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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.${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{-}}\sqrt{\mathbf{x}}{\mathbf{y}}{\mathbf{=}}{\mathbf{sinx}}\phantom{\rule{0ex}{0ex}}{\mathbf{y}}\left(\pi \right){\mathbf{=}}{\mathbf{0}}{\mathbf{,}}{\mathbf{ }}{\mathbf{ }}{\mathbf{y}}{\mathbf{\text{'}}}\left(\pi \right){\mathbf{=}}{\mathbf{11}}{\mathbf{,}}{\mathbf{ }}{\mathbf{ }}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}\left(\pi \right){\mathbf{=}}{\mathbf{3}}$

Hence, the largest interval $\left[\mathbf{0}\mathbf{,} \infty \right).$

See the step by step solution

## Step 1: Solve the given equation

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

Compare with the standard form of a linear differential equation,

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

Therefore,

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

## Step 2:Check the continuity

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

$\mathbf{s}\left(\mathbf{x}\right)\mathbf{=}\mathbf{sinx}$ is continuous in $R$.

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

Now overall r and s is continuous in $\forall \mathbf{x}\in \left[\mathbf{0}\mathbf{,} \infty \right).$

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

And $\mathbf{\pi }\in \left[\mathbf{0}\mathbf{,} \infty \right).$

Hence, the largest interval $\left[\mathbf{0}\mathbf{,} \infty \right).$ ### Want to see more solutions like these? 