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Does the existence/uniqueness theorem guarantee that there is a solution to the differential equation y² - 1 through the given point?

1) True
2) False

User Erodewald
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1 Answer

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Final answer:

The existence/uniqueness theorem cannot be applied to the equation y² - 1 to guarantee a solution, as the equation does not fit the required form dy/dx = f(x, y), and crucial information is missing.

Step-by-step explanation:

The existence/uniqueness theorem concerns ordinary differential equations and states that if a function f(x, y) and its partial derivative ∂f/∂y are both continuous in a region around a point (x0, y0), then there is a unique solution y(x) to the differential equation dy/dx = f(x, y) that passes through the point (x0, y0). However, the differential equation given in the question, y² - 1, does not fit the form dy/dx = f(x, y), rather it seems to be an implicit form of a potential differential equation, but as written, it is not clear. Therefore, based on the incomplete information provided in the question, the existence/uniqueness theorem cannot be applied to guarantee a solution through a given point.

User Ilikeorangutans
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