Second Order Differential Equations

A second-order differential equation involves the second derivative $\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}$dx2d2y​. The linear constant-coefficient case, $a y'' + by' + cy = f(x)$ay′′+by′+cy=f(x), is one of the most important topics in Further Mathematics because it governs every oscillating system — springs, pendulums, electrical circuits, vibrating structures. The solution strategy is elegant and universal: solve the auxiliary equation for the complementary function, find a particular integral for the right-hand side, and add them. This lesson develops all three root-cases, the PI trial table (including the duplication trap), and the modelling of simple harmonic motion, damping and resonance.

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1 Spec mapping (AQA 7367)

Second-order linear ODEs are compulsory Pure content for Paper 1 and Paper 2 (each 2 h, 100 marks, 33⅓%, with per-paper weighting AO1 55% / AO2 25% / AO3 20%), the capstone of the "Further calculus" strand. The standard solve — auxiliary equation, complementary function, particular integral, apply conditions — is AO1, but the topic is heavy with AO2: recognising which CF form the roots demand, knowing when a trial PI must be multiplied by $x$x, and interpreting damping regimes all reward reasoning and communication. A modelling scenario (a mass on a damped spring, an $LCR$LCR circuit) reaching from set-up to physical interpretation is firmly AO3. Method order matters for marks: CF first, then PI, conditions last — applying boundary values before adding the PI is a guaranteed loss.

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2 Core theory — the auxiliary equation

A second-order linear constant-coefficient ODE has the form

$a\frac{\mathrm{d}^2y}{\mathrm{d}x^2} + b\frac{\mathrm{d}y}{\mathrm{d}x} + cy = f(x),$adx2d2y​+bdxdy​+cy=f(x),

with $a,b,c$a,b,c constant. If $f(x)=0$f(x)=0 it is homogeneous; otherwise non-homogeneous. For the homogeneous case $ay''+by'+cy=0$ay′′+by′+cy=0, try $y=e^{mx}$y=emx. Then $y'=me^{mx},\ y''=m^2e^{mx}$y′=memx, y′′=m2emx, and substituting,

$\big(am^2 + bm + c\big)e^{mx} = 0.$(am2+bm+c)emx=0.

Since $e^{mx}\neq0$emx=0, this forces the auxiliary equation

$\boxed{\,am^2 + bm + c = 0\,}.$am2+bm+c=0​.

The nature of its two roots dictates the complementary function (CF) — the general solution of the homogeneous equation, carrying two arbitrary constants (as a second-order equation must):

Roots of the auxiliary equation	Complementary function
Distinct real $m_1, m_2$m1​,m2​	$y = Ae^{m_1 x} + Be^{m_2 x}$y=Aem1​x+Bem2​x
Repeated real $m$m	$y = (A + Bx)e^{mx}$y=(A+Bx)emx
Complex $m = p\pm qi$m=p±qi	$y = e^{px}\big(A\cos qx + B\sin qx\big)$y=epx(Acosqx+Bsinqx)

The repeated-root case needs the extra factor $x$x because $e^{mx}$emx alone supplies only one independent solution; the complex case uses Euler's formula $e^{(p\pm qi)x} = e^{px}(\cos qx \pm i\sin qx)$e(p±qi)x=epx(cosqx±isinqx), whose real and imaginary parts give the $\cos$cos/$\sin$sin pair.

Why exactly two constants? Integrating a second-order equation is, loosely, "integrating twice", so two arbitrary constants appear — and the general solution is therefore a two-parameter family. This is why two pieces of information (typically $y(x_0)$y(x0​) and $y'(x_0)$y′(x0​), or two boundary values) are needed to pin down a particular solution, exactly one more than a first-order equation requires. The two independent functions in the CF ($e^{m_1x}$em1​x and $e^{m_2x}$em2​x, or $e^{mx}$emx and $xe^{mx}$xemx, or $e^{px}\cos qx$epxcosqx and $e^{px}\sin qx$epxsinqx) form a basis for all homogeneous solutions: every solution is a linear combination of them, and the constants $A,B$A,B are the coordinates.

