What are first order linear equations?

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

TASC TCE 2023-style practice: Find the general solution of the differential equation (1 + x^2)dydx + (1 - y^2) = 0.

This equation is separable. Rearrange: (1 + x^2)dydx = -(1 - y^2) = y^2 - 1, so dy1 - y^2 = -dx1 + x^2. Integrate the left side using partial fractions: 11 - y^2 = 12[11 - y + 11 + y], whose integral is 12ln|1 + y1 - y|. Integrate the right side: - x. So the general solution is 12ln|1 + y1 - y| = - x + C, where C is an arbitrary constant. Equivalent forms obtained by exponentiating are acceptable. Markers reward separating the variables correctly, the partial-fraction integral, and including the constant of integration.

TASC TCE 2024-style practice: Solve the differential equation x^2dydx - y = 1 subject to y(1) = 1.

Divide by x^2 to put it in linear form: dydx - 1x^2y = 1x^2. The integrating factor is μ = e^int -1/x^2\,dx = e^1/x. Multiply through: ddx(y e^1/x) = 1x^2e^1/x. Integrate the right side with the substitution u = 1x, du = -1x^2\,dx, giving int -e^u\,du = -e^1/x. So y e^1/x = -e^1/x + C, that is 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 + 2e^(x - 1)/x. Check: at x = 1, y = -1 + 2 = 1.

TASC TCE 2022-style practice: A hot kettle is in a room at temperature θ degrees Celsius. After t minutes its temperature T(t) satisfies dTdt = -k(T - θ), 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.

The solution of Newton's law of cooling is T = θ + (T_0 - θ)e^-kt, where T_0 is the initial temperature. So T - θ decays geometrically: the ratios over equal time steps are equal. Hence 51 - θ69 - θ = 39 - θ51 - θ. Cross-multiply: (51 - θ)^2 = (69 - θ)(39 - θ). Expanding, 2601 - 102θ = 2691 - 108θ, so 6θ = 90 and θ = 15. The room temperature is 15 degrees Celsius. The common ratio over 5 minutes is e^-5k = 51 - 1569 - 15 = 3654 = 23. At t = 5: 69 - 15 = 54 = (T_0 - 15)e^-5k = (T_0 - 15)·23, so T_0 - 15 = 81 and T_0 = 96. The initial temperature of the kettle was 96 degrees Celsius.

§-Syllabus dot point

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 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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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 separable equation

Solve $\dfrac{dy}{dx} = \dfrac{2x}{y}$ with $y(0) = 3$ .

Step 1: Separate the variables

The idea is to get all $y$ terms on the left and all $x$ terms on the right, so each side can be integrated independently. Multiply both sides by $y$ and by $dx$ :

y\, dy = 2x\, dx.

Step 2: Integrate both sides

Each side is now a straightforward power:

\frac{y^2}{2} = x^2 + C,

where $C$ is an arbitrary constant of integration. It is essential to include $C$ here so the initial condition can be satisfied.

Step 3: Apply the initial condition to find $C$

Substituting $x = 0$ and $y = 3$ gives $\tfrac{9}{2} = 0 + C$ , so $C = \tfrac{9}{2}$ .

Step 4: Solve for $y$

Substitute back and rearrange:

\frac{y^2}{2} = x^2 + \frac{9}{2} \quad\Longrightarrow\quad y^2 = 2x^2 + 9 \quad\Longrightarrow\quad y = \sqrt{2x^2 + 9},

taking the positive square root because $y(0) = 3 > 0$ .

Final answer: $y = \sqrt{2x^2 + 9}$ .

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

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

Step 1: Identify $P(x)$ and compute the integrating factor

The equation is already in standard form $\dfrac{dy}{dx} + P(x)y = Q(x)$ with $P(x) = \tfrac{1}{x}$ and $Q(x) = x$ . The integrating factor is chosen so that multiplying through converts the left side into an exact derivative:

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

Step 2: Multiply through by $\mu$ and recognise the left side

Multiplying every term by $\mu = x$ gives:

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

The left side is exactly the product-rule expansion of $\dfrac{d}{dx}(xy)$ , which is why the integrating factor works.

Step 3: Integrate both sides

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

Step 4: Apply the initial condition and solve for $y$

Substituting $x = 1$ and $y = 2$ : $1 \cdot 2 = \tfrac{1}{3} + C$ , so $C = \tfrac{5}{3}$ . Dividing by $x$ :

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

Check at $x = 1$ : $\tfrac{1}{3} + \tfrac{5}{3} = 2$ . Correct.

Final answer: $y = \dfrac{x^2}{3} + \dfrac{5}{3x}$ .

Modelling with differential equations

Many Specialist questions are word problems that you must first turn into a differential equation. The pattern is always "rate of change equals something", and the skill is translating the words.

Exponential growth and decay. "The rate of change is proportional to the amount" becomes $\dfrac{dy}{dt} = ky$ , solving to $y = y_0 e^{kt}$ . A positive $k$ is growth (populations, compound processes); a negative $k$ is decay (radioactivity, discharge).
Newton's law of cooling. "The rate of cooling is proportional to the temperature difference" becomes $\dfrac{dT}{dt} = -k(T - \theta)$ , solving to $T = \theta + (T_0 - \theta)e^{-kt}$ .
Mixing problems. "Rate of change of dissolved substance equals rate in minus rate out" gives a first order linear equation, solved with an integrating factor.

A mixing problem

A tank holds $100$ L of pure water. Brine containing $2$ g/L of salt flows in at $5$ L/min, and the well-stirred mixture flows out at $5$ L/min. Let $Q(t)$ be the grams of salt at time $t$ . Find $Q(t)$ .

