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Q3E

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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 326

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.
$y''' - y'' + \sqrt{x - 1} y = tanx y (5) = y' (5) = y'' (5) = 1$

Hence, the largest interval for the existence of a unique solution to the given initial value problem is:

$(\frac{3 π}{2}, \frac{5 π}{2})$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Solve the given equation

The given equation is $y''' - y'' + \sqrt{x - 1} y = tanx .$

Compare with the standard form of a linear differential equation,

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

We have, $p (x) = - 1, r (x) = \sqrt{x - 1}, s (x) = tanx$

Step 2:Check the continuity

$r (x) = \sqrt{x - 1}$ is continuous for all $x - 1 < 0$

That is r is continuous $x < 1 .$

And

$s (x) = tanx$ is continuous in $((2 n - 1) \frac{π}{2}, (2 n + 1) \frac{π}{2})$

For n = 2,

$s (x) = tanx$ is continuous in $(\frac{3 π}{2}, \frac{5 π}{2})$

Step 3:The largest interval (a, b)

Now p and r continuous for all $x \in (- \infty, 1) .$

And s is continuous in $(\frac{3 π}{2}, \frac{5 π}{2})$

The initial condition is defined at $x_{0} = 5$

And $5 \in (\frac{3 π}{2}, \frac{5 π}{2})$

Hence, the largest interval for the existence of a unique solution on (a, b) to the given initial value problem is: $(\frac{3 π}{2}, \frac{5 π}{2})$

Most popular questions for Math Textbooks

find a general solution to the given equation.

$y''' + y'' - 2 y = x e^{x} + 1$

Given that the function $f (x) = x$ is a solution to $y''' - x^{2} y' + xy = 0$ , show that the substitution $y (x) = v (x) f (x) = v (x) x$ reduces this equation to, $xw'' + 3 w' - x^{3} w = 0$ where $w = v'$ .

Let $P (r_{1}) = a_{n} r_{1}^{n} + a_{n - 1} r_{1}^{n - 1} + . .. + a_{1} r_{1} + a_{0}$ be a polynomialwith real coefficients $a_{n}, . ..., a_{0}$ . Prove that if r₁ is azero of $P (r)$ , then so is its complex conjugate r₁. [Hint:Show that $\bar{P (r)} = P (\bar{r})$ , where the bar denotes complexconjugation.]

find a differential operator that annihilates the given function.

$e^{5 x}$

use the annihilator method to determinethe form of a particular solution for the given equation.

$y'' + 2 y' + 2 y = e^{- x} c o s x + x^{2}$

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