1A (Differential Equations)

The zero function is a solution. On an interval where

y \neq = 0

, put

z = \frac{y ^{'}}{y} .

(1)

Then

z^{'} = \frac{y y ^{''} - ( y ^{'} ) ^{2}}{y ^{2}} .

(2)

The differential equation becomes

x z^{'} = z .

(3)

Hence

z = C x

, and integration gives

lo g ∣ y ∣ = \frac{C}{2} x^{2} + C_{0} .

(4)

Thus all nonzero solutions are

y (x) = A exp (\frac{C}{2} x^{2}), A \neq = 0,

(5)

together with

y \equiv 0

b

 0  0

Solution

 0  0

Multiplying

y^{'} + P (x) y = Q (x) y^{n}

(1)

y^{- n}

and setting

z = y^{1 - n}

gives

\frac{1}{1 - n} z^{'} + P z = Q .

(2)

Therefore the Bernoulli differential equation becomes

z^{'} + (1 - n) P z = (1 - n) Q .

(3)

For the stated equation,

\overset{x}{˙} + 6 t^{2} x = 2 t e^{- t^{3}} x^{1/2} .

(4)

Set

z = x

. Then

\overset{z}{˙} + 3 t^{2} z = t e^{- t^{3}} .

(5)

The integrating factor is

e^{t^{3}}

, so

\frac{d}{d t} (e^{t^{3}} z) = t .

(6)

Hence

e^{t^{3}} x = \frac{t ^{2}}{2} + C .

(7)

The condition

x (1) = 4

gives

C = 2 e - \frac{1}{2}

. Consequently

x (t) = e^{- 2 t^{3}} (\frac{t ^{2}}{2} + 2 e - \frac{1}{2})^{2} .

(8)

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Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 2 / 1A (Differential Equations)

a

Solution

b

Solution

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