Question

In Problems 23-28, determine whether Theorem 1 implies that the given initial value problem has a unique solution.

$\mathbf{y}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{x}\mathbf{,}\mathbf{}\mathbf{y}\left(1\right)\mathbf{=}\mathbf{0}$

Short answer

The hypotheses of Theorem 1 are not satisfied.

The initial value problem does not have a unique solution.

Step-by-step solution

Finding the partial derivative of the given relation concerning y

Here,$\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\frac{\mathrm{x}}{\mathrm{y}}$and$\frac{\partial \mathrm{f}}{\partial \mathrm{y}}=-\frac{\mathrm{x}}{{\mathrm{y}}^2}$

Determining whether Theorem 1 implies the existence of a unique solution or not

From Step 1, we find that$\frac{\partial \mathrm{f}}{\partial y}$is not continuous or even defined when $\mathrm{y}=0$. Consequently, there is no rectangle containing the point $\left(1,0\right)$, in which both$\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)$ and$\frac{\partial \mathrm{f}}{\partial y}$ 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 implies that the given initial value problem does not have a unique solution.