Question: If the function $f(x,y)$ is continuous near the point $(a,b)$, then at least one solution of the differential equation $y' = f(x,y)$ exists on some open interval $i$ containing the point $x = a$ and moreover, that if in addition the partial derivative $\partial f / \partial y$ is continuous near $(a,b)$ then this solution is unique on some (perhaps smaller) interval $J$. Determine whether existence of at least one solution of the given initial value problem is thereby guaranteed and, if so, whether uniqueness of that solution is guaranteed.
$$y \frac{dy}{dx} = x - 17; \quad y(17) = 0$$
Select the correct choice below and fill in the answer box(es) to complete your choice.
(Type an ordered pair.)
A. The theorem implies the existence of at least one solution because $f(x,y)$ is continuous near the point $\square$.
This solution is unique because $\partial f / \partial y = \square$ is also continuous near that same point.
B. The theorem implies the existence of at least one solution because $f(x,y)$ is continuous near the point $\square$.
However, this solution is not necessarily unique because $\partial f / \partial y = \square$ is not continuous near that same point.
C. The theorem does not imply the existence of at least one solution because $f(x,y)$ is not continuous near the point $\square$.
1. **State the problem:**
We have the initial value problem (IVP):
$$y \frac{dy}{dx} = x - 17, \quad y(17) = 0$$
We want to determine if the existence and uniqueness theorem guarantees at least one solution and whether that solution is unique near the point $(a,b) = (17,0)$.
2. **Rewrite the differential equation in standard form:**
Divide both sides by $y$ (assuming $y \neq 0$):
$$\frac{dy}{dx} = \frac{x - 17}{y}$$
So,
$$f(x,y) = \frac{x - 17}{y}$$
3. **Check continuity of $f(x,y)$ near $(17,0)$:**
- The function $f(x,y)$ involves division by $y$.
- At the point $(17,0)$, $y=0$, so $f(x,y)$ is not defined and thus not continuous near $(17,0)$.
4. **Check the partial derivative $\partial f / \partial y$ near $(17,0)$:**
Calculate:
$$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( \frac{x - 17}{y} \right) = -\frac{x - 17}{y^2}$$
This is also not defined at $y=0$.
5. **Apply the existence and uniqueness theorem:**
- The theorem requires $f(x,y)$ to be continuous near $(a,b)$ for existence.
- It requires $\partial f / \partial y$ to be continuous near $(a,b)$ for uniqueness.
Since $f(x,y)$ is not continuous near $(17,0)$, the theorem does **not** guarantee existence of a solution.
**Final conclusion:**
The correct choice is:
C. The theorem does not imply the existence of at least one solution because $f(x,y)$ is not continuous near the point $(17,0)$.
**Answer:** $(17,0)$ for the point where continuity fails.