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Expert-verified Found in: Page 1 ### 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 # (a) For the initial value problem (12) of Example 9. Show that ${{\mathbf{\varphi }}}_{{\mathbf{1}}}\left(x\right){\mathbf{=}}{\mathbf{0}}$ and ${{\mathbf{\varphi }}}_{{\mathbf{2}}}\left(x\right){\mathbf{=}}{\left(x-2\right)}^{{\mathbf{3}}}$ are solutions. Hence, this initial value problem has multiple solutions. (See also Project G in Chapter 2.)(b) Does the initial value problem ${\mathbit{y}}{\mathbf{\text{'}}}{\mathbf{=}}{\mathbf{3}}{{\mathbit{y}}}^{\frac{\mathbf{2}}{\mathbf{3}}}{\mathbf{,}}$ ${\mathbf{y}}\left(0\right){\mathbf{=}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{7}}$ have a unique solution in a neighbourhood of ${\mathbf{x}}{\mathbf{=}}{\mathbf{0}}$?

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

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

Clearly, ${\varphi }_{1}\left(\mathrm{x}\right)=0$ as y is a solution to the given initial value problem.

Now taking $\mathrm{y}={\mathrm{\varphi }}_{2}\left(\mathrm{x}\right)={\left(\mathrm{x}-2\right)}^{3}$

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

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

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

Hence, this initial value problem has multiple solutions.

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

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

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

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

## 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 $\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)$ and $\frac{\partial f}{\partial y}$ are continuous in any rectangle containing the point $\left(0,{10}^{-7}\right)$, 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 $\left(-\delta ,\delta \right)$, where $\delta$ is some positive number.

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

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