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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 $\left(\mathbf{0}\mathbf{,} \mathbf{1}\right).$