MAST10006 · Calculus 2
First-Order Differential Equations
A first-order ODE relates y and its rate of change y′. The chapter works through the standard solving methods and the qualitative tools, in roughly the order of difficulty. Separable equations split the right-hand side into an x-part times a y-part, so each variable integrates independently. Linear equations y′ + P(x)y = Q(x) are cracked by the integrating factor μ = e∫P dx, which folds the left side into a single exact derivative d/dx(μy). When an equation is neither, a well-chosen substitution (v = y/x for homogeneous, u = ax+by+c, Bernoulli) transforms it into one that is. Then come the qualitative tools that describe behaviour without solving — the direction field and the phase line with stability from the sign of f′(y*) — and the applications (logistic growth, mixing tanks), where the marks are in building the rate balance.
What this chapter covers
- 018.1 Separable equations
- 028.2 Linear ODEs and the integrating-factor method
- 038.3 Substitution methods (homogeneous, Bernoulli)
- 048.4 The direction (slope) field
- 05Qualitative analysis and the phase line (stability)
- 068.5 Applications: logistic populations and mixing tanks
Worked example: a linear ODE by integrating factor
- +1Standardise so the coefficient of y′ is 1: y′ + y/x = x, so P(x) = 1/x and Q(x) = x.
- +1Integrating factor: μ = e∫(1/x)dx = elog x = x (no +C in the exponent — it just rescales μ).
- +1Multiply through by μ = x; the left side collapses to an exact derivative: d/dx(xy) = x².
- +1Integrate and divide: xy = x³/3 + C, so y = x²/3 + C/x.
Key terms
- Separable equation
- A first-order ODE of the form dy/dx = f(x)g(y), where the right side factorises into an x-part times a y-part. You divide by g(y), integrate each side independently, and apply the initial condition — remembering to check separately for equilibrium solutions g(y) = 0, which dividing silently drops.
- Integrating factor
- The multiplier μ(x) = e∫P(x)dx for a linear ODE y′ + P(x)y = Q(x). Multiplying the standardised equation by μ makes the left side exactly d/dx(μy), so y = (1/μ)(∫μQ dx + C). No constant is needed in the exponent of μ.
- Direction (slope) field
- The picture obtained by drawing, at a grid of points, a short tick of slope f(x,y) for the ODE y′ = f(x,y). Every solution curve threads through the field tangent to it everywhere, so you can sketch behaviour without integrating; curves of constant slope f = c are isoclines.
- Phase line
- For an autonomous equation y′ = f(y), a single vertical axis marking the equilibria f(y) = 0 and the sign of f between them (arrow up where f > 0, down where f < 0). It shows the long-run fate of every solution without solving.
- Stability test
- At an equilibrium y* of y′ = f(y): f′(y*) < 0 means stable (a nudge decays back), f′(y*) > 0 means unstable (a nudge grows), and f′(y*) = 0 is inconclusive, so fall back on the sign of f on each side.
First-Order Differential Equations FAQ
How do I know which first-order method to use?
Check the form. If the right side factorises into (function of x)(function of y), it is separable. If it can be written y′ + P(x)y = Q(x), it is linear — use the integrating factor. If it is neither but has a recognisable structure (depends only on y/x, or has the form F(ax+by+c), or is Bernoulli y′ + Py = Qyn), a substitution converts it to a separable or linear equation.
Why must I standardise before finding the integrating factor?
Because P(x) is read off the equation only after the coefficient of y′ is 1. If you compute μ from a non-standardised equation your P is wrong, and every step after it is wrong too. Divide through first, then read off P and Q.
What is the most common substitution mistake?
Transforming the variable but not the derivative. Under v = y/x you have y = vx, so by the product rule y′ = v + x v′ — NOT just v′. Substituting only the variable and forgetting to convert dy/dx produces a wrong, often un-separable equation. Always finish by back-substituting to express the answer in the original x and y.
How do I tell whether an equilibrium is stable?
Use the sign of f′ at the equilibrium of an autonomous ODE y′ = f(y): f′(y*) < 0 is stable (attracting), f′(y*) > 0 is unstable (repelling). The marks for a phase-line question are in locating the equilibria, getting the flow arrows right from the sign of f, and labelling each equilibrium correctly — not in solving the ODE.
Exam move
Make method-identification automatic: separable if the right side splits; linear if it fits y′ + Py = Q; otherwise a substitution (v = y/x, u = ax+by+c, or Bernoulli). For separable equations, check for lost equilibrium solutions and fix the +C with the initial condition. For linear equations, the ritual is standardise → μ = e∫P dx with no exponent constant → fold into d/dx(μy) → integrate → divide and add C. For the qualitative tools, practise drawing a slope field by isoclines and reading a phase line off the sign of f, since these ask you to describe behaviour without solving. For applications, the hard part is setting up the equation: translate “rate of change = in − out” into y′ = f, then recognise it as separable or linear. The modelling is the marks; the solving is the routine part.
