Exercise · ZW-EXISTENCE-AND-UNIQUENESS-THEOREM-004 · ★☆☆ Basic
For \(y'=\dfrac{y}{x-3}\), where do \(f\) and \(\partial f/\partial y\) stop being continuous?
Choose one option.
A On the line \(y=3\) B On the line \(y=0\) C On the line \(x=0\) D On the line \(x=3\)
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Explanation Both \(f=y/(x-3)\) and \(\partial f/\partial y=1/(x-3)\) fail only where the denominator vanishes, that is, at \(x=3\).
The existence and uniqueness theorem guarantees that an initial value problem has a unique solution near the initial point, if the equation and its partial derivative with respect to the unknown are continuous there.
\[f,\ \frac{\partial f}{\partial y}\ \text{continuous near }(x_0,y_0)\ \Longrightarrow\ \text{a unique solution}\]
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