Initial-value problems | Year 12

01 / LEARN

The idea, step by step

Two steps: integrate, then use the condition

A derivative gives the changing part of a function. A starting value fixes its vertical position. First find the general antiderivative with $C$ . Then substitute the stated point into the whole formula and solve for $C$ .

The point does not go into the derivative to find $C$ ; the derivative has no $C$ . Substitute into the function you found. Finally check both the derivative and the initial condition.

Initial does not always mean zero

A condition such as $y (2) = 7$ is just as useful as $y (0) = 7$ . Write the independent-variable value precisely before substituting. In a context, also state the time interval for which the model is intended.

Units help you interpret the function. Integrating litres per minute with respect to minutes produces a volume in litres. An initial volume must be added to the accumulated change, and that volume can never be replaced by an arbitrary $C$ in a final numerical prediction.

A common trap

Replacing $C$ by the initial value automatically.

Instead: The antiderivative may already be nonzero at the given input. Solve the complete initial-value equation.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

Given $y^{'} = 3 x^{2} - 2 x$ and $y (1) = 4$ , find $y$ .

Integrate: $y = x^{3} - x^{2} + C$ .
Use $4 = 1 - 1 + C$ , so $C = 4$ .
Check the derivative and the value at 1.

y = x^{3} - x^{2} + 4

Example 2

Worked example

A tank has $V^{'} (t) = 4 e^{t / 2}$ litres/minute and $V (0) = 10$ litres. Find $V (t)$ .

Integrate: $V = 8 e^{t / 2} + C$ .
Use $10 = 8 + C$ , so $C = 2$ .
At $t = 0$ the volume is 10, and its derivative is the stated inflow.

V (t) = 8 e^{t / 2} + 2

litres

Check your understanding

$f^{'} (x) = 2 x$ and $f (1) = 5$ . Find $f$ .

Show answer

$f (x) = x^{2} + 4$

$5 = 1 + C$ gives $C = 4$ .

03 / WORKSHEET

Your turn

6 questions · 18 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

$F^{'} (x) = 6 x$ and $F (0) = 3$ . Find $F$ .
3 marks
$y^{'} = 4 x^{3}$ and $y (1) = 7$ . Find $y$ .
3 marks
$f^{'} (x) = cos x$ and $f (π / 2) = 4$ . Find $f$ .
3 marks
$g^{'} (x) = 2 / x$ for $x > 0$ and $g (1) = 5$ . Find $g$ .
3 marks
A particle has acceleration $a (t) = 6 t$ m/s² and initial velocity 2 m/s. Find its velocity at $t = 3$ s.
3 marks
A quantity satisfies $N^{'} (t) = 3 e^{t}$ and $N (0) = 8$ . Find $N (ln 2)$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

$F^{'} (x) = 6 x$ and $F (0) = 3$ . Find $F$ .
3 marks
$F = 3 x^{2} + C$ ; at 0, $C = 3$ .
$F = 3 x^{2} + 3$
$y^{'} = 4 x^{3}$ and $y (1) = 7$ . Find $y$ .
3 marks
$7 = 1 + C$ gives $C = 6$ .
$y = x^{4} + 6$
$f^{'} (x) = cos x$ and $f (π / 2) = 4$ . Find $f$ .
3 marks
Integrate to $sin x + C$ ; $4 = 1 + C$ .
$f (x) = sin x + 3$
$g^{'} (x) = 2 / x$ for $x > 0$ and $g (1) = 5$ . Find $g$ .
3 marks
$ln 1 = 0$ , so $C = 5$ .
$g (x) = 2 ln x + 5$
A particle has acceleration $a (t) = 6 t$ m/s² and initial velocity 2 m/s. Find its velocity at $t = 3$ s.
3 marks
$v (t) = 3 t^{2} + C$ .
Use $v (0) = 2$ to obtain $C = 2$ , then $v (3) = 29$ .
$29$ m/s
A quantity satisfies $N^{'} (t) = 3 e^{t}$ and $N (0) = 8$ . Find $N (ln 2)$ .
3 marks
$N = 3 e^{t} + 5$ .
$N (ln 2) = 3 (2) + 5 = 11$ .
$11$

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods integration and logarithmic antiderivatives; original learning and assessment material, with by-hand methods and explicitly described numerical models.

Official curriculum source ↗

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