It is worth seeing why the repeated-root case needs the $x$x. If the auxiliary equation has a double root $m$m, then $e^{mx}$emx is one solution, but we need a second, independent one. Substituting the trial $y = xe^{mx}$y=xemx into $ay''+by'+cy$ay′′+by′+cy and using $b = -2am$b=−2am and $c = am^2$c=am2 (which a double root forces, since $am^2+bm+c = a(m')^2$am2+bm+c=a(m′)2 has the repeated root) shows every term cancels — so $xe^{mx}$xemx genuinely solves the equation. The factor $x$x is not a guess but a necessity whenever the two roots coincide.

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3 Worked examples with M1/A1 mark scheme (homogeneous)

Worked Example 1 — distinct real roots

Solve $y'' - 5y' + 6y = 0$y′′−5y′+6y=0. Auxiliary equation $m^2 - 5m + 6 = 0$m2−5m+6=0, i.e. $(m-2)(m-3)=0$(m−2)(m−3)=0, so $m=2$m=2 or $m=3$m=3. (M1 form & solve the AE; A1 roots.) Hence

$y = Ae^{2x} + Be^{3x}.$y=Ae2x+Be3x.

(A1 for the CF.)

Worked Example 2 — repeated root

Solve $y'' - 4y' + 4y = 0$y′′−4y′+4y=0. AE $m^2-4m+4 = (m-2)^2 = 0$m2−4m+4=(m−2)2=0, so $m=2$m=2 (repeated). The CF therefore carries the $x$x-factor:

$y = (A + Bx)e^{2x}.$y=(A+Bx)e2x.

Worked Example 3 — complex roots

Solve $y'' + 2y' + 5y = 0$y′′+2y′+5y=0. AE $m^2+2m+5=0$m2+2m+5=0:

$m = \frac{-2 \pm \sqrt{4-20}}{2} = \frac{-2\pm\sqrt{-16}}{2} = -1 \pm 2i.$m=2−2±4−20​​=2−2±−16​​=−1±2i.

(M1 quadratic formula; A1 for $-1\pm2i$−1±2i.) With $p=-1,\ q=2$p=−1, q=2,

$y = e^{-x}\big(A\cos 2x + B\sin 2x\big).$y=e−x(Acos2x+Bsin2x).

(A1 CF.)

Worked Example 4 — with initial conditions

Solve $y'' - y' - 2y = 0$y′′−y′−2y=0 with $y(0)=1,\ y'(0)=4$y(0)=1, y′(0)=4. AE $m^2-m-2 = (m-2)(m+1)=0$m2−m−2=(m−2)(m+1)=0, so $y = Ae^{2x} + Be^{-x}$y=Ae2x+Be−x and $y' = 2Ae^{2x} - Be^{-x}$y′=2Ae2x−Be−x. Apply the conditions after writing the general solution:

$\begin{aligned} y(0)=1 &: \quad A + B = 1, \\ y'(0)=4 &: \quad 2A - B = 4. \end{aligned}$y(0)=1y′(0)=4​:A+B=1,:2A−B=4.​

Adding, $3A = 5\Rightarrow A = \tfrac53$3A=5⇒A=35​, then $B = -\tfrac23$B=−32​. (M1 set up simultaneous equations; A1 both constants.) Hence

$y = \tfrac53 e^{2x} - \tfrac23 e^{-x}.$y=35​e2x−32​e−x.

