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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.${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}\sqrt{\mathbf{x}\mathbf{-}\mathbf{1}}{\mathbf{y}}{\mathbf{=}}{\mathbf{tanx}}\phantom{\rule{0ex}{0ex}}{\mathbf{y}}\left(\mathbf{5}\right){\mathbf{=}}{\mathbf{y}}{\mathbf{\text{'}}}\left(\mathbf{5}\right){\mathbf{=}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}\left(\mathbf{5}\right){\mathbf{=}}{\mathbf{1}}$

Hence, the largest interval for the existence of a unique solution to the given initial value problem is:

$\left(\frac{\mathbf{3}\mathbf{\pi }}{\mathbf{2}}\mathbf{,} \frac{\mathbf{5}\mathbf{\pi }}{\mathbf{2}}\right)$

See the step by step solution

## Step 1: Solve the given equation

The given equation is $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\sqrt{\mathbf{x}\mathbf{-}\mathbf{1}}\mathbf{y}\mathbf{=}\mathbf{tanx}.$

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, $\mathbf{p}\left(\mathbf{x}\right)\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,} \mathbf{r}\left(\mathbf{x}\right)\mathbf{=}\sqrt{\mathbf{x}\mathbf{-}\mathbf{1}}\mathbf{,} \mathbf{s}\left(\mathbf{x}\right)\mathbf{=}\mathbf{tanx}$

## Step 2:Check the continuity

$\mathbf{r}\left(\mathbf{x}\right)\mathbf{=}\sqrt{\mathbf{x}\mathbf{-}\mathbf{1}}$ is continuous for all $\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{<}\mathbf{0}$

That is r is continuous $\mathbf{x}\mathbf{<}1.$

And

$\mathbf{s}\left(\mathbf{x}\right)\mathbf{=}\mathbf{tanx}$ is continuous in $\left(\left(\mathbf{2}\mathbf{n}\mathbf{-}\mathbf{1}\right)\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{,} \left(\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\right)\frac{\mathbf{\pi }}{\mathbf{2}}\right)$

For n = 2,

$\mathbf{s}\left(\mathbf{x}\right)\mathbf{=}\mathbf{tanx}$ is continuous in $\left(\frac{\mathbf{3}\mathbf{\pi }}{\mathbf{2}}\mathbf{,} \frac{\mathbf{5}\mathbf{\pi }}{\mathbf{2}}\right)$

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

Now p and r continuous for all $\mathbf{x}\in \left(\mathbf{-}\infty \mathbf{,} \mathbf{1}\right).$

And s is continuous in $\left(\frac{\mathbf{3}\mathbf{\pi }}{\mathbf{2}}\mathbf{,} \frac{\mathbf{5}\mathbf{\pi }}{\mathbf{2}}\right)$

The initial condition is defined at ${\mathbf{x}}_{\mathbf{0}}\mathbf{=}\mathbf{5}$

And $\mathbf{5}\in \left(\frac{\mathbf{3}\mathbf{\pi }}{\mathbf{2}}\mathbf{,} \frac{\mathbf{5}\mathbf{\pi }}{\mathbf{2}}\right)$

Hence, the largest interval for the existence of a unique solution on (a, b) to the given initial value problem is: $\left(\frac{\mathbf{3}\mathbf{\pi }}{\mathbf{2}}\mathbf{,} \frac{\mathbf{5}\mathbf{\pi }}{\mathbf{2}}\right)$ ### Want to see more solutions like these? 