Question

Reread example 6 and then discuss why it is technically incorrect to say that the function (10) is a “solution” of the IVP on the interval

Short answer

Proved

Step-by-step solution

Definition of a Differential Equation

Differential Equation is defined as the equation that consists of the derivatives of one or more dependent functions in respect to one or more independent functions.

Proving the solution of IVP is incorrect  

From example 6:

$$\begin{gathered} y\left(x\right)=\left\{1-e^{-x}, 0⩽x⩽1 \\ (e-1)e^{-x},x>1\right. \end{gathered}$$

Then;

$$\begin{gathered} y\text{'}\left(x\right)=\left\{{}^{-x}, 0⩽x⩽1 \\ -(e-1)e^{-x},x>1\right. \end{gathered}$$

In the given function,

The left hand derivative at $x=1$is and the right hand derivative at$x=1$ is$1-e^{-1}$

Both are different and so.

Hence, at $x=1$, y is not differentiable and it is technically incorrect that the y solution of the IVP on the interval$[0,\infty )$