A first-order differential equation links dy/dx with x and y. When it is separable, dy/dx = f(x)g(y), you can gather the y terms with dy and the x terms with dx, then integrate both sides: ∫ 1/g(y) dy = ∫ f(x) dx. One constant of integration is enough, and a boundary condition then fixes it to give the particular solution.
Solving separable differential equations
- Separate: put all y terms with dy and all x terms with dx.
- Integrate both sides, adding one constant c.
- Rearrange for y if asked.
- Use a given condition to find the constant.
dy/dt = ky has solution y = y₀e^(kt): growth if k > 0, decay if k < 0.
What is a separable differential equation?
A differential equation is separable if dy/dx can be written as a function of x multiplied by a function of y. Then the variables can be separated: divide both sides by the y-part and multiply by dx, so each side contains only one variable. Integrating both sides gives the general solution.
General and particular solutions
The general solution contains an arbitrary constant, so it describes a family of curves. A boundary condition, a known pair of values such as y = 3 when x = 0, fixes the constant and gives the particular solution. Find the general solution first, then substitute the condition.
Logarithms and the constant A
Separating often gives ln|y| on one side. To make y the subject, exponentiate: ln|y| = x² + c becomes y = e^(x² + c) = Ae^(x²), where A = e^c is a new constant. Writing the constant as A at this stage keeps the algebra tidy.
Exponential growth and decay
When the rate of change of a quantity is proportional to the quantity itself, dy/dt = ky, the solution is always exponential: y = y₀e^(kt), where y₀ is the value at t = 0. A positive k means growth and a negative k means decay. You can quote this result once you have seen it derived by separating the variables.
Harder separable equations
After separating, the integrals may need earlier techniques: the f′(x)/f(x) pattern, integration by parts, or partial fractions. For example, dy/dx = e^(x − y) separates as e^y dy = eˣ dx once you write e^(x − y) = eˣe^(−y).
Modelling with differential equations
Exam questions often describe a rate in words. “The rate of change of P is proportional to P” becomes dP/dt = kP; “decreases at a rate proportional to” needs a minus sign. Form the equation, solve it, use the conditions to find the constants, then answer the question in context, such as finding a time or a long-term value.
Worked examples
Example 1. Find the general solution of dy/dx = 2xy.
- Separate: (1/y) dy = 2x dx.
- Integrate both sides: ln|y| = x² + c.
- Exponentiate: y = e^(x² + c) = Ae^(x²).
Answer: y = Ae^(x²).
Marks: M1 for separating and integrating, M1 for ln|y| = x² + c, A1 for y = Ae^(x²).
Example 2. Solve dy/dx = 2xy, given that y = 3 when x = 0.
- The general solution is y = Ae^(x²).
- At x = 0: 3 = Ae⁰ = A.
Answer: y = 3e^(x²).
Marks: M1 for using the condition, A1 for the particular solution.
Example 3. A colony grows so that dN/dt = 0.5N, with N = 100 when t = 0. Find N in terms of t.
- This is the growth model dN/dt = kN with k = 0.5, so N = N₀e^(0.5t).
- At t = 0, N = 100, so N₀ = 100.
Answer: N = 100e^(0.5t).
Marks: M1 for the exponential form, A1 for N = 100e^(0.5t).
Example 4. Solve dy/dx = e^(x − y).
- Write e^(x − y) = eˣe^(−y), then separate: e^y dy = eˣ dx.
- Integrate: e^y = eˣ + c.
- Take logs: y = ln(eˣ + c).
Answer: y = ln(eˣ + c).
Marks: M1 for separating, M1 for e^y = eˣ + c, A1 for the answer.
Example 5. dy/dx = 2y, with y = 1 when x = 0. Find y, and its value when x = 1.
- This is exponential growth: y = Ae^(2x).
- At x = 0, y = 1, so A = 1.
- When x = 1, y = e².
Answer: y = e^(2x), and y = e² when x = 1.
Marks: M1 for y = e^(2x), A1 for e².
Common mistakes
- Not separating fully. Every y must be on the dy side and every x on the dx side before integrating.
- Adding a constant to both sides. One constant of integration is enough.
- Writing e^(x² + c) = e^(x²) + c. e^(x² + c) = e^c × e^(x²) = Ae^(x²).
- Using the condition before integrating. Find the general solution first, then substitute the condition.
- Missing the minus sign for decay. A quantity that decreases has a negative rate of change.
How differential equations are examined
First-order separable differential equations are Pure 4 content (WMA14) in Edexcel International A-Level, and appear in second-year Pure for Edexcel A-Level (9MA0). Questions usually:
- ask for the general solution of a separable equation
- use a boundary condition to find a particular solution
- model a situation such as growth, decay, cooling or draining
- need partial fractions or parts to complete the integration
Modelling questions with full mark schemes are among the Pure 4 integration exam questions. For more, try the Pure 4 integration worksheet or the revision engine for integration.
Differential equations often arise from connected rates of change, and their solutions link to exponential growth and decay.
Frequently asked questions
What is a separable differential equation?
One that can be written as dy/dx = f(x)g(y), so the x and y terms can be moved to opposite sides and integrated separately.
What is the difference between a general and a particular solution?
The general solution contains an unknown constant. A particular solution uses a given condition to find that constant.
Why does dy/dt = ky give exponential growth or decay?
Separating gives (1/y) dy = k dt, so ln|y| = kt + c and y = Ae^(kt). The sign of k decides growth or decay.