Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2003 / Paper 1 / 11 / a

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

i

Solution

For a nonzero solution, separation of variables gives

d y / y = k d x / (1 + x^{2})

, so

ln ∣ y ∣ = k arctan x + C

. Restoring the zero solution by allowing an arbitrary real multiplicative constant gives

y (x) = C e^{k a r c t a n x}, C \in R .

It is valid for all real numbers

x

, including the constant solutions when

k = 0

.

ii

Solution

Use integrating factor

e^{- x^{3} /3}

for this first-order linear differential equation. Then

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

. Integrating gives

y (x) = C e^{x^{3} /3} - 1.

This includes the constant particular solution

y = - 1

when

C = 0

.

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