The separation method
Sorting a mixed basket of fruit puts apples on one side and oranges on the other. In a separable equation, all y terms with dy go to one side and all x terms with dx go to the other, and then both sides are integrated.
If dy/dx = f(x) g(y), write it as dy / g(y) = f(x) dx. Integrate both sides, and put a single constant C on one side.
Formula: dy / g(y) = f(x) dx, hence ∫ dy / g(y) = ∫ f(x) dx + C
Rule 1: Separate, then integrate — Divide by g(y) and multiply by dx.
- Easy: dy/dx = xy gives dy/y = x dx, so ln|y| = x²/2 + C and y = A e^(x²/2).
- Medium: dy/dx = 1 + y² gives dy/(1 + y²) = dx, so tan⁻¹y = x + C.
- Exam-type: dy/dx = e^(x + y) gives e^(−y) dy = eˣ dx, so eˣ + e^(−y) = C.
Rule 2: Ratio forms — Use partial fractions or standard integrals when needed.
- Easy: x dy = y dx gives dy/y = dx/x, so y = Cx.
- Medium: (1 + x²) dy/dx = x gives dy = x dx/(1 + x²), so y = ½ ln(1 + x²) + C.
- Exam-type: dy/dx = (1 + y²)/(1 + x²) gives tan⁻¹y = tan⁻¹x + C.
Only one constant C is needed. Write ln|y| = ... + C and then change to y = A e^(...), where A = e^C.
Writing dy/dx = xy as ∫ dy = ∫ xy dx, which does not separate the variables.
y with dy, x with dx, then integrate.
Practice questions
- 1Solve dy/dx = 3x².
- 2Solve dy/dx = y.
- 3Solve dy/dx = xy.
- 4Solve dy/dx = 1 + y².
- 5Solve dy/dx = e^(x + y).
- 6Solve dy/dx = (1 + y²)/(1 + x²).
- 7Solve x dy = y dx.
- 8Solve (1 + x²) dy/dx = x.
Show answers (try them yourself first)
- 1y = x³ + C
- 2y = C eˣ
- 3y = C e^(x²/2)
- 4tan⁻¹y = x + C
- 5eˣ + e^(−y) = C
- 6tan⁻¹y = tan⁻¹x + C
- 7y = Cx
- 8y = ½ ln(1 + x²) + C
Particular solutions and growth
A bacteria culture grows at a speed proportional to its size. This single sentence is a differential equation, and a starting count turns the general solution into one definite formula.
After finding the general solution, use the given condition to find C. For a quantity whose rate of change is proportional to itself, dN/dt = kN, the solution is N = N₀ eᵏᵗ.
Formula: dN/dt = kN gives N = N₀ e^(kt)
Rule 1: Use the condition — Substitute the given point into the general solution.
- Easy: dy/dx = y, y(0) = 2 gives y = 2 eˣ.
- Medium: dy/dx = x/y, y(0) = 3 gives y dy = x dx, y² = x² + 9.
- Exam-type: dy/dx = (1 + y)/(1 + x), y(0) = 2 gives 1 + y = K(1 + x) with K = 3, so y = 3x + 2.
Rule 2: Growth and decay — Positive k grows, negative k decays.
- Easy: dy/dx = −2y, y(0) = 4 gives y = 4 e^(−2x).
- Medium: dN/dt = 0.1 N with N(0) = 100 gives N = 100 e^(0.1t), and N(10) = 100e, about 272.
- Exam-type: dy/dx = y/x, y(1) = 3 gives ln y = ln x + c, so y = 3x.
Find the general solution first, then use the condition. Do not put the numbers in before integrating.
Using the condition before separating, or forgetting to substitute at the end.
General first, then fix C.
Practice questions
- 1Solve dy/dx = y with y(0) = 2.
- 2Solve dy/dx = x/y with y(0) = 3.
- 3Solve dy/dx = (1 + y)/(1 + x) with y(0) = 2.
- 4Solve dy/dx = 2x with y(1) = 5.
- 5Solve dy/dx = −2y with y(0) = 4.
- 6A population obeys dN/dt = 0.1 N with N(0) = 100. Find N(10) in terms of e.
- 7Solve dy/dx = y/x with y(1) = 3.
- 8Solve dy/dx = cos x with y(0) = 1.
Show answers (try them yourself first)
- 1y = 2 eˣ
- 2y dy = x dx, so y² = x² + 9
- 31 + y = K(1 + x) with K = 3, so y = 3x + 2
- 4y = x² + C and 5 = 1 + C give y = x² + 4
- 5y = 4 e^(−2x)
- 6N = 100 e^(0.1t), so N(10) = 100e, about 272
- 7y = 3x
- 8y = sin x + C and 1 = C, so y = sin x + 1