Differential equations

Further Maths · Further Pure Mathematics 2. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

A linear differential equation has y and its derivatives only to the first power. A first order one, dy/dx + Py = Q, is solved with an integrating factor: multiplying by it turns the left side into the derivative of a single product.

When the coefficients are constants, the general solution has two parts. The complementary function (CF) solves the equation with the right side replaced by zero and holds the arbitrary constants. The particular integral (PI) is any one function that fits the full equation. Add them.

A given substitution, such as x = e^t or y = ux, changes a harder equation into one of these types. Initial conditions then fix the constants.

Rules to remember

For dy/dx + Py = Q (coefficient of dy/dx must be 1): integrating factor I = e^{∫P dx}, then y × I = ∫(Q × I) dx + C
General solution = complementary function + particular integral; a second order equation has two arbitrary constants, a first order equation has one
First order dy/dx + ay = f(x): the CF is Ae^−ax
Second order ay′′ + by′ + cy = f(x): solve the auxiliary equation am² + bm + c = 0. Distinct real roots m₁, m₂: Ae^m₁x + Be^m₂x. Repeated root m: (A + Bx)e^mx. Complex roots p ± qi: e^px(A cos qx + B sin qx)
PI trial forms: polynomial of the same degree for a polynomial (include every lower power); ke^bx for ae^bx; λ cos px + μ sin px for a cos px + b sin px (always both terms)
If the trial form is already part of the CF, multiply it by x (by x² if it matches a repeated root)
x = e^t gives x dy/dx = dy/dt and x² d²y/dx² = d²y/dt² − dy/dt; y = ux gives dy/dx = u + x du/dx

Common mistake

Students use the initial conditions on the complementary function alone and then add the particular integral. Always write the full general solution (CF + PI) first, and only then substitute the conditions to find A and B.

Worked example

Find the general solution of y′′ + 2y′ − 8y = 16x.

Auxiliary equation: m² + 2m − 8 = 0, so (m − 2)(m + 4) = 0 and m = 2 or −4. CF: Ae^2x + Be^−4x.
Try the PI y = ax + b, so y′ = a and y′′ = 0. Substituting: 0 + 2a − 8(ax + b) = 16x.
Coefficient of x: −8a = 16, so a = −2. Constant: 2a − 8b = 0, so −4 − 8b = 0 and b = −1/2.
Check: 2(−2) − 8(−2x − 1/2) = −4 + 16x + 4 = 16x.

Answer: y = Ae^2x + Be^−4x − 2x − 1/2

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Differential Equations

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