 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-1/11/b/solution

The initial value is positive, so set

z = y^{- 2}

. The chain rule gives

z^{'} = - 2 y^{- 3} y^{'} = - 2 (y^{- 2} - 1)

, making

z^{'} + 2 z = 2

. Its solution is

z = 1 + C e^{- 2 x}

. Since

y (0) = 1/2

z (0) = 4

and

C = 3

. Choosing the positive branch gives

y (x) = \frac{1}{1 + 3 e ^{- 2 x}}, x \in R .

It satisfies the initial condition and approaches the stable positive equilibrium one as

x \to \infty

 New to topics? Read the docs here!