Differential equations and initial conditions | Year 12

01 / LEARN

The idea, step by step

An equation about a rate

A differential equation relates an unknown function to one or more derivatives. In $d y / d x = 2 x$ , the function is unknown but its slope at each $x$ is specified. Integrating gives a family, $y = x^{2} + C$ , because every constant has derivative zero.

An initial condition specifies a point on the required solution, for example $y (1) = 5$ . It is not necessarily at time zero. Substitute that point into the family to determine its constant.

Verify both requirements

To check a proposed solution, differentiate it and substitute both the function and derivative into the differential equation. Then check the initial condition separately. Passing only one check is not enough.

Record the domain on which a solution is valid. A logarithmic solution may require a restricted interval, while a denominator can prevent continuation through a particular value. A rate model also has contextual limits: a negative population is usually not meaningful.

A common trap

Any antiderivative solves the given initial-value problem.

Instead: All family members may solve the differential equation, but only those satisfying the initial data solve the initial-value problem.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

Solve $d y / d x = 6 x - 2$ with $y (1) = 5$ .

Integrate: $y = 3 x^{2} - 2 x + C$ .
$5 = 3 - 2 + C$ , so $C = 4$ .
The derivative is $6 x - 2$ and $y (1) = 5$ .

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

Example 2

Worked example

Does $y = 4 e^{3 x}$ solve $y^{'} = 3 y$ and $y (0) = 4$ ?

Differentiate: $y^{'} = 12 e^{3 x}$ .
Also $3 y = 12 e^{3 x}$ , so the differential equation holds.
$y (0) = 4 e^{0} = 4$ , so the initial condition holds.

Yes.

Check your understanding

Find $C$ if $y = x^{3} + C$ and $y (2) = 5$ .

Show answer

$C = - 3$

$5 = 8 + C$ , giving $C = - 3$ .

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find the general solution of $y^{'} = 4 x^{3}$ .
2 marks
Solve $y^{'} = cos x$ with $y (0) = 2$ .
2 marks
Verify that $y = 3 e^{- 2 x}$ solves $y^{'} = - 2 y$ .
2 marks
Does $y = x^{2}$ solve $y^{'} = 2 y$ for all real $x$ ?
2 marks
Solve $y^{'} = 2 / (x + 1)$ with $y (0) = 3$ on $x > - 1$ .
2 marks
An amount follows $P^{'} = 0.04 P$ , $P (0) = 800$ . Verify $P = 800 e^{0.04 t}$ .
2 marks

Read the worked solutions

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

Find the general solution of $y^{'} = 4 x^{3}$ .
2 marks
Integrate $4 x^{3}$ to $x^{4}$ .
Include an arbitrary constant.
$y = x^{4} + C$
Solve $y^{'} = cos x$ with $y (0) = 2$ .
2 marks
$y = sin x + C$ .
$2 = 0 + C$ , so $C = 2$ .
$y = sin x + 2$
Verify that $y = 3 e^{- 2 x}$ solves $y^{'} = - 2 y$ .
2 marks
$y^{'} = - 6 e^{- 2 x}$ .
$- 2 y = - 6 e^{- 2 x}$ , equal to the derivative.
It does.
Does $y = x^{2}$ solve $y^{'} = 2 y$ for all real $x$ ?
2 marks
The derivative is $2 x$ , but $2 y = 2 x^{2}$ .
They are not equal for all $x$ ; equality at isolated points is insufficient.
No.
Solve $y^{'} = 2 / (x + 1)$ with $y (0) = 3$ on $x > - 1$ .
2 marks
Integrate to $2 ln (x + 1) + C$ on the stated domain.
At $x = 0$ , $C = 3$ .
$y = 2 ln (x + 1) + 3$
An amount follows $P^{'} = 0.04 P$ , $P (0) = 800$ . Verify $P = 800 e^{0.04 t}$ .
2 marks
$P^{'} = 32 e^{0.04 t} = 0.04 P$ .
$P (0) = 800$ .
It satisfies both requirements.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics topic 6.1–6.2, with the approved school-level rates, differential-equation and motion applications.

Official curriculum source ↗

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