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### 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.${\mathbf{x}}\left(\mathbf{x}\mathbf{+}\mathbf{1}\right){\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{3}}{\mathbf{xy}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{y}}{\mathbf{=}}{\mathbf{0}}\phantom{\rule{0ex}{0ex}}{\mathbf{y}}\left(\frac{\mathbf{-}\mathbf{1}}{\mathbf{2}}\right){\mathbf{=}}{\mathbf{1}}{\mathbf{,}}{ }{ }{\mathbf{y}}{\mathbf{\text{'}}}\left(\frac{\mathbf{-}\mathbf{1}}{\mathbf{2}}\right){\mathbf{=}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}\left(\frac{\mathbf{-}\mathbf{1}}{\mathbf{2}}\right){\mathbf{=}}{\mathbf{0}}$

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

See the step by step solution

## Step 1:Solve the given equation

The given equation is $\mathbf{x}\left(\mathbf{x}\mathbf{+}\mathbf{1}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathbf{xy}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}.$

Divide both sides by x(x+1) in the above equation,

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathbf{x}\left(\frac{\mathbf{1}}{\mathbf{x}\left(\mathbf{x}\mathbf{+}\mathbf{1}\right)}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{+}\left(\frac{\mathbf{1}}{\mathbf{x}\left(\mathbf{x}\mathbf{+}\mathbf{1}\right)}\right)\mathbf{y}\mathbf{=}\mathbf{0}$

Simplify the above equation,

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\left(\frac{\mathbf{3}}{\mathbf{x}\mathbf{+}\mathbf{1}}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{+}\left(\frac{\mathbf{1}}{\mathbf{x}\left(\mathbf{x}\mathbf{+}\mathbf{1}\right)}\right)\mathbf{y}\mathbf{=}\mathbf{0}$

Compare with the standard form of a linear differential equation,

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

One has, $\mathbf{q}\left(\mathbf{x}\right)\mathbf{=}\left(\frac{\mathbf{-}\mathbf{3}}{\mathbf{x}\mathbf{+}\mathbf{1}}\right)\mathbf{,} \mathbf{r}\left(\mathbf{x}\right)\mathbf{=}\left(\frac{\mathbf{1}}{\mathbf{x}\left(\mathbf{x}\mathbf{+}\mathbf{1}\right)}\right)$

## Step 2:Check the continuity

$\mathbf{q}\left(\mathbf{x}\right)\mathbf{=}\left(\frac{\mathbf{-}\mathbf{3}}{\mathbf{x}\mathbf{+}\mathbf{1}}\right)$ is continuous for all $\mathbf{x}\ne \mathbf{-}\mathbf{1}$.

$\mathbf{r}\left(\mathbf{x}\right)\mathbf{=}\left(\frac{\mathbf{1}}{\mathbf{x}\left(\mathbf{x}\mathbf{+}\mathbf{1}\right)}\right)$ is continuous in $\mathbf{x}\ne \mathbf{0}\mathbf{,} \mathbf{-}\mathbf{1}$.

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

Now q and r continuous for all $\mathbf{x}\in \left(\mathbf{-}\infty \mathbf{,} \mathbf{-}\mathbf{1}\right)\cup \left(\mathbf{-}\mathbf{1}\mathbf{,} \mathbf{0}\right)\cup \left(\mathbf{0}\mathbf{,} \infty \right)$

And the initial condition is defined at ${\mathbf{x}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{-}\mathbf{1}}{\mathbf{2}}$

And $\frac{\mathbf{-}\mathbf{1}}{\mathbf{2}}\in \left(\mathbf{-}\mathbf{1}\mathbf{,} \mathbf{0}\right)$

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