Two students are asked if the solution to the initial value problem. dy/dx=3(x^2

Q: Two students are asked if the solution to the initial value problem. dy/dx=3(x^2

y is negative for x>1. This problem arises since x=1 is point of discountinuity of the solution y=1/(1-x^3). So we can not say about thye nature of solution y in [0,2] inetrval. Function is increasing but it is negative also at x=2 due to discountinuity at x=1. So Riley is right.

ID: 2857211 • Letter: T

Question

Two students are asked if the solution to the initial value problem. dy/dx=3(x^2)(y^2), y(0)=1 can ever take a negative value for.x>0. Kasey argues that y can't be negative because 3(x^2)(y^2)>0, so the derivative is always non-negative, hence y is always increasing or flat, and, since we increase from y=1, y must always be positive. On the other hand, Riley solves the problem to get y=1/(1x^3) and computes y(2)=1/12^3 =1/7<0 hence y can be negative. Is Kasey or Riley (or both or neither) correct.

Explanation / Answer

y is negative for x>1. This problem arises since x=1 is point of discountinuity of the solution y=1/(1-x^3). So we can not say about thye nature of solution y in [0,2] inetrval. Function is increasing but it is negative also at x=2 due to discountinuity at x=1. So Riley is right.

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Two students are asked if the solution to the initial value problem. dy/dx=3(x^2

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