Question

In Example 7 the largest interval I over which the explicit solutions$y={\varphi }_1\left(x\right)$ and $y={\varphi }_2\left(x\right)$are defined is the open interval (-5,5).Why can’t the interval I of definition be the closed interval [-5,5]?

Short answer

The function${\varphi }_1$is not differentiable at$5$and${\varphi }_2$is also not differentiable at$-5$or$5$.

Step-by-step solution

Definition of differential equation.

A differential equation is defined as the derivative function of one or more unknown functions (dependable variables) with respect to one or more undependable variables.

Apply the definition of differentially of a function.

Look at the case when ${\varphi }_1\left(x\right)=\sqrt{25-x^2}$.

Apply the definition of differentially of a function at the point $c=-5$:

$$\begin{gathered} \underset{h\to 0}{\lim }\frac{{\varphi }_1(C+h)-\varphi \left(C\right)}{h}=\underset{h\to 0}{\lim }\frac{{\varphi }_1(-5+h)-\varphi \left(-5\right)}{h} \\ =\underset{h\to 0}{\lim }\frac{\sqrt{25-{\left(-5+h\right)}^2}-0}{h} \end{gathered}$$

Simplify,

$\underset{h\to 0}{\lim }\frac{\sqrt{25-{(-5+h)}^2}-0}{h}$

Use L’Hospitals Rule.

Now we have two cases. The limit as$h\to 0^{-}:\underset{h\to 0}{\lim }\frac{\sqrt{25-{\left(-5+h\right)}^2}}{h}$

The limit does not exist because$25-(-5+h{)}^2<0$

Further, as$h\to 0^{+}:$the limit is:$\underset{h\to 0}{\lim }\frac{\sqrt{25-{\left(-5+h\right)}^2}}{h}=\infty$

Use L’Hospitals Rule to find the limit.

Therefore, the function${\varphi }_1$is not differentiable at$-5$.

Similarly, the function${\varphi }_1$is not differentiable at5androle="math" localid="1664172535830" ${\varphi }_2$is also not differentiable at-5or5.

Since${\varphi }_1$and${\varphi }_2$are not differentiable in-5or5, the interval cannot include them.

Therefore, Functions${\varphi }_1$and${\varphi }_2$are not differentiable in-5or5.