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Q4E

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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.
$x (x + 1) y''' - 3 xy' + y = 0 y (\frac{- 1}{2}) = 1, y' (\frac{- 1}{2}) = y'' (\frac{- 1}{2}) = 0$

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

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 $x (x + 1) y''' - 3 xy' + y = 0 .$

Divide both sides by x(x+1) in the above equation,

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

Simplify the above equation,

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

Compare with the standard form of a linear differential equation,

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

One has, $q (x) = (\frac{- 3}{x + 1}), r (x) = (\frac{1}{x (x + 1)})$

Step 2:Check the continuity

$q (x) = (\frac{- 3}{x + 1})$ is continuous for all $x \neq - 1$ .

$r (x) = (\frac{1}{x (x + 1)})$ is continuous in $x \neq 0, - 1$ .

Step 3:The largest interval (a, b)

Now q and r continuous for all $x \in (- \infty, - 1) \cup (- 1, 0) \cup (0, \infty)$

And the initial condition is defined at $x_{0} = \frac{- 1}{2}$

And $\frac{- 1}{2} \in (- 1, 0)$

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

Most popular questions for Math Textbooks

find a general solution to the given equation.

$y''' + 4 y'' + y' - 26 y = e^{- 3 x} s i n 2 x + x$

Find a general solution for the given

linear system using the elimination method of Section 5.2.

$\begin{array}{l} \frac{d^{3} x}{d t^{3}} - x + \frac{d y}{d t} + y = 0 \\ \frac{d x}{d t} - x + y = 0 \end{array}$

find a differential operator that annihilates the given function.

$x e^{3 x} c o s 5 x$

Determine the intervals for which Theorem guarantees the existence of a solution in that

(a) $y^{(4)} - (l n x) y^{''} + x y^{'} + 2 y = c o s 3 x$

(b) $(x^{2} - 1) y^{'''} + (s i n x) y^{''} + \sqrt{x + 4} y^{'} + e^{x} y = x^{2} + 3$

Determine whether the given functions are linearly dependent or linearly independent on the specified interval. Justify your decisions.

${e^{3 x}, e^{5 x}, e^{- x}}$ on $(- \infty, \infty)$

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