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Q42Section B (Part B)42 of 120

Differential equations — uniqueness and the range of a solution

Consider the following statements:

(P) The initial value problem y′=f(y)y'=f(y), where f(y)=ycos⁡1yf(y)=y\cos\frac1y for y≠0y\neq0 and f(0)=0f(0)=0, with y(0)=0y(0)=0, has at most one solution.

(Q) If z(x)z(x) is the solution of the initial value problem z′=(1−z4)1001+z2z'=\dfrac{(1-z^4)^{100}}{1+z^2}, z(0)=0z(0)=0, then for each α∈(0,2)\alpha\in(0,2) there exists xα∈Rx_{\alpha}\in\mathbb{R} such that z(xα)=1−αz(x_{\alpha})=1-\alpha.

Then which of the following statements is true?

Options

Pick an option to check your answer.