Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

Question: Coupled Equations. In analyzing coupled equations of the form
$\begin{array}{l} \frac{dy}{dt} = ax + by \\ \frac{dx}{dt} = αx + βy \end{array}$
where a, b, $α and y$ are constants, we may wish to determine the relationship between x and y rather than the individual solutions x(t), y(t). For this purpose, divide the first equation by the second to obtain
$\frac{dy}{dx} = \frac{ax + by}{αx + βy}$
This new equation is homogeneous, so we can solve it via the substitution $v = \frac{x}{y}$ . We refer to the solutions of (17) as integral curves. Determine the integral curves for the system
$\begin{array}{l} \frac{dy}{dt} = - 4 x - y \\ \frac{dx}{dt} = 2 x - y \end{array}$

${(y + x)}^{3} {(y - 4 x)}^{2} = C$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given equation is a homogeneous differential equation

The given equations are,

$\begin{array}{l} \frac{dy}{dt} = - 4 x - y \\ \frac{dx}{dt} = 2 x - y \end{array}$

Divide the first equation by the second equation,

$\frac{dy}{dx} = \frac{- 4 x - y}{2 x - y}$

Substitute the $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ in the above equation,

$v + x \frac{dv}{dx} = \frac{- 4 x - vx}{2 x - vx}$

Simplify the above equation,

$\begin{array}{l} x \frac{dv}{dx} = \frac{- 4 - v}{2 - v} - v \\ x \frac{dv}{dx} = \frac{- 4 - v - 2 v + v^{2}}{2 - v} \\ x \frac{dv}{dx} = \frac{v^{2} - 3 v - 4}{2 - v} \end{array}$

Cross multiplication on both sides in the above equation,

$\frac{2 - v}{v^{2} - 3 v - 4} dv = \frac{dx}{x}$

Step 2: Integrating

Taking integration both sides in the above equation

$\int \frac{2 - v}{v^{2} - 3 v - 4} dv = \int \frac{dx}{x}$

Solve the above equation,

$\begin{array}{l} \int \frac{2 - v}{v^{2} - 3 v - 4} dv = \int \frac{dx}{x} \\ \int \frac{2}{v^{2} - 3 v - 4} dv - \int \frac{v}{v^{2} - 3 v - 4} dv = \int \frac{dx}{x} \\ 2 (- \frac{1}{5} \ln |v + 1| + \frac{1}{5} \ln |v - 4|) - (\frac{1}{5} \ln |v + 1| + \frac{4}{5} \ln |v - 4|) = \ln |x| + \ln |c| \\ - \frac{3}{5} \ln |v + 1| - \frac{2}{5} \ln |v - 4| - \ln |x| = \ln |c| \end{array}$

Step 3: Finding the solution

Substitute the $v = \frac{y}{x}$ in the above equation,

$\begin{array}{l} - \frac{3}{5} \ln |\frac{y}{x} + 1| - \frac{2}{5} \ln |\frac{y}{x} - 4| - \ln |x| = \ln |c| \\ - 3 \ln |\frac{y}{x} + 1| - 2 \ln |\frac{y}{x} - 4| - 5 \ln |x| = 5 \ln |c| \\ 3 \ln |\frac{y}{x} + 1| + 2 \ln |\frac{y}{x} - 4| + 5 \ln |x| = - 5 \ln |c| \\ {(\frac{y}{x} + 1)}^{3} {(\frac{y}{x} - 4)}^{2} x^{5} = c^{- 5} \end{array}$

Simplify the above equation,

$\begin{matrix} \frac{{(y + x)}^{3} {(y - 4 x)}^{2} x^{5}}{x^{3} \cdot x^{2}} = c^{- 5} \\ {(y + x)}^{3} {(y - 4 x)}^{2} = c^{- 5} \end{matrix}$

Substitute $c^{- 5} = C$ in above equation,

${(y + x)}^{3} {(y - 4 x)}^{2} = C$

Hence the solution is ${(y + x)}^{3} {(y - 4 x)}^{2} = C$

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Step by Step Solution

Step 1: The given equation is a homogeneous differential equation

Step 2: Integrating

Step 3: Finding the solution

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Fundamentals Of Differential Equations And Boundary Value Problems

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