Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 1 / Section B / 5 / a

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii

i

Solution

Writing the differential one-form as

M d x + N d y

, its coefficients satisfy

M_{y} = e^{y} cos x = N_{x}

. It is an exact differential form, with potential

F = (1 + e^{y}) sin x

, because

d F = (1 + e^{y}) cos x d x + e^{y} sin x d y .

The implicit solution is therefore

(1 + e^{y}) sin x = C .

ii

Solution

On the real domain

y > 0

,

M = y (x^{2} + ln y)

and

N = x

have

M_{y} = x^{2} + ln y + 1

and

N_{x} = 1

, so this differential one-form is not exact. An integrating factor for a differential one-form is

μ = 1/ y

, which is nonzero on this domain. Multiplication gives

(x^{2} + ln y) d x + \frac{x}{y} d y = d (\frac{x ^{3}}{3} + x ln y) .

Hence the implicit solution is

\frac{x ^{3}}{3} + x ln y = C, y > 0.

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