Q5E - Fundamentals Of Differential Equations And Boundary Value Problems

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.

$$\begin{gathered} \mathbf{x}\sqrt{\mathbf{x}\mathbf{+}\mathbf{1}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{xy}\mathbf{=}\mathbf{0} \\ \mathbf{y}\left(\frac{\mathbf{1}}{\mathbf{2}}\right)\mathbf{=}\mathbf{y}\mathbf{\text{'}}\left(\frac{\mathbf{1}}{\mathbf{2}}\right)\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\frac{\mathbf{1}}{\mathbf{2}}\right)\mathbf{=}\mathbf{1} \end{gathered}$$

Step-by-Step Solution

Answer

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

Step 1: Solve the given equation

The given equation is $\mathbf{x}\sqrt{\mathbf{x}\mathbf{+}\mathbf{1}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{xy}\mathbf{=}\mathbf{0}.$

 

Both sides divide by $\mathbf{x}\sqrt{\mathbf{x}\mathbf{+}\mathbf{1}}$ in the above equation,

 

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

 

Simplify the above equation,

 

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

 

Compare with the standard form of a linear differential equation,

 

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

 

We have,

$\mathbf{q}\left(\mathbf{x}\right)\mathbf{=}\left(\frac{\mathbf{1}}{\mathbf{x}\sqrt{\mathbf{x}\mathbf{+}\mathbf{1}}}\right)\mathbf{,} \mathbf{r}\left(\mathbf{x}\right)\mathbf{=}\left(\frac{\mathbf{1}}{\sqrt{\mathbf{x}\mathbf{+}\mathbf{1}}}\right)$

Step 2: Check the continuity

$\mathbf{q}\left(\mathbf{x}\right)\mathbf{=}\left(\frac{\mathbf{1}}{\mathbf{x}\sqrt{\mathbf{x}\mathbf{+}\mathbf{1}}}\right)$ is continuous for all $\mathbf{x}\ne \mathbf{0}, \mathbf{-}\mathbf{1}$.

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

Step 3:The largest interval (a, b)

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

The initial condition is defined at ${\mathbf{x}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}.$

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

 

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