Key takeaways
- A differential equation is separable if dy/dx = g(x)·h(y).
- Move all y's to one side, all x's to the other, then integrate both sides.
- dy/dx = ky has the solution y = y₀·eᵏˣ.
- The constant multiplies: y = A·eᵏˣ, not eᵏˣ + C.
In this guide
What is a separable differential equation?
A differential equation is an equation with a derivative in it, such as dy/dx = 2y. It describes change: 'the growth rate depends on how much there is'. Solving it means finding the function y(x) itself.
It is separable if you can move all the y's to one side and all the x's to the other: (something with y) dy = (something with x) dx. Then you integrate each side on its own. This calculator solves the most important example, dy/dx = ky: the rate of change is proportional to the amount.
Why do separable equations matter?
The rule 'the more there is, the faster it changes' is everywhere. More rabbits have more babies. A bigger balance earns more interest. Hotter coffee cools faster. All of these start as dy/dx = ky, and separation of variables solves them in a few lines.
The result is the exponential curve, one of the most important shapes in science.
With k = 0.05 per month, a population doubles about every 13.9 months, whether it starts with 200 rabbits or 2 million.
The formula
| Symbol | What it means | Example |
|---|---|---|
| y | The quantity that changes | number of rabbits |
| x | The input, often time | 10 months |
| k | The rate constant: k > 0 is growth, k < 0 is decay | 0.05 per month |
| y₀ | The starting value, y(0) | 200 |
| e | Euler's number, see the exponential calculator | 2.71828 |
How to solve a separable ODE step by step
- 1Separate. Divide by y and multiply by dx:
dy / y = k dx. - 2Integrate both sides.
∫ dy/y = ∫ k dxgivesln|y| = kx + C. - 3Solve for y. Raise e to both sides:
y = A·eᵏˣ, where A is a constant. - 4Use the initial condition. At x = 0,
y = A, soA = y₀. - 5Evaluate. Put in your x:
y(x) = y₀·eᵏˣ.
Now try your own growth or decay problem. Enter k, the starting value and x in the calculator below.
Try it yourself
Pre-filled with the example — change any value.
Common mistakes to avoid
- Forgetting the constant C. Add it right after integrating, or you lose the family of solutions.
- Treating dy/dx = ky + x as separable. You cannot split y and x apart there. Use the first-order linear ODE calculator instead.
- Dividing by y without checking y = 0. y = 0 is also a solution of dy/dx = ky.
- Mixing time units. If k is per month, x must be in months.
Tips and tricks
- Doubling time: for growth, it is
ln 2 ÷ k ≈ 0.693 ÷ k. - Check your answer: differentiate
y₀eᵏˣand you getk·y₀eᵏˣ = ky. It works. - Half-life and doubling problems: the exponential growth and decay calculator does these directly.
Where you'll use it in real life
- Biology: population growth and bacteria in a lab.
- Finance: continuously compounded interest.
- Medicine: how a drug level in the blood falls over time.
- Physics: radioactive decay and Newton's law of cooling.
Quick summary
To solve a separable equation, put all the y's on one side and all the x's on the other, integrate both sides and use the initial condition. For dy/dx = ky this gives y = y₀·eᵏˣ. Use the calculator above to see every step for your own numbers.
Worked example
A town has 200 rabbits, and the population grows at a rate proportional to its size with k = 0.05 per month, so dy/dx = 0.05y. How many rabbits will there be after 10 months? Use k = 0.05, y₀ = 200 and x = 10.
- 01ODE: dy/dx = 0.05y, y(0) = 200
- 02Separate variables: dy/y = 0.05dx
- 03Integrate both sides: ln|y| = 0.05x + C
- 04Apply initial condition y(0) = 200: C = ln(200)
- 05Solution: y(x) = 200·e^(0.05x)
- 06At x = 10: y = 200 × e^(0.5)
- 07= 200 × 1.6487
- 08= 329.7443
These numbers are pre-filled in the calculator above.
Frequently asked questions
How do I know if a differential equation is separable?
Check whether dy/dx can be written as a function of x times a function of y, like g(x)·h(y). If so, you can separate it into dy/h(y) = g(x) dx.
What is the solution of dy/dx = 2y with y(0) = 1?
y = e²ˣ. At x = 1, that gives e² ≈ 7.389.
What does the initial condition do?
Integration gives a whole family of solutions with an unknown constant. The initial condition, such as y(0) = 200, picks the one solution that starts at the right value.
What happens if k is negative?
Then the quantity decays towards 0 instead of growing. For example, with k = −0.5 and y₀ = 100, y(2) = 100·e⁻¹ ≈ 36.8.
