Most differential equations cannot be solved with pencil and paper — but some can, and the separable type is the most important for the HSC. If you can rearrange the equation so that all $y$ terms are on one side and all $x$ terms on the other, you can integrate both sides independently. This simple idea unlocks models of population growth, radioactive decay, chemical reactions, and cooling objects. The technique is not just a mathematical procedure; it is how scientists extract predictions from the laws of nature.
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Consider $\frac{dy}{dx} = xy$. Can you rearrange this so that all $y$ terms are on the left and all $x$ terms on the right? What would integrating both sides give?
Core Content
A differential equation is separable if it can be written in the form:
dy/dx = f(x) · g(y)
To solve:
Example: Solve $\frac{dy}{dx} = xy$ with $y(0) = 2$.
Step 1: dy/y = x dx
Step 2: ln|y| = x^2/2 + C
Step 3: |y| = e^{x^2/2 + C} = e^C · e^{x^2/2} = A e^{x^2/2}
Step 4: y(0) = 2: A = 2
Solution: y = 2e^{x^2/2}
The differential equation $\frac{dy}{dx} = ky$ is the prototype for exponential growth ($k > 0$) or decay ($k < 0$).
Separation: $\frac{dy}{y} = k\,dx$
Integration: $\ln|y| = kx + C$
Solution: $y = Ae^{kx}$ where $A = e^C$
Interpretation:
Half-life: For decay, the half-life is $t_{1/2} = \frac{\ln 2}{|k|}$.
Doubling time: For growth, the doubling time is $t_{double} = \frac{\ln 2}{k}$.
Solve $\frac{dy}{dx} = \frac{x}{y}$ with $y(0) = 3$.
The particular solution.
Separate: y dy = x dx
Integrate: y^2/2 = x^2/2 + C
Multiply by 2: y^2 = x^2 + 2C = x^2 + A
y(0) = 3: 9 = 0 + A, so A = 9
y^2 = x^2 + 9
y = √(x^2 + 9) (taking positive root since y(0) = 3 > 0)
$y = \sqrt{x^2 + 9}$.
Solve $\frac{dy}{dx} = 2x y$ with $y(0) = 1$.
Answer:
$\frac{dy}{y} = 2x \, dx$. $\ln|y| = x^2 + C$. $y = Ae^{x^2}$.
$y(0) = 1$: $A = 1$. Solution: $y = e^{x^2}$.
$\frac{dy}{dx} = f(x)g(y)$
$\frac{dy}{g(y)} = f(x)\,dx$
$\int \frac{dy}{g(y)} = \int f(x)\,dx$
$\frac{dy}{dx} = ky \Rightarrow y = Ae^{kx}$
Rearranging $\frac{dy}{dx} = xy$ gives $\frac{dy}{y} = x\,dx$. Integrating: $\ln|y| = \frac{x^2}{2} + C$. Exponentiating: $y = Ae^{x^2/2}$. This works because each side depends on only one variable, allowing independent integration. The constant $A$ absorbs $e^C$ and is determined by the initial condition. The solution $y = Ae^{kx}$ appears everywhere in science because the assumption "rate of change proportional to current amount" applies to populations, radioactive nuclei, chemical concentrations, temperatures, and financial investments.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. Solve $\frac{dy}{dx} = \frac{x^2}{y}$ with $y(0) = 2$. Show all working. 3 MARKS
9. A radioactive substance decays according to $\frac{dN}{dt} = -kN$. Its half-life is 5 years. (a) Find $k$. (b) If there are initially 100 grams, how much remains after 15 years? Show all working. 4 MARKS
10. Explain why $\frac{dy}{dx} = x + y$ is not separable. Then describe a real-world situation where a non-separable differential equation might arise, and discuss what methods mathematicians use when separation fails. 3 MARKS
1. $\frac{dy}{y} = \frac{dx}{x}$, $\ln|y| = \ln|x| + C$, $y = Ax$. $y(1) = 2$: $A = 2$. $y = 2x$.
2. $e^y dy = dx$, $e^y = x + C$, $y = \ln(x + C)$. $y(0) = 0$: $C = 1$. $y = \ln(x + 1)$.
3. $P(t) = 1000e^{0.03t}$. $P(10) = 1000e^{0.3} \approx 1349.9$.
1. No — $x + y$ cannot be factored as $f(x)g(y)$. The variables are added, not multiplied.
2. $N(t) = N_0 e^{-kt}$. Half-life: $\frac{\ln 2}{k} = 10$, so $k = \frac{\ln 2}{10} \approx 0.0693$ per year.
3. $k$ is the proportionality constant. It measures how quickly the population changes relative to its size. Larger $k$ means faster growth.
Q8 (3 marks): $y\,dy = x^2\,dx$ [0.5]. $\frac{y^2}{2} = \frac{x^3}{3} + C$ [1]. $y(0) = 2$: $2 = 0 + C$, so $C = 2$ [0.5]. $y^2 = \frac{2x^3}{3} + 4$, so $y = \sqrt{\frac{2x^3}{3} + 4}$ (positive since $y(0) > 0$) [1].
Q9 (4 marks): (a) $\frac{\ln 2}{k} = 5$, so $k = \frac{\ln 2}{5} \approx 0.1386$ per year [2]. (b) $N(t) = 100e^{-kt}$ [0.5]. $N(15) = 100e^{-3\ln 2} = 100 \cdot 2^{-3} = 100 \cdot \frac{1}{8} = 12.5$ grams [1.5].
Q10 (3 marks): $x + y$ cannot be written as $f(x)g(y)$ because it's a sum, not a product [1]. A real-world example: predator-prey models where $\frac{dx}{dt} = ax - bxy$ (the interaction term $xy$ is not separable in the usual sense) [1]. When separation fails, mathematicians use integrating factors, series solutions, numerical methods (Euler's method, Runge-Kutta), or qualitative analysis (phase portraits) [1].
Climb platforms by separating variables, integrating both sides, and applying initial conditions. Pool: lesson 12.
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