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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

Decide whether the statement made is True or False. The function $y (x) = - \frac{1}{3} (x + 1)$ is a solution to $\frac{dy}{dx} = \frac{y - 1}{x + 3}$ .

The statement is false.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Differentiating y(x)=-13(x+1) concerning x.

The given differential equation is,

$y (x) = - \frac{1}{3} (x + 1) \frac{dy}{dx} = - \frac{1}{3} (1 + 0) = - \frac{1}{3}$

Substituting the value of y(x) in dydx=y-1x+3.

Right hand side of the equation becomes,

$\frac{y - 1}{x + 3} = \frac{- \frac{1}{3} (x + 1) - 1}{x + 3} = \frac{- x - 1 - 3}{3 (x + 3)} = \frac{- x - 4}{3 (x + 3)}$

Left Hand Side of the equation becomes,

$\frac{dy}{dx} = - \frac{1}{3}$

LHS ≠ RHS

Hence, the statement is not true.

Most popular questions for Math Textbooks

Question: Let f(x) and g(x) be analytic at x₀. Determine whether the following statements are always true or sometimes false:

(a) 3f(x)+g(x) is analytic at x₀ .

(b) f(x)/g(x) is analytic at x₀ .

(c) f'(x) is analytic at x₀ .

(d) ${[f (x)]}^{3} - \int_{x_{0}}^{x} g (t) d t$ is analytic at x₀ .

Verify that the function $ϕ (x) = c_{1} e^{x} + c_{2} e^{- 2 x}$ is a solution to the linear equation $\frac{d^{2} y}{{dx}^{2}} + \frac{dy}{dx} - 2 y = 0$ for any choice of the constants $c_{1}$ and $c_{2}$ . Determine $c_{1}$ and $c_{2}$ so that each of the following initial conditions is satisfied.

(a) $y (0) = 2, y' (0) = 1$

(b) $y (1) = 1, y' (1) = 0$

In Problems 9–20, determine whether the equation is exact.

If it is, then solve it.

$(\frac{t}{y}) dy + (1 + \ln y) dt = 0$

In Problems 13-19, find at least the first four nonzero terms in a power series expansion of the solution to the given initial value problem.

$(x^{2} + 1) y'' - e^{x} y' + y = 0; y (0) = 1, y' (0) = 1$

Decide whether the statement made is True or False. The function $x (t) = t^{- 3} \sin t + 4 t^{- 3}$ is a solution to $t^{3} \frac{dx}{dt} = \cos t - 3 t^{2} x$ .

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