Exercise · ZW-EXISTENCE-AND-UNIQUENESS-THEOREM-003 · ★☆☆ Basic
The problem \(y'=\sqrt{y},\ y(0)=0\) has the solution \(y=0\). Which of these functions, for \(x\ge 0\), is another solution?
Choose one option.
Show answer
- A. \(y=x^2/4\)
Explanation
If \(y=x^2/4\), then \(y'=x/2=\sqrt{x^2/4}\) and \(y(0)=0\). There are two solutions because \(\partial f/\partial y=1/(2\sqrt{y})\) is not continuous at \(y=0\).
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
- ZW-EXISTENCE-AND-UNIQUENESS-THEOREM-001 For y'=xy^2, what is ∂ f/∂ y?
- ZW-EXISTENCE-AND-UNIQUENESS-THEOREM-002 For which initial value problem does the theorem guarantee a unique solution?
- ZW-EXISTENCE-AND-UNIQUENESS-THEOREM-004 For y'=(y)/(x-3), where do f and ∂ f/∂ y stop being continuous?
- ZW-EXISTENCE-AND-UNIQUENESS-THEOREM-005 If f and ∂ f/∂ y are continuous near (x_0,y_0), the theorem ensures that the initial value problem:
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- ZW-EXISTENCE-AND-UNIQUENESS-THEOREM-007 The solution of y'=y^2,\ y(0)=1 is y=(1)/(1-x). What is the largest interval containing 0 on which…
- ZW-EXISTENCE-AND-UNIQUENESS-THEOREM-008 For y'=3y^2/3, the set of initial points (x_0,y_0) where the theorem guarantees a unique solution i…
- ZW-EXISTENCE-AND-UNIQUENESS-THEOREM-009 The hypotheses of the theorem fail for an initial value problem. What can be concluded?
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