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.

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  • D. On the line \(x=3\)

Explanation

Both \(f=y/(x-3)\) and \(\partial f/\partial y=1/(x-3)\) fail only where the denominator vanishes, that is, at \(x=3\).

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Review the concept: Existence and uniqueness theorem

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.

\[f,\ \frac{\partial f}{\partial y}\ \text{continuous near }(x_0,y_0)\ \Longrightarrow\ \text{a unique solution}\]

More practice problems: existence and uniqueness theorem

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