (SDEs).md|SDE]] of the form: $$ f(y'', y', y) = 0 $$

• A linear homogeneous SDE has the following form: $$ a\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}} + b\frac{\mathrm{d}y}{\mathrm{d}x} + cy = 0,\ \text{where constants } a, b, c \in \mathbb{R} \tag{1} $$

To find a general solution of a linear homogeneous SDE, let us assume that $y = Ae^{\alpha x} + Be^{\beta x}$ is a solution, where $A$ and $B$ are arbitrary constants, and $\alpha$ and $\beta$ are constants to be determined. By substituting $y, y', y''$ into (1) and simplifying we get: ^408c5f $$ Ae^{\alpha x}(a \alpha^{2} + b \alpha + c) + Be^{\beta x }(a \beta^{2} + b \beta + c) = 0 \tag{2} $$ This implies that in order for $\alpha$ and $\beta$ to satisfy (2), and consequently (1), they have to be the solutions of the auxiliary equation of the form: $$ am^{2} + bm + c = 0 \tag{3} $$^auxiliary-equation

• To find the general solution of a linear homogeneous SDE (1), first find the discriminant ($\Delta$) of the auxiliary equation (3) that has roots $\alpha$ and $\beta$, and then if:
  1. $\Delta > 0$ ($\alpha, \beta \in \mathbb{R}, \alpha \neq \beta$): $\qquad \boxed{y = Ae^{\alpha x} + Be^{\beta x}}$
  2. $\Delta = 0$ ($\alpha, \beta \in \mathbb{R}, \alpha = \beta$): $\qquad \boxed{y = (A + Bx)e^{\alpha x}}$
  3. $\Delta < 0$ ($\alpha, \beta \in \mathbb{C} \text{ and } \alpha = z = p + qi,\ \beta = z^{*} = p - qi$): $\qquad \boxed{y = e^{px}(A \cos{qx} + B \sin{qx})}$ where $A$ and $B$ are arbitrary constants. ^solve-homogeneous-sde

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Q: Find the particular solution of $y'' - y' - 6y = 0$ given that if $x = 0$, then $y = 0$ and $y' = -1$. A: The auxiliary equation is $m^{2} - m - 6 = 0 \implies \Delta = 25 > 0 \implies m_1 = -2, m_2 = 3$

$$ \begin{align} y &= Ae^{3x} + Be^{-2x} \tag{1}\\ y' &= 3Ae^{3x} - 2B^{-2x} \tag{2} \end{align} $$

After substitutin the given values into (1) and (2) respectively we get:

$$ \begin{cases} A + B = 0 \\ 3A - 2B = -1 \end{cases} \iff \begin{cases} A = -\frac{1}{5} \\ B = \frac{1}{5} \end{cases} $$

Answer: $y = -\frac{1}{5}e^{3x} + \frac{1}{5}e^{-2x}$