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4 Non-homogeneous equations: CF + PI

For $ay'' + by' + cy = f(x)$ay′′+by′+cy=f(x) the general solution is

$y = y_{\text{CF}} + y_{\text{PI}},$y=yCF​+yPI​,

where $y_{\text{CF}}$yCF​ solves the homogeneous equation (as above) and $y_{\text{PI}}$yPI​ is any one solution of the full equation, found by trying a form that matches $f(x)$f(x):

$f(x)$f(x)	Trial $y_{\text{PI}}$yPI​
constant $k$k	$\lambda$λ
$px+q$px+q	$\lambda x + \mu$λx+μ
$p\,e^{kx}$pekx	$\lambda e^{kx}$λekx
$p\sin kx$psinkx or $p\cos kx$pcoskx	$\lambda\cos kx + \mu\sin kx$λcoskx+μsinkx
polynomial of degree $n$n	general polynomial of degree $n$n

The duplication rule. If your trial PI already appears in the CF, multiply it by $x$x (and by $x^2$x2 if $xe^{mx}$xemx is also in the CF, i.e. for a repeated root). Otherwise the trial gives $0=f(x)$0=f(x), a contradiction.

Worked Example 5 — exponential RHS

Solve $y'' - 3y' + 2y = e^{4x}$y′′−3y′+2y=e4x. CF: $m^2-3m+2=(m-1)(m-2)=0\Rightarrow y_{\text{CF}} = Ae^{x} + Be^{2x}$m2−3m+2=(m−1)(m−2)=0⇒yCF​=Aex+Be2x. PI: $e^{4x}$e4x is not in the CF, so try $y_{\text{PI}} = \lambda e^{4x}$yPI​=λe4x. Then $y'_{\text{PI}} = 4\lambda e^{4x},\ y''_{\text{PI}} = 16\lambda e^{4x}$yPI′​=4λe4x, yPI′′​=16λe4x; substituting,

$16\lambda - 12\lambda + 2\lambda = 1 \;\implies\; 6\lambda = 1 \;\implies\; \lambda = \tfrac16.$16λ−12λ+2λ=1⟹6λ=1⟹λ=61​.

(M1 differentiate & substitute; A1 $\lambda=\tfrac16$λ=61​.) So

$y = Ae^{x} + Be^{2x} + \tfrac16 e^{4x}.$y=Aex+Be2x+61​e4x.

Worked Example 6 — trigonometric RHS

Solve $y'' + y = \sin 2x$y′′+y=sin2x. CF: $m^2+1=0\Rightarrow m=\pm i\Rightarrow y_{\text{CF}} = A\cos x + B\sin x$m2+1=0⇒m=±i⇒yCF​=Acosx+Bsinx. PI: try $y_{\text{PI}} = \lambda\cos 2x + \mu\sin 2x$yPI​=λcos2x+μsin2x (not in the CF). Then $y''_{\text{PI}} = -4\lambda\cos 2x - 4\mu\sin 2x$yPI′′​=−4λcos2x−4μsin2x, so

$y''_{\text{PI}} + y_{\text{PI}} = -3\lambda\cos 2x - 3\mu\sin 2x = \sin 2x.$yPI′′​+yPI​=−3λcos2x−3μsin2x=sin2x.

Comparing coefficients: $-3\lambda = 0\Rightarrow\lambda=0$−3λ=0⇒λ=0 and $-3\mu = 1\Rightarrow\mu = -\tfrac13$−3μ=1⇒μ=−31​. (M1 compare coefficients; A1 both.) Hence

$y = A\cos x + B\sin x - \tfrac13\sin 2x.$y=Acosx+Bsinx−31​sin2x.

Worked Example 7 — when the trial PI duplicates the CF

Solve $y'' - 2y' + y = e^{x}$y′′−2y′+y=ex. CF: $m^2-2m+1=(m-1)^2=0\Rightarrow y_{\text{CF}} = (A+Bx)e^{x}$m2−2m+1=(m−1)2=0⇒yCF​=(A+Bx)ex. PI: the natural trial $\lambda e^{x}$λex is in the CF; so is $\lambda x e^{x}$λxex (the repeated root gives both $e^x$ex and $xe^x$xex). So go up to $y_{\text{PI}} = \lambda x^2 e^{x}$yPI​=λx2ex. Differentiating,

$y'_{\text{PI}} = \lambda e^{x}(x^2 + 2x), \qquad y''_{\text{PI}} = \lambda e^{x}(x^2 + 4x + 2).$yPI′​=λex(x2+2x),yPI′′​=λex(x2+4x+2).

