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

Determine the largest interval (a, b) for which Theorem 1 guarantees the existence of a unique solution on (a, b) to the given initial value problem.
$(x^{2} - 1) y''' + e^{x} y = \ln (x) y (\frac{3}{4}) = 1, y' (\frac{3}{4}) = y'' (\frac{3}{4}) = 0$

Hence, the largest interval for the existence of a unique solution on (a, b) to the given initial value problem is $(0, 1) .$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Solve the given equation,

The given equation is $(x^{2} - 1) y''' + e^{x} y = \ln (x) .$

Both sides divide by $x^{2} - 1$ in the above equation,

$y''' + (\frac{1}{x^{2} - 1}) e^{x} y = (\frac{1}{x^{2} - 1}) \ln (x)$

Compare with the standard form of a linear differential equation,

$y''' + p (x) y'' + q (x) y' + r (x) y = s (x)$

We have,

$r (x) = (\frac{e^{x}}{x^{2} - 1}), s (x) = (\frac{\ln (x)}{x^{2} - 1})$

Step 2:Check the continuity

$r (x) = (\frac{e^{x}}{x^{2} - 1})$ is continuous for all $x \neq \pm 1$ .

$s (x) = (\frac{\ln (x)}{x^{2} - 1})$ is continuous in $x \neq \pm 1, x > 0$ .

Step 3:The largest interval (a, b)

Now combined on p, q, r, and s continuous for all $x \in (0, 1) \cup (1, \infty) .$

The initial condition is defined at $x_{0} = \frac{3}{4} .$

And $\frac{3}{4} \in (0, 1)$

Hence, the largest interval for the existence of a unique solution on (a, b) to the given initial value problem is $(0, 1) .$

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Determine the largest interval (a, b) for which Theorem 1 guarantees the existence of a unique solution on (a, b) to the given initial value problem.
$(x^{2} - 1) y''' + e^{x} y = \ln (x) y (\frac{3}{4}) = 1, y' (\frac{3}{4}) = y'' (\frac{3}{4}) = 0$

Step by Step Solution

Step 1: Solve the given equation,

Step 2:Check the continuity

Step 3:The largest interval (a, b)

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Determine the largest interval (a, b) for which Theorem 1 guarantees the existence of a unique solution on (a, b) to the given initial value problem.x2-1y'''+exy=lnxy34=1, y'34=y''34=0

Step by Step Solution

Step 1: Solve the given equation,

Step 2:Check the continuity

Step 3:The largest interval (a, b)

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Determine the largest interval (a, b) for which Theorem 1 guarantees the existence of a unique solution on (a, b) to the given initial value problem.
$(x^{2} - 1) y''' + e^{x} y = \ln (x) y (\frac{3}{4}) = 1, y' (\frac{3}{4}) = y'' (\frac{3}{4}) = 0$