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Maths › Differential Equations ›

Linear Simultaneous Equations

Linear Simultaneous equations using the D factor

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Overview
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Overview

Simultaneous Differential Equations occur in both mechanics and Physics. This analysis concentrates on linear equations with Constant Coefficients. For example find the General Solution to the equations:-

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

(1)

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

(2)

The equations may be written as:-

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

(3)

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

(4)

Eliminating y from equations (3) and (4)

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

(5)

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

(6)

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

(7)

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

(8)

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

(9)

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

(10)

From equation (3)

$y\;=\;(D^2\;-\;4D\;+\;4)[A\,e^{3t}\;+\;B\,e^{-3t}\;+\;E\,cos\,t\;+\;F\,sin\,t\;-\;e^t]$

(11)

$\matbhf{\therefore\;\;\;\;y\;=\;A\,e^{3t}\;+\;25\,B\,e^{-3t}\;+\;E\,(3\,cos\,t\;+\;4\,sin\,t)\;+\;F\,(3\,sin\,t\;-\,4\,cos\,t)\;-\;e^t}$

(12)

Example 1:

Solve the following Equations subject:-

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

(13)

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

(14)

Give that x = y = 0 at t = 0

Show also that y is negative for all values of t and that its minimum value is :-

$y\,(min.)\;=\;-\,\frac{1}{12}\left(\frac{2}{9} \right)^{\frac{2}{7}}$

(15)

The equations may be written as:-

$(2D\;+\;90)\;+\;Dy\;=\;45$

(16)

$D\;+\;(2D\;+\;240)y\;=\;0$

(17)

Multiply equation (16) by $\displaystyle (2D\;+\;240)$ Equation (17) by D and then subtract.

$\displaystyle \left[(2D\;+\;90)(2D\;+\;240)\;-\;D^2 \right]x\;=\;(2D\;+\;240)\,45$

$\therefore\;\;\;\;(D^2\;+\;220\,D\;+\;7200)x\;=\;3600$

(18)

$Or\;\;\;\;(D\;+\;180)(D\;+\;40)\,x\;=\;3600$

(19)

$\therefore\;\;\;\;x\;=\;A\,e^{-180\,t}\;+\;B\,e^{-40\,t}\;+\;\frac{1}{2}$

(20)

But when t = 0 x = 0 and so B = (A + 1/2)

From Equation (16)

$Dy\;=\;45\;-\;(2D\;+\;90)\,x$

(21)

$=\;45\;-\;(2D\;+\;90)\,\left[A\,e^{-180\,t}\;-\;(A\;+\;\frac{1}{2})\,e^{-40\,t}\;+\;\frac{1}{2} \right]$

(22)

$\therefore\;\;\;\;Dy\;=\;45\;-\;\left[-\;270\,A\,e^{-180\,t}\;-\;10(A\;+\;\frac{1}{2})\,e^{-40\,t}\;+\;45 \right]$

(23)

$Thus\;\;\;\;y\;=\;\frac{270}{180}\timesA\,e^{-180\,t}\;-\;\frac{10}{40}\;(A\;+\;\frac{1}{2})\,e^{-40\,t}\;+\;C$

(24)

Putting y = 0 when t = 0

$0\;=\;-\,\frac{3}{2}A\;-\;\frac{1}{4}(A\;+\;\frac{1}{2})\;+\;C$

(25)

$\therefore\;\;\;\;C\;=\;\frac{7}{4}A\;+\;\frac{1}{8}$

(26)

$Thus\;\;\;\;A\;=\;-\,\frac{1}{14}$

(27)

$\therefore\;\;\;\;C\;=\;\frac{7}{4}\,A\;+\;\frac{1}{8}$

(28)

The Solutions are;_

$x\;=\;A\,e^{-180\,t}\;-\;(A\;+\;\frac{1}{2})\,e^{-40\,t}\;+\;\frac{1}{2}$

(29)

$y\;=\;-\,\frac{3}{2}\,A\,e^{-180\,t}\;-\;\frac{1}{4}(A\;+\;\frac{1}{2})\,e^{-40\,t}\;+\;\frac{7}{4}A\;+\;\frac{1}{8}$

(30)

Substitute these values into equation (17)

$-\,180\,A\,e^{-180\,t}\;+\;40(A\;+\;\frac{1}{2})\,e^{-40\,t}\;+\;180\;A\,e^{-180\,t}\;-\;40(A\;+\;\frac{1}{2})\,e^{-40\,t}\;+\;240\;\left(\frac{7}{4}\,A\;+\;\frac{1}{8} \right)\;=\;0}$

(31)

$\therefore\;\;\;\;\frac{7}{4}\,A\;+\;\frac{1}{8}\;=\;0$

(32)

Substituting these values into equations (29) and (30), the required solutions are:-

$x\;=\;-\,\frac{1}{14}\,e^{-180\,t}\;-\;\frac{3}{7}\,e^{-40\,t}\,+\;\frac{1}{2}$

(33)

$y\;=\;\frac{3}{28}\,e^{-180\,t}\;-\;\frac{3}{28}\,e^{-40\,t}$

(34)

Hence y is negative for all values of t > 0

To find the minimum value of y Dy = 0

$Dy\;=\;\frac{3}{28}\left( -180\,e^{-180\,t}\;+\;40\,e^{-40\,t}\;\right)\;=\;0$

(35)

$i.e.\;\;\;\;e^{180\,t}\;=\;\frac{9}{2}$

(36)

$Or\;\;\;\;e^{-20}\;=\;\left( \frac{2}{9} \right)^{\frac{1}{7}}$

(37)

$\therefore\;\;\;\;y\;=\;\frac{3}{8}\left(\frac{2}{9} \right)^{\frac{9}{7}}\;=\;-\,\frac{3}{8}\left(\frac{2}{9}} \right)^{\frac{9}{7}}\;=\;-\,\frac{1}{12}\left(\frac{2}{9} \right)^{\frac{2}{7}}$

(38)