Suggested languages for you:

Americas

Europe

Q29 E

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 ${\mathbf{x}}{\mathbf{=}}{\mathbf{0}}$.

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.