What are first order linear equations?

A first order linear equation has the standard form $(dy)/(dx) + P(x)\, y = Q(x).$

TASSpecialist MathematicsSyllabus dot point

How do we solve equations that relate a function to its own rate of change?

Solve separable and linear first order differential equations and apply them to models.

Solving separable and first order linear differential equations using the integrating factor, with applications to growth and mixing, worked examples and pitfalls, for TCE Mathematics Specialised.

Generated by Claude Opus 4.79 min answerUpdated 2026-05-29

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What this dot point is asking

A differential equation involves an unknown function and its derivatives. A solution is the function itself. The general solution contains an arbitrary constant; a particular solution uses an initial condition to fix that constant.

Separable equations

An equation is separable if it can be written $\tfrac{dy}{dx} = f(x)g(y)$ . Rearrange so each variable is with its own differential, then integrate both sides:

\int \frac{1}{g(y)}\, dy = \int f(x)\, dx.

A key application is exponential growth and decay, $\tfrac{dy}{dt} = ky$ , which separates to $y = y_0 e^{kt}$ .

First order linear equations

A first order linear equation has the standard form

\frac{dy}{dx} + P(x)\, y = Q(x).

A first order linear equation

Worked example

Solve $\dfrac{dy}{dx} + \dfrac{1}{x}\, y = x$ for $x > 0$ , with $y(1) = 2$ .

Here $P(x) = \tfrac{1}{x}$ and $Q(x) = x$ . The integrating factor is

\mu = e^{\int \frac{1}{x}\, dx} = e^{\ln x} = x.

Multiply the equation by

\mu = x

x\frac{dy}{dx} + y = x^2.

The left side is

\tfrac{d}{dx}(xy)

, so

\frac{d}{dx}(xy) = x^2 \quad\Longrightarrow\quad xy = \frac{x^3}{3} + C.

Apply

y(1) = 2

1\cdot 2 = \tfrac{1}{3} + C

, so

C = \tfrac{5}{3}

. Therefore

y = \frac{x^2}{3} + \frac{5}{3x}.

Check at

x = 1

\tfrac{1}{3} + \tfrac{5}{3} = 2

. Correct.

Linear equations model mixing problems, where the rate of change of dissolved substance equals rate in minus rate out, and Newton's law of cooling.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 TASC7 marksFind the general solution of the differential equation (1 + x^2) dy/dx + (1 - y^2) = 0.

Show worked answer →

This equation is separable. Rearrange: (1 + x^2) dy/dx = -(1 - y^2) = y^2 - 1, so dy/(1 - y^2) = -dx/(1 + x^2).

Integrate the left side using partial fractions: 1/(1 - y^2) = 1/2 [1/(1 - y) + 1/(1 + y)], whose integral is (1/2) ln|(1 + y)/(1 - y)|.

Integrate the right side: -arctan(x).

So the general solution is (1/2) ln|(1 + y)/(1 - y)| = -arctan(x) + C, where C is an arbitrary constant. Equivalent forms (for example writing y = tanh(arctan(x) ... ) or exponentiating) are acceptable. Markers reward separating the variables correctly, the partial-fraction integral, and including the constant of integration.

2024 TASC5 marksSolve the differential equation x^2 dy/dx - y = 1 subject to y(1) = 1.

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Divide by x^2 to put it in linear form: dy/dx - (1/x^2) y = 1/x^2.

The integrating factor is e^(integral of -1/x^2 dx) = e^(1/x).

Multiply through: d/dx (y e^(1/x)) = (1/x^2) e^(1/x). Integrate the right side with the substitution u = 1/x, du = -1/x^2 dx, giving integral of -e^u du = -e^(1/x). So y e^(1/x) = -e^(1/x) + C, i.e. y = -1 + C e^(-1/x).

Apply y(1) = 1: 1 = -1 + C e^(-1), so C = 2e. The particular solution is y = -1 + 2e . e^(-1/x) = -1 + 2 e^((x - 1)/x). (Check: at x = 1, y = -1 + 2 = 1.)

2022 TASC7 marksA hot kettle is in a room at temperature theta degrees Celsius. After t minutes its temperature T(t) satisfies dT/dt = -k(T - theta), k > 0. After 5 minutes T = 69, after 10 minutes T = 51, after 15 minutes T = 39. Find the initial temperature of the kettle and the room temperature.

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The solution of Newton's law of cooling is T = theta + (T0 - theta) e^(-kt), where T0 is the initial temperature.

So T - theta decays geometrically: the ratios over equal time steps are equal. Hence (51 - theta)/(69 - theta) = (39 - theta)/(51 - theta).

Cross-multiply: (51 - theta)^2 = (69 - theta)(39 - theta). Expanding, 2601 - 102 theta = 2691 - 108 theta, so 6 theta = 90 and theta = 15. The room temperature is 15 degrees C.

The common ratio over 5 minutes is e^(-5k) = (51 - 15)/(69 - 15) = 36/54 = 2/3. At t = 5: 69 - 15 = 54 = (T0 - 15) e^(-5k) = (T0 - 15)(2/3), so T0 - 15 = 81 and T0 = 96. The initial temperature of the kettle was 96 degrees C.