Q4E - 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.

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

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

Step 1:Solve the given equation

The given equation is $\mathbf{x}\left(\mathbf{x}\mathbf{+}\mathbf{1}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathbf{xy}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}.$

 

Divide both sides by x(x+1) in the above equation,

 

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathbf{x}\left(\frac{\mathbf{1}}{\mathbf{x}\left(\mathbf{x}\mathbf{+}\mathbf{1}\right)}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{+}\left(\frac{\mathbf{1}}{\mathbf{x}\left(\mathbf{x}\mathbf{+}\mathbf{1}\right)}\right)\mathbf{y}\mathbf{=}\mathbf{0}$

 

Simplify the above equation,

 

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\left(\frac{\mathbf{3}}{\mathbf{x}\mathbf{+}\mathbf{1}}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{+}\left(\frac{\mathbf{1}}{\mathbf{x}\left(\mathbf{x}\mathbf{+}\mathbf{1}\right)}\right)\mathbf{y}\mathbf{=}\mathbf{0}$

 

Compare with the standard form of a linear differential equation,

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

One has, $\mathbf{q}\left(\mathbf{x}\right)\mathbf{=}\left(\frac{\mathbf{-}\mathbf{3}}{\mathbf{x}\mathbf{+}\mathbf{1}}\right)\mathbf{,} \mathbf{r}\left(\mathbf{x}\right)\mathbf{=}\left(\frac{\mathbf{1}}{\mathbf{x}\left(\mathbf{x}\mathbf{+}\mathbf{1}\right)}\right)$

 

Step 2: Check the continuity

 $\mathbf{q}\left(\mathbf{x}\right)\mathbf{=}\left(\frac{\mathbf{-}\mathbf{3}}{\mathbf{x}\mathbf{+}\mathbf{1}}\right)$ is continuous for all $\mathbf{x}\ne \mathbf{-}\mathbf{1}$.

$\mathbf{r}\left(\mathbf{x}\right)\mathbf{=}\left(\frac{\mathbf{1}}{\mathbf{x}\left(\mathbf{x}\mathbf{+}\mathbf{1}\right)}\right)$ is continuous in $\mathbf{x}\ne \mathbf{0}\mathbf{,} \mathbf{-}\mathbf{1}$.

Step 3:The largest interval (a, b)

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

And the initial condition is defined at ${\mathbf{x}}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{-}\mathbf{1}}{\mathbf{2}}$

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

 

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