Step 1: Write the rate equation

The rate of change of salt equals the rate in minus the rate out. Because inflow and outflow volumes are equal, the tank volume stays fixed at $100$ L. Salt enters at $2 \times 5 = 10$ g/min. The concentration in the tank is $\dfrac{Q}{100}$ g/L, so salt leaves at $\dfrac{Q}{100} \times 5 = \dfrac{Q}{20}$ g/min:

\frac{dQ}{dt} = 10 - \frac{Q}{20}.

Step 2: Put in standard linear form and find the integrating factor

Rearrange to standard form $\dfrac{dQ}{dt} + P(t)Q = R(t)$ :

\frac{dQ}{dt} + \frac{1}{20}Q = 10.

The integrating factor is $\mu = e^{\int (1/20)\, dt} = e^{t/20}$ .

Step 3: Multiply through and integrate

Multiplying both sides by $\mu$ converts the left side to an exact derivative:

\frac{d}{dt}\!\left(Q\, e^{t/20}\right) = 10\, e^{t/20}.

Integrating both sides: $Q\, e^{t/20} = 200\, e^{t/20} + C$ , so $Q = 200 + C e^{-t/20}$ .

Step 4: Apply the initial condition

At $t = 0$ the tank holds pure water, so $Q(0) = 0$ . Substituting: $0 = 200 + C$ , giving $C = -200$ . Therefore:

Q(t) = 200\!\left(1 - e^{-t/20}\right).

As $t \to \infty$ , $Q \to 200$ g, the steady state at which the tank reaches the inflow concentration of $2$ g/L over $100$ L.

Final answer: $Q(t) = 200\!\left(1 - e^{-t/20}\right)$ grams.

Slope fields

A slope field (direction field) sketches the gradient $\dfrac{dy}{dx} = f(x, y)$ as a short line segment at a grid of points. A solution curve threads through the field tangent to every segment it passes. Slope fields let you see the qualitative behaviour, equilibrium solutions (where $\dfrac{dy}{dx} = 0$ ) and long-term trends, without solving the equation. An equilibrium is stable if nearby solutions move toward it and unstable if they move away.

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.

Why this matters

Differential equations are how Specialist Mathematics models the real world: anything that changes at a rate depending on its current value is described by one. Mastering the two solution methods (separation and the integrating factor) and the habit of translating a word problem into a rate equation covers nearly every TASC differential-equations question, and the partial-fraction and substitution skills from the integration topic feed directly into solving them.

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.

TCE 20237 marksFind the general solution of the differential equation

(1 + x^2)\dfrac{dy}{dx} + (1 - y^2) = 0

Show worked answer →

This equation is separable. Rearrange: $(1 + x^2)\dfrac{dy}{dx} = -(1 - y^2) = y^2 - 1$ , so $\dfrac{dy}{1 - y^2} = -\dfrac{dx}{1 + x^2}$ .

Integrate the left side using partial fractions: $\dfrac{1}{1 - y^2} = \dfrac{1}{2}\left[\dfrac{1}{1 - y} + \dfrac{1}{1 + y}\right]$ , whose integral is $\dfrac{1}{2}\ln\left|\dfrac{1 + y}{1 - y}\right|$ .

Integrate the right side: $-\arctan x$ .

So the general solution is $\dfrac{1}{2}\ln\left|\dfrac{1 + y}{1 - y}\right| = -\arctan x + C$ , where $C$ is an arbitrary constant. Equivalent forms obtained by exponentiating are acceptable. Markers reward separating the variables correctly, the partial-fraction integral, and including the constant of integration.

TCE 20245 marksSolve the differential equation

x^2\dfrac{dy}{dx} - y = 1

subject to

y(1) = 1

Show worked answer →

Divide by $x^2$ to put it in linear form: $\dfrac{dy}{dx} - \dfrac{1}{x^2}y = \dfrac{1}{x^2}$ .

The integrating factor is $\mu = e^{\int -1/x^2\,dx} = e^{1/x}$ .

Multiply through: $\dfrac{d}{dx}\left(y e^{1/x}\right) = \dfrac{1}{x^2}e^{1/x}$ . Integrate the right side with the substitution $u = \dfrac{1}{x}$ , $du = -\dfrac{1}{x^2}\,dx$ , giving $\int -e^u\,du = -e^{1/x}$ . So $y e^{1/x} = -e^{1/x} + C$ , that is $y = -1 + C e^{-1/x}$ .

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

TCE 20227 marksA hot kettle is in a room at temperature

\theta

degrees Celsius. After

t

minutes its temperature

T(t)

satisfies

\dfrac{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.

Show worked answer →

The solution of Newton's law of cooling is $T = \theta + (T_0 - \theta)e^{-kt}$ , where $T_0$ is the initial temperature.

So $T - \theta$ decays geometrically: the ratios over equal time steps are equal. Hence $\dfrac{51 - \theta}{69 - \theta} = \dfrac{39 - \theta}{51 - \theta}$ .

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 Celsius.

The common ratio over 5 minutes is $e^{-5k} = \dfrac{51 - 15}{69 - 15} = \dfrac{36}{54} = \dfrac{2}{3}$ . At $t = 5$ : $69 - 15 = 54 = (T_0 - 15)e^{-5k} = (T_0 - 15)\cdot\dfrac{2}{3}$ , so $T_0 - 15 = 81$ and $T_0 = 96$ . The initial temperature of the kettle was $96$ degrees Celsius.