Past exam of the mathematics course of the University of Cambridge / 2005 / ia / Paper 2 / 1B (Differential Equations)

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- Solution 1B

Solution

An integrating factor is

μ (x) = exp (\int 3 x^{2} d x) = e^{x^{3}} .

Multiplication by this integrating factor turns the first-order linear differential equation into

(e^{x^{3}} y)^{'} = x^{2} e^{x^{3}}

. Integrating from zero to

x

gives

e^{x^{3}} y (x) - a = \frac{1}{3} (e^{x^{3}} - 1) .

Thus the initial value problem has the solution

y (x) = \frac{1}{3} + (a - \frac{1}{3}) e^{- x^{3}}, x \to + \infty lim y (x) = \frac{1}{3} .

Indeed, direct differentiation gives

y^{'} = - 3 x^{2} (a - 1/3) e^{- x^{3}}

, which verifies both the differential equation and the initial condition.

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