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Linear Simultaneous Equations

Linear simultaneous differential equations
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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:

Example - First example
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:

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:

Example - Apply Laplace
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}

(sX(s)-x(0)) + 2(sY(s)-y(0)) + 240\,Y(s) = 0
Therefore Then we multiply the equation (3) with (s+120) : We subtract equation (4) from equation (5) :
((2s+90)(s+120)-s)X(s)=(s+120)\frac{45}{s}
Solution
Therefore
X(s)=\frac{45(s+120)}{s(2s^2+329s+10800)}
From equation (4) we get :
Y(s)=\frac{2s+240}{s}X(s)