Question

(a) Explain why the interval of definition of the explicit solution $\mathbf{y}={\mathrm{\varphi }}_2\mathbf{\left(}\mathbf{x}\mathbf{\right)}$of the initial-value problem in Example 2 is the open interval$\mathbf{(}\mathbf{-}\mathbf{5}\mathbf{,}\mathbf{5}\mathbf{)}$.

(b) Can any solution of the differential equation cross the$\mathit{x}$-axis? Do you think that is${\mathbf{x}}^2\mathbf{+}{\mathbf{y}}^2\mathbf{=}\mathbf{1}$solution of the initial-value problem$\mathbf{dy}\mathbf{/}\mathbf{dx}\mathbf{=}\mathbf{-}\mathbf{x}\mathbf{/}\mathbf{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0}$?

Short answer

1. $y\text{'}$is not defined at these values, so the interval of definition is the open interval.
2. No solution can cross $x-$axis.

Step-by-step solution

Step 1:

An explicit solution is any solution that is given in the form$\mathbf{y}\mathbf{=}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$.

Implicit solution

(a)

In example 2, the implicit solution of the initial - value problem is $y=-\sqrt{25-x^2}$.

$y$is defined at $x=-5$and $x=5$, but $y\text{'}$is not defined at these values, so the interval of definition is the open interval $(-5,5)$.

Implicit solution(b)

At any point on the$x$-axis; $(y=0)$the derivative of $y\left(x\right)$is undefined. Hence, no solution curve can cross the $x$-axis.

While setting,$y\left(1\right)=0$ we notice that$-\frac{x}{y}$is undefined when $y=0$.

Thus, the initial - value problem has no solution.