Substituting into $y''-2y'+y$y′′−2y′+y:

$\lambda e^{x}\big[(x^2+4x+2) - 2(x^2+2x) + x^2\big] = \lambda e^{x}\cdot 2 = e^{x} \;\implies\; \lambda = \tfrac12.$λex[(x2+4x+2)−2(x2+2x)+x2]=λex⋅2=ex⟹λ=21​.

(M1 recognise both $e^x,xe^x$ex,xex lie in the CF and try $x^2e^x$x2ex; A1 $\lambda=\tfrac12$λ=21​.) The $x$x-terms cancel identically (a good check), leaving

$y = (A + Bx)e^{x} + \tfrac12 x^2 e^{x}.$y=(A+Bx)ex+21​x2ex.

Worked Example 8 — polynomial RHS

Solve $y'' + 3y' + 2y = 4x + 6$y′′+3y′+2y=4x+6. CF: $m^2+3m+2=(m+1)(m+2)=0\Rightarrow y_{\text{CF}} = Ae^{-x} + Be^{-2x}$m2+3m+2=(m+1)(m+2)=0⇒yCF​=Ae−x+Be−2x. PI: try $y_{\text{PI}} = \lambda x + \mu$yPI​=λx+μ, so $y'_{\text{PI}}=\lambda,\ y''_{\text{PI}}=0$yPI′​=λ, yPI′′​=0:

$0 + 3\lambda + 2(\lambda x + \mu) = 4x + 6 \;\implies\; 2\lambda x + (3\lambda + 2\mu) = 4x + 6.$0+3λ+2(λx+μ)=4x+6⟹2λx+(3λ+2μ)=4x+6.

Comparing: $2\lambda = 4\Rightarrow\lambda=2$2λ=4⇒λ=2; $3(2)+2\mu = 6\Rightarrow\mu=0$3(2)+2μ=6⇒μ=0. So $y_{\text{PI}} = 2x$yPI​=2x and

$y = Ae^{-x} + Be^{-2x} + 2x.$y=Ae−x+Be−2x+2x.

Worked Example 8b — the full process with initial conditions

This is the most common complete-question format: CF, then PI, then conditions applied to the sum. Solve $y'' - y' - 6y = e^{2x}$y′′−y′−6y=e2x with $y(0)=0,\ y'(0)=0$y(0)=0, y′(0)=0. CF: $m^2-m-6 = (m-3)(m+2)=0\Rightarrow y_{\text{CF}} = Ae^{3x} + Be^{-2x}$m2−m−6=(m−3)(m+2)=0⇒yCF​=Ae3x+Be−2x. PI: $e^{2x}$e2x is not in the CF, so try $\lambda e^{2x}$λe2x; substituting, $(4-2-6)\lambda = 1\Rightarrow -4\lambda = 1\Rightarrow\lambda = -\tfrac14$(4−2−6)λ=1⇒−4λ=1⇒λ=−41​. (M1 CF; M1 PI substitution; A1 $\lambda=-\tfrac14$λ=−41​.) So the general solution and its derivative are

$y = Ae^{3x} + Be^{-2x} - \tfrac14 e^{2x}, \qquad y' = 3Ae^{3x} - 2Be^{-2x} - \tfrac12 e^{2x}.$y=Ae3x+Be−2x−41​e2x,y′=3Ae3x−2Be−2x−21​e2x.

Now apply the conditions to the full solution (never to the CF alone):

$\begin{aligned} y(0)=0 &: \quad A + B - \tfrac14 = 0 \;\Rightarrow\; A+B = \tfrac14, \\ y'(0)=0 &: \quad 3A - 2B - \tfrac12 = 0 \;\Rightarrow\; 3A - 2B = \tfrac12. \end{aligned}$y(0)=0y′(0)=0​:A+B−41​=0⇒A+B=41​,:3A−2B−21​=0⇒3A−2B=21​.​

Solving: from the first, $A = \tfrac14 - B$A=41​−B; substituting, $3(\tfrac14 - B) - 2B = \tfrac12 \Rightarrow \tfrac34 - 5B = \tfrac12 \Rightarrow B = \tfrac{1}{20}$3(41​−B)−2B=21​⇒43​−5B=21​⇒B=201​, so $A = \tfrac15$A=51​. (M1 simultaneous equations; A1 both constants.) Hence

$y = \tfrac15 e^{3x} + \tfrac1{20} e^{-2x} - \tfrac14 e^{2x}.$y=51​e3x+201​e−2x−41​e2x.

