Pure Mathematics Practice: Differential Equations — Answers

Question 1

Separating variables gives

\frac{1}{1 + y} d y = 3 x^{2} d x .

Since the initial value has $1 + y > 0$ , the relevant solution branch satisfies

ln (1 + y) = x^{3} + C .

Using $y (0) = 2$ gives $C = ln 3$ . Hence

y = 3 e^{x^{3}} - 1.

(a) Separation and integration give

ln N = - 0.04 t + C .

Since $N (0) = 500$ ,

N = 500 e^{- 0.04 t} .

(b)

300 = 500 e^{- 0.04 t} ⟹ t = - \frac{ln 0.6}{0.04} = 12.7706 \dots .

Therefore

t = 12.8

to 3 significant figures.

For $0 < y < 4$ ,

\frac{1}{y ( 4 - y )} = \frac{1}{4} (\frac{1}{y} + \frac{1}{4 - y}) .

After separation and integration,

\frac{1}{4} ln (\frac{y}{4 - y}) = x + C .

Thus

\frac{y}{4 - y} = A e^{4 x} .

Using $y (0) = 1$ gives $A = 1/3$ , so

y = \frac{4 e ^{4 x}}{3 + e ^{4 x}} .

This solution remains between the equilibrium values 0 and 4, as required by the initial condition.

While $h > 0$ ,

\frac{d h}{h} = - k d t .

Therefore

2 h = - k t + C .

Using $h (0) = 16$ gives $C = 8$ , so

2 h = 8 - k t .

Using $h (10) = 9$ gives

6 = 8 - 10 k,

and hence $k = 0.2$ . The depth is therefore

h = (4 - 0.1 t)^{2}

while the tank contains liquid. It becomes empty when $4 - 0.1 t = 0$ , giving

t = 40 s .

The physically valid interval for this expression is

0 \leq t \leq 40.

For $t > 40$ , the physical model should use $h = 0$ rather than continuing the squared algebraic expression.