The Equation That Describes Its Own Change Differential Equations

Every earlier chapter of this thread has used differentiation and integration as tools applied to a function that was already fully known and written down. A differential equation flips this around entirely: it is an equation involving an unknown function y and one or more of its derivatives, and solving it means finding that unknown function itself, not just a number. The simplest possible example, dy/dx = 2x, is something already familiar from the very idea of integration: y must be x squared + C, for some constant C, found by asking which functions have derivative 2x. A differential equation can be far less obviously connected to a known integral than that, such as dy/dx = x + y, where the unknown function y appears mixed together with x on the right-hand side, and no direct integration of the right side alone will produce the answer. The techniques in this chapter are, in effect, a toolkit of specific patterns, variables that can be separated, right-hand sides with a specific symmetric structure, and equations that are 'linear' in a particular sense, each with its own reliable method for recovering the hidden function y.

Before attempting to solve any differential equation, two quick descriptive labels are worth reading off first. The order is the order of the highest derivative that appears anywhere in the equation: d squared y/dx squared + 3(dy/dx) + 2y = 0 has order 2, since the highest derivative present is the second derivative. The degree is the power to which that highest-order derivative is raised, but only once the whole equation has been cleared of any fractional powers or roots involving derivatives. For [d squared y/dx squared] cubed + (dy/dx) + y = 0, the order is 2 (the second derivative is the highest one present), and the degree is 3 (that second derivative appears cubed). Degree is undefined, and simply not assigned, whenever an equation cannot be written as a polynomial in its derivatives at all, such as one containing sin(dy/dx). Order and degree together give a quick sense of how structurally complicated an equation is before any solving method is even chosen.

A differential equation of order n typically has a general solution containing exactly n arbitrary constants, representing an entire family of curves all satisfying the same equation. The equation d squared y/dx squared = 6x, of order 2, has general solution y = x cubed + Ax + B, containing two arbitrary constants A and B, matching the order. Verify this by substitution: differentiating once gives dy/dx = 3x squared + A, and differentiating again gives d squared y/dx squared = 6x, exactly matching the original equation, for any choice of A and B whatsoever. A particular solution fixes those constants to specific values, usually using extra given information such as the value of y (and possibly dy/dx) at a specific point, called initial conditions. If it is additionally known that y = 5 when x = 0, and dy/dx = 2 when x = 0, substitute into the general solution: y(0) = 0 + 0 + B = 5, so B = 5; dy/dx(0) = 0 + A = 2, so A = 2. The particular solution is therefore y = x cubed + 2x + 5, one single specific curve picked out from the entire family the general solution described.

Every example so far started with the equation already given and asked for the solution. It is worth seeing the process run the other way too: starting from a known family of curves, and finding the differential equation they all satisfy. This is called forming a differential equation, and the method is to differentiate the family's equation enough times to eliminate every arbitrary constant, leaving behind a single equation relating x, y, and derivatives only. Take the family of parabolas y = A x squared, for any constant A, all opening upward from the origin but with different steepness. Differentiate once: dy/dx = 2Ax. Since this still contains the constant A, solve the original equation for A: A = y/(x squared), then substitute into the differentiated equation: dy/dx = 2(y/x squared)(x) = 2y/x. Rearranged, this gives the differential equation x(dy/dx) = 2y, containing no trace of A at all, satisfied by every single parabola in the entire family y = Ax squared, regardless of which specific A was chosen. One differentiation was needed because the family had exactly one arbitrary constant; a family with two constants would generally need two differentiations to eliminate both.

The easiest type of differential equation to solve is one where dy/dx can be rearranged so every term involving y (including dy) sits on one side, and every term involving x (including dx) sits on the other, called variables separable. Solve dy/dx = xy. Rearrange by dividing both sides by y and multiplying both sides by dx: (1/y) dy = x dx. Integrate both sides independently: the integral of (1/y) dy = the integral of x dx, giving log|y| = (x squared)/2 + C. Solving for y by exponentiating both sides: |y| = e to the power of [(x squared)/2 + C] = e to the power C times e to the power [(x squared)/2], and since e to the power C is itself just some positive constant, rename it k, giving the general solution y = k times e to the power [(x squared)/2], for an arbitrary constant k (allowed to be positive or negative once the absolute value is dropped, matching the equation's actual solutions). Only one integration constant is written even though both sides were integrated separately, since combining the two separate constants from each side algebraically produces just one overall arbitrary constant in the end.

