Question

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.

$$\begin{gathered} \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\sqrt{\mathbf{x}}\mathbf{y}\mathbf{=}\mathbf{sinx} \\ \mathbf{y}\left(\pi \right)\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\text{'}}\left(\pi \right)\mathbf{=}\mathbf{11}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\pi \right)\mathbf{=}\mathbf{3} \end{gathered}$$

Short answer

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

Step-by-step solution

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