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.

$$\begin{gathered} \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\sqrt{\mathbf{x}\mathbf{-}\mathbf{1}}\mathbf{y}\mathbf{=}\mathbf{tanx} \\ \mathbf{y}\left(\mathbf{5}\right)\mathbf{=}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{5}\right)\mathbf{=}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{5}\right)\mathbf{=}\mathbf{1} \end{gathered}$$

Short answer

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

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

Step-by-step solution

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