Differential equations and initial conditions | Year 12

01 / LEARN

Understand the idea.

An equation about a rate

A differential equation specifies a relationship involving an unknown function and its derivatives. For $y^{'} = 2 x$ , integration gives the family $y = x^{2} + C$ : all vertical shifts share the same derivative because a constant differentiates to zero.

An initial condition identifies a point on the required solution, even when its input is not zero. Substitute both coordinates into the integrated family and determine the constant. The condition selects a function, rather than replacing the process of integration.

Verify both requirements

A solution must satisfy two separate requirements. Differentiate the proposed function and substitute into the original differential equation to establish equality on its domain. Then evaluate it at the initial input to check the stated output.

Record the interval where the formula and the original equation are defined. Logarithms or denominator zeros may restrict it. In a physical model, also distinguish the mathematical domain from the time or quantity range where the model's assumptions are meaningful.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example

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

Worked solution

1
Integrate the given derivative.
$y = \int (6 x - 2) d x = 3 x^{2} - 2 x + C$
Direct integration gives all functions with the prescribed derivative.
2
Use the initial condition.
$5 = 3 {(1)}^{2} - 2 (1) + C = 1 + C, C = 4$
The entire function must pass through the point (1,5).
3
State and verify the solution.
$\begin{array}{c} y = 3 x^{2} - 2 x + 4 \\ y^{'} = 6 x - 2, y (1) = 3 - 2 + 4 = 5 \end{array}$
The polynomial satisfies both requirements for every real x.

Answer

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

EXAMPLE 02

Worked example

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

Worked solution

1
Differentiate the proposed function.
$y = 4 e^{3 x}, y^{'} = 12 e^{3 x}$
The chain rule supplies the inner derivative 3.
2
Compare both sides of the differential equation.
$3 y = 3 (4 e^{3 x}) = 12 e^{3 x} = y^{'}$
The identity holds for every real x, rather than only at the initial point.
3
Check the initial condition.
$y (0) = 4 e^{0} = 4$
The proposed function also passes through the required initial point, so it is a solution to the initial-value problem.

Answer

Yes.

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$C = - 3$

Substituting the point gives $5 = 2^{3} + C = 8 + C$ , so $C = - 3$ . Differentiation alone cannot determine this constant.

Keep practising

6 worksheet questions · 12 marks

Open worksheet 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 $y^{'} = \frac{2}{x + 1}$ 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 $y^{'} = \frac{2}{x + 1}$ 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 ↗

In this lesson

Understand the idea.

An equation about a rate

Verify both requirements

See the method in action.

Worked example

Integrate the given derivative.

Use the initial condition.

State and verify the solution.

Worked example

Differentiate the proposed function.

Compare both sides of the differential equation.

Check the initial condition.

Try it yourself.

Check your understanding

Keep practising