One of the most widely used applications of variables separable is any situation where a quantity's rate of change is proportional to the quantity's own current size, written dP/dt = kP, for some constant k. This single equation describes both growth (k positive) and decay (k negative). Suppose a bacterial culture starts with 500 cells and, being closely monitored, is found to be growing at a rate proportional to its current population. Separating variables: (1/P) dP = k dt, integrating to log(P) = kt + C, so P = P0 times e to the power (kt), where P0 is the population at t=0. Here P0 = 500. If the culture is observed to double after 4 hours, then 1000 = 500 times e to the power (4k), giving e to the power (4k) = 2, so k = (log 2)/4. The population after any time t is then P(t) = 500 times e to the power [t times (log 2)/4]; after 10 hours, P(10) = 500 times e to the power [10(log2)/4] = 500 times 2 to the power (2.5), which works out to approximately 2828 cells. The exact same equation, with k negative instead, describes radioactive decay: a radioactive sample's mass falls as M(t) = M0 times e to the power (-kt), and the time taken for the mass to fall to exactly half its starting value, called the half-life, is found the same way, by setting M(t) = M0/2 and solving for t.

Some equations cannot be separated directly, but yield to a specific substitution once recognised as homogeneous, meaning the right-hand side can be written purely as a function of the ratio y/x, with x and y never appearing separately, only in that one combined ratio. Solve dy/dx = (x squared + y squared)/(xy). Divide numerator and denominator by x squared: dy/dx = [1 + (y/x) squared] / (y/x), which depends only on the ratio y/x, confirming the equation is homogeneous. Substitute v = y/x, so y = vx, and by the product rule, dy/dx = v + x(dv/dx). The original equation becomes v + x(dv/dx) = (1 + v squared)/v. Rearranging: x(dv/dx) = (1+v squared)/v - v = (1 + v squared - v squared)/v = 1/v. This is now variables separable in v and x: v dv = (1/x) dx, integrating to (v squared)/2 = log|x| + C. Substitute back v = y/x: (y squared)/(2x squared) = log|x| + C, or equivalently y squared = 2x squared [log|x| + C]. The substitution v = y/x is the single trick that turns an otherwise unmanageable mixed-variable equation into one where the variables genuinely separate.

A first-order linear differential equation has the form dy/dx + Py = Q, where P and Q are functions of x alone (never of y). Unlike most other methods in this chapter, the trick for solving this type, multiplying by an integrating factor, can be derived rather than simply asserted. Look for a function I(x) to multiply the whole equation by, chosen specifically so the left side collapses into the derivative of a single product, d/dx(Iy). Expanding that target using the product rule: d/dx(Iy) = I(dy/dx) + y(dI/dx). Comparing this to I times the left side of the original equation, I(dy/dx) + IPy, these two expressions match exactly if dI/dx = IP, which is itself a variables-separable equation in I: (1/I) dI = P dx, integrating to log(I) = the integral of P dx, so I = e to the power of [the integral of P dx]. This is the integrating factor. Multiplying the original equation through by this I turns the left side into exactly d/dx(Iy), so the whole equation becomes d/dx(Iy) = IQ, which integrates directly to Iy = the integral of (IQ) dx + C, giving y = (1/I) times [the integral of (IQ) dx + C]. Solve dy/dx + (1/x)y = x, for x greater than 0. Here P = 1/x, so the integrating factor is I = e to the power [the integral of (1/x) dx] = e to the power [log(x)] = x. Multiply through by x: x(dy/dx) + y = x squared, and the left side is exactly d/dx(xy). Integrating both sides: xy = the integral of x squared dx = (x cubed)/3 + C, so y = (x squared)/3 + C/x.