A-Level Further Maths 9231 / Topic 2.6 / AS & A Level
Guess an exponential, solve a quadratic.
A linear differential equation with constant coefficients and zero on the right-hand side is solved by guessing \(y = e^{mx}\). The guess turns calculus into algebra: an auxiliary quadratic whose roots, real, repeated or complex, decide the shape of every solution, from steady growth to a dying oscillation.
For \(a\dfrac{d^{2}y}{dx^{2}} + b\dfrac{dy}{dx} + cy = 0\), try \(y = e^{mx}\) to get the auxiliary equation \(am^{2} + bm + c = 0\). Distinct real roots \(m_{1}, m_{2}\): \(y = Ae^{m_{1}x} + Be^{m_{2}x}\). Repeated root \(m\): \(y = (A + Bx)e^{mx}\). Complex roots \(p \pm qi\): \(y = e^{px}(A\cos qx + B\sin qx)\). This is the complementary function (CF); for an equation with a non-zero right-hand side, the general solution is CF + PI, where the particular integral (PI) is any one solution of the full equation.
Fig. 1: the solution of 4y'' + 4y' + 17y = 0 with y(0) = 2, y'(0) = 1 is y = e^(-x/2)(2 cos 2x + sin 2x). The auxiliary roots are -1/2 ± 2i: the real part -1/2 makes the amplitude decay inside the envelope ±√5 e^(-x/2), and the imaginary part 2 fixes the period at π
Section 01
From calculus to algebra.
A linear equation with constant coefficients, such as \(\dfrac{d^{2}y}{dx^{2}} - 5\dfrac{dy}{dx} + 6y = 0\), asks for a function whose derivatives are multiples of itself. Exponentials do exactly that: if \(y = e^{mx}\) then \(y' = me^{mx}\) and \(y'' = m^{2}e^{mx}\). Substituting gives \(e^{mx}(m^{2} - 5m + 6) = 0\), and since \(e^{mx}\) is never zero, \(m^{2} - 5m + 6 = 0\). This quadratic is the auxiliary equation.
Reading it off. \(ay'' + by' + cy = 0\) has auxiliary equation \(am^{2} + bm + c = 0\): \(y''\) becomes \(m^{2}\), \(y'\) becomes \(m\) and \(y\) becomes \(1\). Keep every sign.
First order. \(ay' + by = 0\) gives \(am + b = 0\), so \(y = Ae^{-bx/a}\): one root, one arbitrary constant.
Why two constants. A second order equation needs two integrations to undo, so its general solution has two arbitrary constants. If \(y_{1}\) and \(y_{2}\) are solutions, so is \(Ay_{1} + By_{2}\) (the equation is linear), and two different roots give two independent solutions.
CF, PI and general solution. The solution of the equation with zero on the right is the complementary function. For \(y'' - 5y' + 6y = 12\), the constant \(y = 2\) works (\(6 \times 2 = 12\)); it is a particular integral. The general solution is CF + PI: \(y = Ae^{2x} + Be^{3x} + 2\). Finding PIs is the next lesson.
Worked example Distinct real roots
Find the general solution of \(2\dfrac{d^{2}y}{dx^{2}} + 5\dfrac{dy}{dx} - 3y = 0\).
Auxiliary equation: \(2m^{2} + 5m - 3 = 0\).
Factorise: \((2m - 1)(m + 3) = 0\), so \(m = \tfrac{1}{2}\) or \(m = -3\).
Two distinct real roots give two exponentials, each with its own constant.
Answer: \(y = Ae^{\frac{1}{2}x} + Be^{-3x}\)
Worked example First order, and a missing y term
Find the general solutions of (a) \(3\dfrac{dy}{dx} + 2y = 0\) (b) \(\dfrac{d^{2}y}{dx^{2}} - 9\dfrac{dy}{dx} = 0\).
(a) Auxiliary equation \(3m + 2 = 0\), so \(m = -\tfrac{2}{3}\) and \(y = Ae^{-\frac{2}{3}x}\).
(b) There is no \(y\) term, so \(c = 0\): \(m^{2} - 9m = 0\), giving \(m(m - 9) = 0\).
\(m = 0\) gives \(e^{0x} = 1\), so that part of the solution is the constant \(A\).
The roots go straight into the exponents with their own signs. \(m^{2} - 5m + 6 = (m - 2)(m - 3)\) has roots \(2\) and \(3\), so the CF is \(Ae^{2x} + Be^{3x}\), not \(Ae^{-2x} + Be^{-3x}\). Read the roots from the factors, not the numbers inside the brackets.
Section 02
Repeated and complex roots.
When the auxiliary equation has a repeated root, \(Ae^{mx} + Be^{mx}\) is really \((A + B)e^{mx}\): only one constant. The second independent solution is \(xe^{mx}\). When the roots are complex, \(m = p \pm qi\), Euler's formula \(e^{iqx} = \cos qx + i\sin qx\) turns the two complex exponentials into a real cosine and sine, multiplied by \(e^{px}\).
