1The idea in one paragraph

A differential equation is an equation that contains a derivative. A first-order one contains dy/dx and no higher derivative, and it tells you the gradient of the unknown curve at every point: dy/dx = x + y says that at (1, 2) the curve is climbing with gradient 3. Solving it means finding the curve, y as a function of x. Because many curves can obey the same gradient rule, the answer at first contains an arbitrary constant: that is the general solution, a whole family of curves. An initial condition such as y(0) = 1 picks out one member, the particular solution. This page gives you three exact methods, each for a particular shape of equation (separable, homogeneous, linear), and one numerical method, Euler's, which works on any equation but gives numbers, not a formula.

2What a solution looks like

Figure 1 draws the gradient rule dy/dx = x + y as a field of short slope marks: at each point, a mark with the gradient the equation demands there. Any curve that follows the marks is a solution. There are infinitely many, y = Ceˣ − x − 1 for every constant C, and the condition y(0) = 1 picks out C = 2.

You can always check a claimed solution by substituting it. For y = 2eˣ − x − 1: dy/dx = 2eˣ − 1, and x + y = x + 2eˣ − x − 1 = 2eˣ − 1. The two sides match for every x, and y(0) = 2 − 0 − 1 = 1. That is a complete verification, and a "verify" or "show that y = … is a solution" question wants exactly those lines.

3Euler's method

When no exact method works, you can still walk along the slope field. Start at the known point. The equation tells you the gradient there. Take a small step of width h along the tangent line, arrive at a new point, work out the new gradient, step again. The formula booklet writes each step as

Euler's method: y_n+1 = y_n + h × f(x_n, y_n), with x_n+1 = x_n + h, for dy/dx = f(x, y). New y = old y + step × gradient at the old point.

Example 1 (Paper 1). For dy/dx = x + y with y(0) = 1, use Euler's method with h = 0.1 to estimate y(0.3).

Three steps reach x = 0.3, because 0.3 ÷ 0.1 = 3. Lay the working out as a table like this, one row per step; a mistake in one row then costs one mark, not all of them.

Is the estimate too big or too small? The true value, from the exact solution in Figure 1, is y(0.3) = 2e^0.3 − 1.3 ≈ 1.400. Euler's estimate is too low. Figure 2 shows why: each step follows the tangent at the start of the step, and this curve is concave up (it bends upwards), so the tangent always lies below it. For a concave-down curve the steps overshoot instead. That is the reasoning an "explain" question wants.

Making it better. A smaller step keeps each tangent line closer to the curve. Figure 2 compares two step sizes out to x = 1: h = 0.25 gives 2.88, h = 0.1 gives 3.19, and the true value is 2e − 2 ≈ 3.44. Roughly, halving h halves the error. On Paper 2 you run many steps with the GDC's sequence mode or a spreadsheet: set x and y as two recursive sequences, x_n+1 = x_n + 0.1 and y_n+1 = y_n + 0.1(x_n + y_n), and read off the row you need.

4Separating the variables

An equation is separable when dy/dx can be written as a function of x times a function of y: dy/dx = g(x)h(y). Move all the y to the left with the dy, all the x to the right with the dx, and integrate both sides.

Separable: dy/dx = g(x) h(y) → ∫ 1/h(y) dy = ∫ g(x) dx, with one constant, added on the x side.

Example 2. Solve dy/dx = 3x²y, given that y = 2 when x = 0.

Two points of technique. One constant is enough: a constant on each side combines into one. And turning ln|y| = x³ + c into y = Ae^x³ is where many students slip; write e^{x³ + c} = e^c × e^x³ as a line.

Exponential growth and decay, and carbon dating. The simplest separable model is dN/dt = kN: the rate of change is proportional to the amount present. Separating gives N = N₀e^kt. With k > 0 it is growth; with k = −λ < 0 it is decay. Carbon-14 decays this way with a half-life of 5730 years, so e^−5730λ = ½ and λ = ln 2 ÷ 5730 ≈ 1.21 × 10⁻⁴ per year. A bone that has 30% of the carbon-14 it had when alive satisfies e^−λt = 0.3, so t = ln(10/3) ÷ λ ≈ 9950 years (Figure 4). The same equation describes decay curves in physics and first-order reactions in chemistry.

