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

(a) For the initial value problem (12) of Example 9. Show that $ϕ_{1} (x) = 0$ and $ϕ_{2} (x) = {(x - 2)}^{3}$ are solutions. Hence, this initial value problem has multiple solutions. (See also Project G in Chapter 2.)
(b) Does the initial value problem $y' = 3 y^{\frac{2}{3}},$ $y (0) = 10^{- 7}$ have a unique solution in a neighbourhood of $x = 0$ ?

For the given initial value problem (12), $ϕ_{1} (x) = 0$ and $ϕ_{2} (x) = {(x - 2)}^{3}$ are solutions. Hence, this initial value problem has multiple solutions.
The initial value problem $y' = 3 y^{\frac{2}{3}}, y (0) = 10^{- 7}$ has a unique solution in a neighbourhood of $x = 0$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1(a): Showing that the given initial value problem (12) has multiple solutions

Clearly, $ϕ_{1} (x) = 0$ as y is a solution to the given initial value problem.

Now taking $y = ϕ_{2} (x) = {(x - 2)}^{3}$

$\begin{array}{l} \frac{dy}{dx} = 3 {(x - 2)}^{2} \\ \frac{dy}{dx} = 3 y^{\frac{2}{3}} \end{array}$

(Because, $y = {(x - 2)}^{3}$ , i.e., $y^{\frac{1}{3}} = (x - 2)$ )

which is identical to the given differential equation. So, $ϕ_{2} (x) = {(x - 2)}^{3}$ is a solution to the differential equation $\frac{dy}{dx} = 3 y^{\frac{2}{3}}$ .

Hence, this initial value problem has multiple solutions.

Step 2: Finding the partial derivative of the given relation with respect to y

Here, $f (x, y) = 3 y^{\frac{2}{3}}$

$\begin{array}{l} \frac{\partial f}{\partial y} = 2 y^{\frac{- 1}{3}} \\ \frac{\partial f}{\partial y} = \frac{2}{y^{\frac{1}{3}}} \end{array}$

which is continuous in any rectangle containing the point $(0, 10^{- 7})$ .

Step 3(b): Verify whether the given initial value problem y'=3y23, y(0)=10-7 has a unique solution in a neighbourhood of x=0 or not

Now from Step 1, we find that both of the functions $f (x, y)$ and $\frac{\partial f}{\partial y}$ are continuous in any rectangle containing the point $(0, 10^{- 7})$ , so the hypotheses of Theorem 1 are satisfied.

It then follows from the theorem that the given initial value problem has a unique solution in an interval about $x = 0$ of the form $(- δ, δ)$ , where $δ$ is some positive number.

Therefore, the initial value problem $y' = 3 y^{\frac{2}{3}}, y (0) = 10^{- 7}$ have a unique solution in a neighbourhood of $x = 0$ .

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Step by Step Solution

Step 1(a): Showing that the given initial value problem (12) has multiple solutions

Step 2: Finding the partial derivative of the given relation with respect to y

Step 3(b): Verify whether the given initial value problem y'=3y23, y(0)=10-7 has a unique solution in a neighbourhood of x=0 or not

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Step by Step Solution

Step 1(a): Showing that the given initial value problem (12) has multiple solutions

Step 2: Finding the partial derivative of the given relation with respect to y

Step 3(b): Verify whether the given initial value problem y'=3y23, y(0)=10-7 has a unique solution in a neighbourhood of x=0 or not

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