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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 259
Fundamentals Of Differential Equations And Boundary Value Problems

Fundamentals Of Differential Equations And Boundary Value Problems

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 1–7, convert the given initial value problem into an initial value problem for a system in normal form.

y4(t)-y3(t)+7y(t)=cost;y(0)=y'(0)=1,y''(0)=0,y3(0)=2

x'4(t)=x4-7x1+cost

See the step by step solution

Step by Step Solution

Step 1: express the equation in form of x

Here given

y4(t)-y3(t)+7y(t)=cost

Denote,

x1(t)=y(t)x2(t)=y'(t)x3(t)=y''(t)x4(t)=y'''(t)

The equation transforms as;

x'1(t)=x2(t)x'2(t)=y''(t)=x3(t)x'3(t)=y'''(t)=x4(t)x'4(t)=y4(t)x'4(t)=x4-7x1+cost

Step 2: the initial conditions

The given initial conditions are y(0)=y'(0)=1,y''(0)=0,y3(0)=2

Initial conditions after transformations;

x1(0)=1x2(0)=1x3(0)=0x4(0)=2

This is the required result.

Most popular questions for Math Textbooks

A Problem of Current Interest. The motion of an ironbar attracted by the magnetic field produced by a parallel current wire and restrained by springs (see Figure 5.17) is governed by the equation\(\frac{{{{\bf{d}}^{\bf{2}}}{\bf{x}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}{\bf{ = - x + }}\frac{{\bf{1}}}{{{\bf{\lambda - x}}}}\) for \({\bf{ - }}{{\bf{x}}_{\bf{o}}}{\bf{ < x < \lambda }}\)where the constants \({{\bf{x}}_{\bf{o}}}\) and \({\bf{\lambda }}\) are, respectively, the distances from the bar to the wall and to the wire when thebar is at equilibrium (rest) with the current off.

  1. Setting\({\bf{v = }}\frac{{{\bf{dx}}}}{{{\bf{dt}}}}\), convert the second-order equation to an equivalent first-order system.
  2. Solve the related phase plane differential equation for the system in part (a) and thereby show that its solutions are given by\({\bf{v = \pm }}\sqrt {{\bf{C - }}{{\bf{x}}^{\bf{2}}}{\bf{ - 2ln(\lambda - x)}}} \), where C is a constant.
  3. Show that if \({\bf{\lambda < 2}}\) there are no critical points in the xy-phase plane, whereas if \({\bf{\lambda > 2}}\) there are two critical points. For the latter case, determine these critical points.
  4. Physically, the case \({\bf{\lambda < 2}}\)corresponds to a current so high that the magnetic attraction completely overpowers the spring. To gain insight into this, use software to plot the phase plane diagrams for the system when \({\bf{\lambda = 1}}\) and when\({\bf{\lambda = 3}}\).
  5. From your phase plane diagrams in part (d), describe the possible motions of the bar when \({\bf{\lambda = 1}}\) and when\({\bf{\lambda = 3}}\), under various initial conditions.

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