Newton's law of cooling. An object cools at a rate proportional to the difference between its temperature T and the surrounding temperature. In symbols, dT/dt = −k(T − T_s), with k > 0.

Example 3 (Paper 2). Tea at 90 °C is left in a room at 20 °C. After 5 minutes it is at 60 °C. Assuming Newton's law of cooling, find when it reaches 40 °C.

Figure 3 shows the curve: fast cooling at first, when the gap is large, then slower and slower as T approaches 20 °C. The model never reaches 20 °C exactly; say so if asked about the long-term temperature.

5The logistic equation

Exponential growth goes on for ever. Real populations run out of food or space. The logistic equation, named in the guide,

says that growth is proportional both to the population n and to the room left, a − n. When n is small it behaves like exponential growth; as n approaches the carrying capacity a, growth stops. It is separable, and the integral on the n side needs partial fractions (1.11).

Example 4. A lake is stocked with 200 fish. The population N after t years is modelled by dN/dt = 0.0002N(2000 − N). Find N in terms of t.

Watch the sign in the partial fractions: the integral of 1/(2000 − N) is −ln(2000 − N), because the inside has derivative −1. Figure 5 shows the S-shaped logistic curve. The population passes 1000, half the carrying capacity, when 9e^−0.4t = 1, at t = ln 9 ÷ 0.4 ≈ 5.49 years. That is also where it grows fastest: the rate 0.0002N(2000 − N) is a quadratic in N with its maximum at N = 1000, where it is 200 fish per year.

6Homogeneous equations: the substitution y = vx

Some equations cannot be separated as they stand, but the right-hand side depends on x and y only through the ratio y/x. Such an equation, dy/dx = f(y/x), is called homogeneous. Figure 6 shows what that means: on every straight line through the origin, y/x is constant, so the gradient is the same all along the line.

The fix is to make that ratio the new variable. Put v = y/x, so y = vx. Differentiate with the product rule, remembering that v is a function of x:

Homogeneous: for dy/dx = f(y/x), put y = vx, so dy/dx = v + x dv/dx. The equation becomes v + x dv/dx = f(v), which separates.

Example 5. Solve dy/dx = (x² + y²)/(xy), for x > 0, given that y = 2 when x = 1.

Check it is homogeneous: dividing top and bottom by x² gives (1 + (y/x)²)/(y/x), a function of y/x.

The step that loses the most marks is the second line of the working: dy/dx is not dv/dx. It is v + x dv/dx, from the product rule, and without the extra v the equation will not separate properly.

7Linear equations: the integrating factor

A linear first-order equation can be written

with y and dy/dx appearing only to the first power. Multiply the whole equation by the integrating factor I(x) = e^{∫P(x) dx}, from the formula booklet. The left-hand side then becomes the derivative of a product, I(x) × y, which you can integrate directly. It works because I′ = P × I, so by the product rule d/dx(Iy) = Iy′ + PIy = I(y′ + Py).

Integrating factor: for y′ + P(x)y = Q(x), multiply by I = e^{∫P dx}. Then d/dx(I y) = I Q, so I y = ∫ I Q dx. Add the constant before you divide by I.

Example 6. Solve dy/dx + 2y = e^−x, given y(0) = 3.

Example 7: get it into the standard form first. Solve x dy/dx + 2y = x² for x > 0, given y(1) = 1. The coefficient of dy/dx must be 1, so divide by x before finding P.

Simplify e^{2 ln x} to x² every time; it is where the integrating factor usually comes out neat.

Back to Euler. The equation of section 3, dy/dx = x + y, is linear: dy/dx − y = x, with P = −1 and I = e^−x. Then d/dx(ye^−x) = xe^−x, and integration by parts (5.16) gives ye^−x = −xe^−x − e^−x + C, so y = Ceˣ − x − 1, with C = 2 from y(0) = 1. That is the exact solution drawn in Figures 1 and 2, and the reason we could measure Euler's error.

8Which method?

Figure 7 turns the choice into four questions. Try them in order: separation is the quickest when it applies, and Euler is the fallback, or the method a question names.

A question often names the method ("use the substitution y = vx", "use an integrating factor"); then use it. The same equation can suit more than one method: dy/dx = −2y is separable, linear and homogeneous at once, and every method gives y = Ae^−2x.