First-order differential equations
The idea
A differential equation relates an unknown function to its own derivative, and first-order equations — involving only y and y' — are where modeling begins: population growth, radioactive decay, cooling, mixing tanks, RC circuits. Solving one means finding every function whose rate behaves as prescribed; the general solution carries an arbitrary constant, and an initial condition like y(0) = 2 pins down the one trajectory passing through that point.
The first tool is separation of variables: when the equation can be written so that all y-dependence sits with dy and all x-dependence with dx, integrate both sides independently. When that fails, linear first-order equations yield to the integrating factor method. Reading the equation's type before computing is the actual skill — separable, linear, or neither determines the entire plan of attack.
Resist the temptation to bolt the constant on at the end. The + C must appear at the moment of integration, before any exponentiation or rearrangement, because subsequent algebra transforms the constant — in exponential solutions, an additive C becomes a multiplicative one, which is why the final form is A e^(x³) rather than e^(x³) + C.
Worked example
Solve the initial value problem dy/dx = 3x²y with y(0) = 2.
- Classify first: the right side factors as (3x²)(y), a pure function of x times a pure function of y, so the equation is separable.
- Separate: dy/y = 3x² dx, valid where y ≠ 0 (and y = 0 is itself an equilibrium solution, noted and set aside since our initial value is 2, not 0).
- Integrate both sides: ln|y| = x³ + C, attaching the constant now, at the integration step.
- Exponentiate: |y| = e^(x³ + C) = e^C e^(x³), and absorbing the sign and e^C into one constant gives y = A e^(x³) — the additive constant has become multiplicative.
- Apply the initial condition: y(0) = A e⁰ = A = 2, so y = 2e^(x³).
- Verify by substitution: y' = 2e^(x³) × 3x² = 3x²y, and y(0) = 2 — both the equation and the initial condition check out exactly.
Answer. The solution is y = 2e^(x³), the unique trajectory through (0, 2).
Check your understanding
- Why does the additive constant of integration turn into a multiplicative constant in exponential-type solutions?
- What information does an initial condition add to the general solution, and what does it select geometrically?
- How do you recognize a separable equation quickly, and what structural feature rules separation out?
- What is special about the constant solution y = 0 here, and why does it deserve a separate check before you divide by y?
Build the foundations first
First-order differential equations builds on these concepts. If any feel shaky, start there.