Two-Dimensional Differential Equations

A first-order differential equation involves the first derivative $\displaystyle \frac{\mathrm dy}{\mathrm dx}$ . A second-order differential equation involves the second derivative $\displaystyle \frac{\mathrm d^2 y}{\mathrm dx^2}$ and even possibly the first.

Example 1. Given $\alpha \neq \beta$ , find the general solution to the equation

$\displaystyle \frac{\mathrm d^2 y}{\mathrm dx^2} - (\alpha + \beta) \frac{\mathrm d y}{\mathrm dx} + (\alpha \beta) y = 0.$

Solution. We will leave the solution as a guided exercise. Use the substitution $\displaystyle u = \frac{\mathrm dy}{\mathrm dx} - \beta y$ to obtain the differential equation

$\displaystyle \frac{\mathrm du}{\mathrm dx} - \alpha u = 0.$

By the method of separable variables, $u = Ae^{\alpha x}$ , so that

$\displaystyle \frac{\mathrm dy}{\mathrm dx} - \beta y = Ae^{\alpha x}.$

By the method of integrating factors,

$y = C_1 e^{\alpha x} + C_2 e^{\beta x}$ .

This yields the first result for our discussion on second-order differential equations.

Theorem 1. Given real constants $a, b, c$ , the general solution to the differential equation

$\displaystyle a \frac{\mathrm d^2 y}{\mathrm dx^2} + b \frac{\mathrm dy}{\mathrm dx} + c y = 0$

is determined by the roots of its characteristic equation or auxiliary equation $am^2 + bm + c = 0$ :

$y = C_1 e^{\alpha x} + C_2 e^{\beta x}$ if the roots $\alpha \neq \beta$ are real and distinct.
$y = (C_1x + C_2) e^{\alpha x}$ if the roots $\alpha = \beta$ are real and repeated.
$y = e^{px}(C_1 \cos qx + C_2 \sin qx)$ if the roots $p \pm iq$ are complex conjugates.

Proof. We have proven the first case in Example 1. For the third case, we apply the first case to the observation

$e^{(p \pm iq)x} = e^{px}e^{\pm i(qx)} = e^{px} (\cos qx \pm i \sin qx),$

where the second equality follows from Euler’s formula. The solutions are then given by

$\displaystyle \begin{aligned}y &= C_1' e^{(p+iq)x} + C_2' e^{(p-iq)x} \\ &= e^{px} ((C_1' + C_2') \cos qx + i(C_1' - C_2') \sin qx) \\ &= e^{px} (C_1 \cos qx +C_2 \sin qx),\end{aligned}$

where $C_1 := C_1'+C_2'$ and $C_2 := i(C_1'-C_2')$ are constants.

For the second case, the derivation of the first case leads to the differential equation

$\displaystyle \frac{\mathrm dy}{\mathrm dx} - \alpha y = Ae^{\alpha x}.$

By the method of integrating factors, we have an integrating factor $\mu(x) = e^{\int -\alpha\,\mathrm dx} = e^{-\alpha x}$ so that the general solution is given by

$\displaystyle \begin{aligned} y \mu(x) &= \int \mu(x) \cdot Ae^{\alpha x}\, \mathrm dx\\ y e^{-\alpha x} &= \int e^{-\alpha x} \cdot Ae^{\alpha x}\, \mathrm dx \\ &= \int A \, \mathrm dx \\ &= Ax + C \\ y &= (Ax + C)e^{\alpha x} \\ &= (C_1 x + C_2) e^{\alpha x}, \end{aligned}$

where $C_1:= A, C_2 := C$ are constants.

For naming purposes:

Since the left-hand side is of the form $\displaystyle (*) \frac{\mathrm d^2 y}{\mathrm dx^2} + (*) \frac{\mathrm d y}{\mathrm dx} + (*) y$ , the ODE is called linear.
Since the highest derivative is $\displaystyle \frac{\mathrm d^2 y}{\mathrm dx^2}$ , the ODE is of second-order.
Since the coefficients in each $(*)$ are constants $a, b, c$ , the ODE is said to have constant coefficients.
Since the right-hand side is $0$ , the ODE is said to be homogeneous.

Putting it all together, differential equations of the form

$\displaystyle a \frac{\mathrm d^2 y}{\mathrm dx^2} + b \frac{\mathrm dy}{\mathrm dx} + cy = 0$

are called homogeneous second-order linear ODEs with constant coefficients (what a mouthful…).

Example 2. Given $\omega > 0$ , an object is said to obey simple harmonic motion if its acceleration $\displaystyle \frac{\mathrm d^2 x}{\mathrm dt^2}$ is directly proportional to its displacement $x$ with proportionality constant $-\omega^2$ .

Given that the object obeying simple harmonic motion has initial displacement $x_0$ and initial velocity $v_0$ , obtain an expression for $x$ in terms of $\omega$ , $t$ , $x_0$ , $v_0$ .

Solution. Since the object obeys simple harmonic motion,

$\displaystyle \frac{\mathrm d^2 x}{\mathrm d t^2} = -\omega^2 x \quad \Rightarrow \quad \frac{\mathrm d^2 x}{\mathrm d t^2} + \omega^2 x = 0.$

The characteristic equation is given by $m^2 + \omega^2 = 0$ and its roots are $m = 0\pm i\omega$ . Thus, the general solution of the differential equation is given by

$\displaystyle x = C_1 \cos \omega t + C_2 \sin \omega t.$

Since $x(0) = x_0$ yields $C_1 = x_0$ and $x'(0) = v_0$ yields $C_2 = v_0/\omega$ , we have

$\displaystyle x = x_0 \cos \omega t + \frac{v_0}{\omega} \sin \omega t.$

Things start getting rather challenging when the right-hand side of our differential equation is not $0$ ; how do we solve the following differential equation?

$\displaystyle a \frac{\mathrm d^2 y}{\mathrm dx^2} + b \frac{\mathrm dy}{\mathrm dx} + c y = f(x)$

In fact, we haven’t even considered the cases where $a,b,c$ are replaced by functions themselves. This is itself a nontrivial area of study, which we may explore if time and space permits.

Nevertheless, working with constant $a$ , $b$ , $c$ , we have several approaches to solve the corresponding differential equation. We will explore both approaches in due time; the Laplace approach in its traditional presentation, and the undetermined coefficients in a rather compact presentation involving inverse- $\mathcal D$ operators.

These approaches by no means solve every differential equation out there, but it does suffice to analytically solve many equations that are in use in the STEM fields. Most differential equations are solved numerically anyway, and then we invoke the pure mathematicians’ insanity to verify that these solutions genuinely exist. But that’s a discussion for another day.

—Joel Kindiak, 28 Jan 25, 1616H

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Two-Dimensional Differential Equations

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