The order is the whole discipline of the topic: CF, then PI, then conditions on the sum — applying the conditions before adding the PI is the single most common and costly mistake.

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5 Specimen-style exam question

(Specimen-style — not a real past paper.)

A mass on a spring satisfies $\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} + 4\dfrac{\mathrm{d}x}{\mathrm{d}t} + 3x = 6$dt2d2x​+4dtdx​+3x=6, with $x(0)=0$x(0)=0 and $\dfrac{\mathrm{d}x}{\mathrm{d}t}(0)=0$dtdx​(0)=0. (a) Find the complementary function and a particular integral. (b) Hence find $x(t)$x(t), and state the value approached by $x$x as $t\to\infty$t→∞.

(a) CF: auxiliary equation $m^2+4m+3 = (m+1)(m+3)=0$m2+4m+3=(m+1)(m+3)=0, so $m=-1,-3$m=−1,−3 and $x_{\text{CF}} = Ae^{-t} + Be^{-3t}$xCF​=Ae−t+Be−3t. PI: the RHS is the constant $6$6; try $x_{\text{PI}} = \lambda$xPI​=λ, giving $3\lambda = 6\Rightarrow\lambda = 2$3λ=6⇒λ=2. So $x_{\text{PI}} = 2$xPI​=2.

(b) General solution $x = Ae^{-t} + Be^{-3t} + 2$x=Ae−t+Be−3t+2, with $\dot x = -Ae^{-t} - 3Be^{-3t}$x˙=−Ae−t−3Be−3t. Apply conditions:

$\begin{aligned} x(0)=0 &: \quad A + B + 2 = 0 \;\Rightarrow\; A + B = -2, \\ \dot x(0)=0 &: \quad -A - 3B = 0 \;\Rightarrow\; A = -3B. \end{aligned}$x(0)=0x˙(0)=0​:A+B+2=0⇒A+B=−2,:−A−3B=0⇒A=−3B.​

Then $-3B + B = -2\Rightarrow B = 1$−3B+B=−2⇒B=1, so $A=-3$A=−3. Hence

$x(t) = -3e^{-t} + e^{-3t} + 2.$x(t)=−3e−t+e−3t+2.

As $t\to\infty$t→∞, both exponentials decay, so $x \to 2$x→2 — the steady-state equilibrium displacement. (The transient $-3e^{-t}+e^{-3t}$−3e−t+e−3t dies away; this is an over-damped approach to equilibrium, with no oscillation since the roots are real.)

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6 Modelling: SHM, damping and resonance

Second-order ODEs are the mathematics of oscillation.

Simple harmonic motion. $\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} = -\omega^2 x$dt2d2x​=−ω2x has AE $m^2 = -\omega^2$m2=−ω2, $m = \pm\omega i$m=±ωi, so $x = A\cos\omega t + B\sin\omega t = R\cos(\omega t - \varphi)$x=Acosωt+Bsinωt=Rcos(ωt−φ): undamped oscillation of angular frequency $\omega$ω.

Damped oscillations. $\dfrac{\mathrm{d}^2x}{\mathrm{d}t^2} + 2k\dfrac{\mathrm{d}x}{\mathrm{d}t} + \omega^2 x = 0$dt2d2x​+2kdtdx​+ω2x=0, AE $m^2 + 2km + \omega^2 = 0$m2+2km+ω2=0, discriminant $4k^2 - 4\omega^2$4k2−4ω2. The damping regime is decided by the roots: