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Q27 E

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Found in: Page 14

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# In Problems 23-28, determine whether Theorem 1 implies that the given initial value problem has a unique solution.${\mathbf{y}}\frac{\mathbf{dy}}{\mathbf{dx}}{\mathbf{=}}{\mathbf{x}}{\mathbf{,}}{\mathbf{}}{\mathbf{y}}\left(1\right){\mathbf{=}}{\mathbf{0}}$

The hypotheses of Theorem 1 are not satisfied.

The initial value problem does not have a unique solution.

See the step by step solution

## Step 1: Finding the partial derivative of the given relation concerning y

Here, $\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\frac{\mathrm{x}}{\mathrm{y}}$ and $\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=-\frac{\mathrm{x}}{{\mathrm{y}}^{2}}$

## Step 2: Determining whether Theorem 1 implies the existence of a unique solution or not

From Step 1, we find that $\frac{\partial \mathrm{f}}{\partial y}$ is not continuous or even defined when $\mathrm{y}=0$. Consequently, there is no rectangle containing the point $\left(1,0\right)$, in which both $\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)$ and $\frac{\partial \mathrm{f}}{\partial y}$ are continuous. Because the hypotheses of Theorem 1 do not hold, we cannot use Theorem 1 to determine whether the given initial value problem does or does not have a unique solution. It turns out that this initial value problem has more than one solution.

Hence, Theorem 1 implies that the given initial value problem does not have a unique solution.

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