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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 1

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 1-6, identify the independent variable, dependent variable, and determine whether the equation is linear or nonlinear.
$3 r - c o s θ \frac{d r}{d θ} = s i n θ$

⦁ The Independent variable is $θ$ .

⦁ The dependent variable is r.

⦁ The equation is linear.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Identifying the dependent and independent variables.

Clearly, the independent variable is $θ$ while the dependent variable is r.

Determining whether the equation is linear or nonlinear

Since the dependent variable r and its derivatives appear in additive combinations of their first powers, the equation is linear.

Hence, the Independent variable is $θ$ .The dependent variable is r. And the equation is linear.

Most popular questions for Math Textbooks

Let c >0. Show that the function $ϕ (x) = {(c^{2} - x^{2})}^{- 1}$ is a solution to the initial value problem $\frac{dy}{dx} = 2 {xy}^{2}, y (0) = \frac{1}{c^{2}},$ on the interval $- c < x < c$ . Note that this solution becomes unbounded as x approaches $\pm c$ . Thus, the solution exists on the interval $(- δ, δ)$ with $δ = c$ , but not for larger $δ$ . This illustrates that in Theorem 1, the existence interval can be quite small (IFC is small) or quite large (if c is large). Notice also that there is no clue from the equation $\frac{dy}{dx} = 2 {xy}^{2}$ itself, or from the initial value, that the solution will “blow up” at $x = \pm c$ .

In problems 1-6, identify the independent variable, dependent variable, and determine whether the equation is linear or nonlinear.

$5 \frac{dx}{dt} + 5 x^{2} + 3 = 0$

In Problems 10–13, use the vectorized Euler method with h = 0.25 to find an approximation for the solution to the given initial value problem on the specified interval.

$(1 + t^{2}) y'' + y' - y = 0; y (0) = 1, y' (0) = - 1 on [0, 1]$

(a) Show that $y^{2} + x - 3 = 0$ is an implicit solution to $\frac{dy}{dx} = - \frac{1}{2 y}$ on the interval $(- \infty, 3)$ .

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