Q1E - 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.

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

Step-by-Step Solution

Answer

Thus, the largest interval is $\left(-\infty ,0\right)$.

Step 1 : Solve the given equation

The given equation is $\mathbf{xy}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{+}{\mathbf{e}}^{\mathbf{x}}\mathbf{y}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}.$

 

Divide both sides by x in the above equation,

 

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\left(\frac{\mathbf{1}}{\mathbf{x}}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{+}{\mathbf{e}}^{\mathbf{x}}\left(\frac{\mathbf{1}}{\mathbf{x}}\right)\mathbf{y}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\left(\frac{\mathbf{1}}{\mathbf{x}}\right)$

 

Compare with the standard form of a linear differential equation,

 

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

 

Therefore,

$\mathbf{q}\left(\mathbf{x}\right)\mathbf{=}\frac{\mathbf{-}\mathbf{3}}{\mathbf{x}}\mathbf{,} \mathbf{r}\left(\mathbf{x}\right)\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{x}}}{\mathbf{x}}\mathbf{,} \mathbf{s}\left(\mathbf{x}\right)\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}{\mathbf{x}}$

Step 2: Check the continuity

$\mathbf{q}\left(\mathbf{x}\right)\mathbf{=}\frac{\mathbf{-}\mathbf{3}}{\mathbf{x}}$ is continuous whenever $\mathbf{x}\ne \mathbf{0}.$

$\mathbf{r}\left(\mathbf{x}\right)\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{x}}}{\mathbf{x}}$ is continuous whenever $\mathbf{x}\ne \mathbf{0}.$ and

$\mathbf{s}\left(\mathbf{x}\right)\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}}{\mathbf{x}}$ is continuous whenever $\mathbf{x}\ne \mathbf{0}.$

Step 3 The largest interval (a, b)

Now overall q, r, and s are continuous in $\forall \mathbf{x}\in \left(\mathbf{-}\infty \mathbf{,} \mathbf{0}\right)\cup \left(\mathbf{0}\mathbf{,} \infty \right).$

The initial condition is defined as ${\mathbf{x}}_{\mathbf{0}}\mathbf{=}\mathbf{-}\mathbf{2}$ and $\mathbf{-}\mathbf{2}\in \left(\mathbf{-}\infty \mathbf{,} \mathbf{0}\right).$

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