91Ó°ÊÓ

Here,$f\left(x,y\right)=\frac{4x}{y}$and$\frac{∂f}{∂y}=-\frac{4x}{y^2}$.

From Step 1,we find that $\frac{∂f}{∂y}$ is notcontinuousor even defined when $\mathrm{y}=0$. Consequently, there is no rectangle containing the point $({\mathrm{x}}_0=0)$, in which both $\mathrm{f}(\mathrm{x},\mathrm{y})$and $\frac{∂f}{∂y}$are continuous.

Because the hypotheses of Theorem 1 do not hold, we cannot use Theorem 1 to determine whether the given initial value problem does or does not have a unique solution. It turns out that this initial value problem has more than one solution.

Hence, Theorem 1 does not imply the existence of a unique solution that satisfies$\mathbf{y}\left(x_0\right)\mathbf{=}\mathbf{0}$.

Because y comes in the denominator, the function is not defined at $\mathrm{y}=0$.

Hence, when ${\mathit{x}}_{\mathbf{0}}\mathbf{≠}\mathbf{0}$ equation (13) can’t possibly have a solution in a neighbourhood of $\mathbf{x}\mathbf{=}{\mathbf{x}}_{\mathbf{0}}$that satisfies $\mathbf{y}\left(x_0\right)\mathbf{=}\mathbf{0}$.

In Figure 1.4, the implicit solutions for $\mathrm{c}=0,Â\pm 1,Â\pm 4$are sketched.

The curves are hyperbolas with common asymptotes $y=Â\pm 2x$. Notice that the implicit solution curves (with c arbitrary) fill the entire plane and are non-intersecting for $\mathrm{c}≠0$. For $\mathrm{c}=0$, the implicit solution gives rise to the two explicit solutions $\mathrm{y}=2\mathrm{x}$and $y=-2x$, both of which pass through the origin.

Thus, the two distinct solutions to (13) satisfying $\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$are $\mathit{y}\mathbf{=}\mathbf{2}\mathit{x}$and $\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{x}$.