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Solve the initial value problem y'=5y^2 xy^2

User Casablanca
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Final answer:

To solve for the product of the potential energy function and the square of the time-independent wave function, one must simply multiply the potential energy function U(x) with the square of the time-independent part of the wave function, y(x).

Step-by-step explanation:

The given potential energy function is U(x) = 0.5mw²x² where m is the mass of the particle, w is a constant, and x is the position along the x-axis. The task is to compute the product of this potential energy function with the square of a time-independent wave function, Y(x, t). Since the time-dependency factors out of the wave function, we can express this in terms of y(x), which is the time-independent part of the complete wave function Y(x, t).

Therefore, the product is Y(x, t) * U(x) * Y(x, t) = u(x) * y(x)², where u(x) = 0.5mw²x², and y(x) is the time-independent wave function. The result is a new function that describes the potential energy distribution over space based on the probability amplitude described by the square of the wave function y(x).

User Bennick
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