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Linear simultaneous differential equations

Definition

A simultaneous differential equation is one of the mathematical equations for an indefinite function of one or more than one variables that relate the values of the function. Differentiation of an equation in various orders. Differential equations play an important function in engineering, physics, economics, and other disciplines.This analysis concentrates on linear equations with Constant Coefficients.

Using the D Operator

The D operator is a linear operator applied to functions and which is defined as $\displaystyle \frac{d}{dx}\equiv D$ .
For example $D(3x+4)=3$

Example 1

Problem

$\frac{d^2x}{dt^2} - 4\frac{dy}{dx} + 4x = y}$

$\frac{d^2y}{dt^2} + 4\frac{dy}{dt} + 4y = 25x + 16\,e^t$

Workings

The equations may be written as:

$(D^2 - 4D + 4)\,x = y$

(1)

$(D^2 + 4D + 4)\,y = 25\,x + 16\,e^t$

(2)

Eliminating y from equations (#1) and (#2)

$(D^2 + 4D + 4)(D^2 - 4D + 4)\,x = 25x + 16\,e^t$ i.e. $(D^2 + 4)^2 - 16D^2 - 25\,x = 16\,e^t$

$\therefore\;\;\;\;(D^4 - 8D^2 - 9) = 16\,e^t$

$\therefore\;\;\;\;(D^2 - 9)(D^2 + 1) = 16\,e^t$

$\therefore\;\;\;x = A\,e^{3t} + B\,e^{-3t} + E\,cos\,t + F\,sin\,t + \frac{1}{(D^2\,-\,9)(D^2\,+\,1)}\times16\,e^t$

$\mathbf{Thus\;\;\;\;x = A\,e^{3t} + B\,e^{-3t} + E\,cos\,t + F\,sin\,t - e^t}$

Solution

From equation (#1) $y = (D^2 - 4D + 4)[A\,e^{3t} + B\,e^{-3t} + E\,cos\,t + F\,sin\,t - e^t]$ Therefore $\matbhf{y = A\,e^{3t} + 25\,B\,e^{-3t} + E\,(3\,cos\,t + 4\,sin\,t) + F\,(3\,sin\,t\;-\,4\,cos\,t) - e^t}$

Using Laplace transform

Laplace transform or the Laplace operator is a linear operator applied to functions and which is defined as $\mathhf{L}\left\{f(x)\right\}=\int_{0}^{\infty}e^{-st}f(t)dt$ where $s\in\mathhf{C}$

Example 1

Problem

Apply the Laplace Transform and find $X(s)$ and $Y(s)$ :

$2\frac{dx}{dt} + \frac{dy}{dt} + 90\,x = 45$

$\frac{dx}{dt} + 2\frac{dy}{dt} + 240\,y = 0$

Give that $x = y = 0$ at $t = 0$

Workings

The equations may be written as:

$2\mathhf{L\{\frac{dx}{dt}\}} + \mathhf{L\{\frac{dy}{dt}\}} + \mathhf{L\{90\,x\}} = \mathhf{L\{45\}}$

$\mathhf{L\{\frac{dx}{dt}\}} + 2\mathhf{L\{\frac{dy}{dt}\}} + \mathhf{L\{240\,y\}} = 0$ Hence $2(sX(s)-x(0)) + (sY(s)-y(0)) + 90\,X(s) = \frac{45}{s}$