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Q15E

Expert-verified

Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 271

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

In Problems 15–18, find all critical points for the given system. Then use a software package to sketch the direction field in the phase plane and from this description the stability of the critical points (i.e., compare with Figure 5.12).

This is unstable saddle point is (-2,1).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Find critical points

Here the system is;

$\frac{d x}{d t} = 2 x + y + 3 \frac{d y}{d t} = - 3 x - 2 y - 4$

For critical points equate the system equal to zero.

$2 x + y + 3 = 0 - 3 x - 2 y - 4 = 0$

Solve for x and y by eliminating the method.

The values of x=-2 and y=1.

So, this is the unstable saddle point is (-2,1).

Step 2: Sketch

This is the required result.

Most popular questions for Math Textbooks

The motion of a pair of identical pendulums coupled with a spring is modeled by the system

${mx}_{1}^{}'' = - \frac{mg}{l} x_{1} - k (x_{1} - x_{2}) {, mx}_{2}^{}'' = - \frac{mg}{l} x_{2} + k (x_{1} - x_{2})$

for small displacements (see Figure 5.36). Determine the two normal frequencies for the system.

In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$(D^{2} - 1) [u] + 5 v = e^{t}, 2 u + (D^{2} + 2) [v] = 0$

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$(m_{1} + m_{2}) l_{1}^{2} θ_{1}^{}'' + m_{2} l_{1} l_{2} θ_{2}^{}'' + (m_{1} + m_{2}) l_{1} {gθ}_{1} = 0 m_{2} l_{2}^{2} θ_{2}^{}'' + m_{2} l_{1} l_{2} θ_{1}^{}'' + m_{2} l_{2} {gθ}_{2} = 0$

When $θ_{1}$ and $θ_{2}$ are small angles. Solve the system when

$m_{1} = 3 kg, m_{2} = 2 kg, l_{1} = l_{2} = 5 m, θ_{1} (0) = π / 6, θ_{2} {(0) = θ}_{1}^{} {' (0) = θ}_{2}^{}' (0) = 0$ .

In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$\frac{dw}{dt} = 5 w + 2 z + 5 t, \frac{dz}{dt} = 3 w + 4 z + 17 t$

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$x' + 2 y = 0, x' - y' = 0$

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