Q6E - 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.

$$\begin{gathered} \left({\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}{\mathbf{e}}^{\mathbf{x}}\mathbf{y}\mathbf{=}\mathbf{ln}\left(\mathbf{x}\right) \\ \mathbf{y}\left(\frac{\mathbf{3}}{\mathbf{4}}\right)\mathbf{=}\mathbf{1}\mathbf{,} \mathbf{y}\mathbf{\text{'}}\left(\frac{\mathbf{3}}{\mathbf{4}}\right)\mathbf{=}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\frac{\mathbf{3}}{\mathbf{4}}\right)\mathbf{=}\mathbf{0} \end{gathered}$$

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 $\left(\mathbf{0}\mathbf{,} \mathbf{1}\right).$

Step 1: Solve the given equation,

The given equation is $\left({\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}{\mathbf{e}}^{\mathbf{x}}\mathbf{y}\mathbf{=}\mathbf{ln}\left(\mathbf{x}\right).$

Both sides divide by ${\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}$ in the above equation,

 

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\left(\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}\right){\mathbf{e}}^{\mathbf{x}}\mathbf{y}\mathbf{=}\left(\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}\right)\mathbf{ln}\left(\mathbf{x}\right)$

 

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,

 

$\mathbf{r}\left(\mathbf{x}\right)\mathbf{=}\left(\frac{{\mathbf{e}}^{\mathbf{x}}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}\right)\mathbf{,} \mathbf{s}\left(\mathbf{x}\right)\mathbf{=}\left(\frac{\mathbf{ln}\left(\mathbf{x}\right)}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}\right)$

Step 2: Check the continuity

$\mathbf{r}\left(\mathbf{x}\right)\mathbf{=}\left(\frac{{\mathbf{e}}^{\mathbf{x}}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}\right)$ is continuous for all $\mathbf{x}\ne \mathbf{\pm }\mathbf{1}$.

$\mathbf{s}\left(\mathbf{x}\right)\mathbf{=}\left(\frac{\mathbf{ln}\left(\mathbf{x}\right)}{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}\right)$ is continuous in  $\mathbf{x}\ne \mathbf{\pm }\mathbf{1}, \mathbf{x}\mathbf{>}\mathbf{0}$.

Step 3:The largest interval (a, b)

Now combined on p, q, r, and s continuous for all $\mathbf{x}\in \left(\mathbf{0}\mathbf{,} \mathbf{1}\right)\cup \left(\mathbf{1}\mathbf{,} \infty \right).$

The initial condition is defined at  ${\mathbf{x}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{4}}.$

And $\frac{\mathbf{3}}{\mathbf{4}}\in \left(\mathbf{0}\mathbf{,} \mathbf{1}\right)$

 

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