Question

In Example $7$we saw thatrole="math" localid="1664251885176" $y={\varphi }_1\left(x\right)=\sqrt{25-x^2}$ androle="math" localid="1664251899985" $y={\varphi }_2\left(x\right)=-\sqrt{25-x^2}$are solutions ofrole="math" localid="1664251917599" $dy/dx=-x/y$ on the intervalrole="math" localid="1664251930017" $(-5,5)$. Explain why the piecewise-defined functionrole="math" localid="1664251941449" $y=\left\{\begin{array}{l}\sqrt{25-x^2}-5<x<0\\ -\sqrt{25-x^2},0⩽x<5\end{array}\right.$ is not a solution of the differential equation on the interval$(-5,5)$.

Short answer

The piecewise-defined function is not a solution of the differential equation because of the discontinuity.

Step-by-step solution

Determine the left-hand limit of the function.

To check the continuity of the function at$x=0$, find the limits of the function.

The left-hand limit of the function is given by,

$$\begin{gathered} \underset{x\to 0^{-}}{\lim }=\underset{x\to 0^{-}}{\lim }\sqrt{25-0^2} \\ =\sqrt{25} \\ =5 \end{gathered}$$

Determine the right-limit of the function.

The right-hand limit of the function is given by,

$$\begin{gathered} \underset{x\to 0^{+}}{\lim }=\underset{x\to 0^{+}}{\lim }-\sqrt{25-0^2} \\ =-\sqrt{25} \\ =-5 \end{gathered}$$

Thus, the left-hand limit is not equal to the right-hand limit, so the function is not continuous at $x=0$. Then, the solution is not differential at $x=0$.

Hence, the piecewise-defined function is not a solution of the differential equation on the interval $(-5,5)$.