Roots of \(am^{2} + bm + c = 0\)
Complementary function
distinct real \(m_{1}, m_{2}\) (\(b^{2} > 4ac\))
\(Ae^{m_{1}x} + Be^{m_{2}x}\)
repeated real \(m\) (\(b^{2} = 4ac\))
\((A + Bx)e^{mx}\)
complex \(p \pm qi\) (\(b^{2} < 4ac\))
\(e^{px}(A\cos qx + B\sin qx)\)
imaginary \(\pm ki\) (\(b = 0\), \(ac > 0\))
\(A\cos kx + B\sin kx\)
Worked example A repeated root
Find the general solution of \(4\dfrac{d^{2}y}{dx^{2}} - 4\dfrac{dy}{dx} + y = 0\).
Auxiliary equation: \(4m^{2} - 4m + 1 = 0\), that is \((2m - 1)^{2} = 0\).
For roots \(p \pm qi\) the real part goes in the exponential and the imaginary part inside the trig functions: \(-\tfrac{1}{2} \pm 2i\) gives \(e^{-\frac{1}{2}x}(A\cos 2x + B\sin 2x)\), not \(e^{2x}(A\cos\tfrac{1}{2}x + B\sin\tfrac{1}{2}x)\). And \(m^{2} + 16 = 0\) gives \(\pm 4i\), not \(\pm 4\): the CF is trigonometric, not \(Ae^{4x} + Be^{-4x}\).
Stage 1 · Learn
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What is the auxiliary equation of \(2\dfrac{d^{2}y}{dx^{2}} - 3\dfrac{dy}{dx} + y = 0\)?
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Find the complementary function of \(\dfrac{d^{2}y}{dx^{2}} + 6\dfrac{dy}{dx} + 9y = 0\).
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Find the general solution of \(\dfrac{d^{2}y}{dx^{2}} + 9y = 0\).
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The auxiliary equation has roots \(-2 \pm 5i\). What is the complementary function?
Section 03
Fixing the constants, reading the behaviour.
Two conditions, usually \(y\) and \(\dfrac{dy}{dx}\) at \(x = 0\), fix \(A\) and \(B\) and pick out one curve from the family. For an equation with only a CF (zero on the right), differentiate the CF, substitute the conditions and solve the pair of simultaneous equations. The roots also tell you, before any working, how the solution behaves for large \(x\).
Auxiliary roots
Behaviour as \(x\) increases
both real and negative
decays to 0 without oscillating
a positive real root
grows exponentially (unless that constant is 0)
\(p \pm qi\) with \(p < 0\)
damped oscillation: amplitude shrinks like \(e^{px}\) (Fig. 1)
\(\pm qi\) (\(p = 0\))
undamped oscillation: constant amplitude, period \(\dfrac{2\pi}{q}\)
\(p \pm qi\) with \(p > 0\)
oscillation with growing amplitude
Worked example Real roots and two conditions
Find the solution of \(\dfrac{d^{2}y}{dx^{2}} - \dfrac{dy}{dx} - 2y = 0\) for which \(y = 3\) and \(\dfrac{dy}{dx} = 0\) when \(x = 0\).
\(m^{2} - m - 2 = (m - 2)(m + 1) = 0\), so \(y = Ae^{2x} + Be^{-x}\).
\(x = 0,\ y = 3\): \(A + B = 3\).
\(\dfrac{dy}{dx} = 2Ae^{2x} - Be^{-x}\); at \(x = 0\): \(2A - B = 0\).
Adding: \(3A = 3\), so \(A = 1\) and \(B = 2\).
Answer: \(y = e^{2x} + 2e^{-x}\)
Worked example A damped oscillation
Find the solution of \(4\dfrac{d^{2}y}{dx^{2}} + 4\dfrac{dy}{dx} + 17y = 0\) with \(y = 2\) and \(\dfrac{dy}{dx} = 1\) when \(x = 0\), and describe it for large \(x\).
From the previous section, \(y = e^{-\frac{1}{2}x}(A\cos 2x + B\sin 2x)\). At \(x = 0\): \(A = 2\).
At \(x = 0\): \(-\tfrac{1}{2}A + 2B = 1\), so \(-1 + 2B = 1\) and \(B = 1\).
The real part \(-\tfrac{1}{2}\) is negative, so \(e^{-\frac{1}{2}x} \to 0\): the oscillation is damped and \(y \to 0\).
Answer: \(y = e^{-\frac{1}{2}x}(2\cos 2x + \sin 2x)\), a damped oscillation of period \(\pi\) that dies away (Fig. 1)
Examiner trap
Differentiate the whole product. With \(y = e^{-\frac{1}{2}x}(A\cos 2x + B\sin 2x)\), writing \(y'(0) = 2B\) forgets the \(-\tfrac{1}{2}A\) from differentiating the exponential and gives \(B = \tfrac{1}{2}\) instead of \(1\). For an equation with a non-zero right-hand side, find CF + PI first and only then use the conditions.
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Find the general solution of \(\dfrac{d^{2}y}{dx^{2}} - 2\dfrac{dy}{dx} - 15y = 0\).
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Find the general solution of \(\dfrac{d^{2}y}{dx^{2}} + 4\dfrac{dy}{dx} + 13y = 0\).
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Find the solution of \(\dfrac{d^{2}y}{dx^{2}} - 6\dfrac{dy}{dx} + 9y = 0\) for which \(y = 1\) and \(\dfrac{dy}{dx} = 5\) when \(x = 0\).
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For which equation is every non-zero solution an oscillation of constant amplitude?
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