• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q6E

Expert-verified Found in: Page 326 ### 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 # 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.$\left({\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\right){\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{{\mathbf{e}}}^{{\mathbf{x}}}{\mathbf{y}}{\mathbf{=}}{\mathbf{ln}}\left(\mathbf{x}\right)\phantom{\rule{0ex}{0ex}}{\mathbf{y}}\left(\frac{\mathbf{3}}{\mathbf{4}}\right){\mathbf{=}}{\mathbf{1}}{\mathbf{,}}{ }{ }{\mathbf{y}}{\mathbf{\text{'}}}\left(\frac{\mathbf{3}}{\mathbf{4}}\right){\mathbf{=}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}\left(\frac{\mathbf{3}}{\mathbf{4}}\right){\mathbf{=}}{\mathbf{0}}$

Hence, the largest interval for the existence of a unique solution on (a, b) to the given initial value problem is $\left(\mathbf{0}\mathbf{,} \mathbf{1}\right).$

See the step by step solution

## Step 1: Solve the given equation,

The given equation is $\left({\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}{\mathbf{e}}^{\mathbf{x}}\mathbf{y}\mathbf{=}\mathbf{ln}\left(\mathbf{x}\right).$

Both sides divide by ${\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}$ in the above equation,

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\left(\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}\right){\mathbf{e}}^{\mathbf{x}}\mathbf{y}\mathbf{=}\left(\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}\right)\mathbf{ln}\left(\mathbf{x}\right)$

Compare with the standard form of a linear differential equation,

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{p}\left(\mathbf{x}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{q}\left(\mathbf{x}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{r}\left(\mathbf{x}\right)\mathbf{y}\mathbf{=}\mathbf{s}\left(\mathbf{x}\right)$

We have,

$\mathbf{r}\left(\mathbf{x}\right)\mathbf{=}\left(\frac{{\mathbf{e}}^{\mathbf{x}}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}\right)\mathbf{,} \mathbf{s}\left(\mathbf{x}\right)\mathbf{=}\left(\frac{\mathbf{ln}\left(\mathbf{x}\right)}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}\right)$

## Step 2:Check the continuity

$\mathbf{r}\left(\mathbf{x}\right)\mathbf{=}\left(\frac{{\mathbf{e}}^{\mathbf{x}}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}\right)$ is continuous for all $\mathbf{x}\ne \mathbf{±}\mathbf{1}$.

$\mathbf{s}\left(\mathbf{x}\right)\mathbf{=}\left(\frac{\mathbf{ln}\left(\mathbf{x}\right)}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}\right)$ is continuous in $\mathbf{x}\ne \mathbf{±}\mathbf{1}, \mathbf{x}\mathbf{>}\mathbf{0}$.

## Step 3:The largest interval (a, b)

Now combined on p, q, r, and s continuous for all $\mathbf{x}\in \left(\mathbf{0}\mathbf{,} \mathbf{1}\right)\cup \left(\mathbf{1}\mathbf{,} \infty \right).$

The initial condition is defined at ${\mathbf{x}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{4}}.$

And $\frac{\mathbf{3}}{\mathbf{4}}\in \left(\mathbf{0}\mathbf{,} \mathbf{1}\right)$

Hence, the largest interval for the existence of a unique solution on (a, b) to the given initial value problem is $\left(\mathbf{0}\mathbf{,} \mathbf{1}\right).$ ### Want to see more solutions like these? 