Systems of differential equations

A rocket's altitude affects its fuel burn rate, which in turn affects its altitude; a predator population affects the prey population, which affects the predator population back. Whenever two or more quantities change together and feed into each other's rates of change, a single differential equation is not enough — we need a system of them, solved simultaneously. This closing topic of the MA10001 ODE unit shows that such systems, at least the linear constant-coefficient kind in two unknown functions of one variable, are not a new subject at all. Writing $D \equiv d/dt$ turns the two coupled equations into two algebraic-looking ones, and the same elimination trick used to solve pairs of simultaneous linear equations in school algebra collapses the system to a single higher-order ODE — one we already know how to solve.

The core insight

Writing $D \equiv d/dt$ and treating it as an algebraic symbol lets elimination reduce a linear constant-coefficient system in two unknowns to a single higher-order ODE solvable by the auxiliary-equation method, after which back-substitution into an original first-order equation recovers the second unknown without extra constants.

The core idea

Setting up the system. A linear system with constant coefficients in dependent variables $x(t), y(t)$ and independent variable $t$ looks like

\frac{dx}{dt} = a_1x + b_1y + f_1(t), \qquad \frac{dy}{dt} = a_2x + b_2y + f_2(t),

where: $a_1, b_1, a_2, b_2$ are constants, and $f_1(t), f_2(t)$ are given forcing functions (zero for a homogeneous system). Two coupled first-order equations describe one combined process with a single clock $t$ .

Operator form. Replacing $d/dt$ by the symbol $D$ turns the system into

(D-a_1)x - b_1y = f_1(t), \qquad -a_2x + (D-b_2)y = f_2(t).

where: $D$ acts on whatever follows it ( $Dx = dx/dt$ ), but between operations it can be added, subtracted and multiplied just like an ordinary algebraic symbol — this is the entire trick that makes elimination possible.

Elimination. Exactly as you would eliminate a variable from two simultaneous linear algebraic equations, combine the two operator equations — by substitution or by multiplying through by suitable operator expressions and subtracting — to remove one dependent variable, say $y$ . What remains is a single higher-order linear ODE in $x(t)$ alone, with constant coefficients, solvable by the familiar auxiliary-equation method for the complementary function, plus undetermined coefficients for a particular integral if the system is nonhomogeneous.

Back-substitution, not re-solving. Once $x(t)$ is known, find $y(t)$ by substituting back into one of the original first-order equations and solving algebraically — never by separately eliminating $x$ and solving its own higher-order equation for $y$ . The total number of independent arbitrary constants in the full solution must equal the order of the original system (here, two), and solving each variable independently would wrongly produce four.

Figure

Two coupled tanks labelled x and y with arrows showing each rate of change depending on both levels, next to the operator equations after elimination collapses to one higher order equation in x alone

Derivation AReducing a linear 2x2 system to a single second-order ODE

The goal is to start from the general constant-coefficient system $dx/dt = a_1x + b_1y$ , $dy/dt = a_2x + b_2y$ (homogeneous, so the mechanics of elimination are visible without the extra bookkeeping of a forcing term) and show, purely by treating $D$ as an algebraic symbol, that $x(t)$ obeys a single second-order linear ODE whose coefficients come directly from $a_1, b_1, a_2, b_2$ — the same reduction used, worked-example by worked-example, throughout this topic.

Step 1. Write the system in operator form. With $D \equiv d/dt$ , the two equations become $(D-a_1)x - b_1y = 0$ and $-a_2x + (D-b_2)y = 0$ . Both are now algebraic-looking statements relating $x$ , $y$ and the symbol $D$ .

Step 2. Solve the first equation for $y$ in terms of $x$ (assuming $b_1 \neq 0$ ). From $(D-a_1)x = b_1y$ , $y = \dfrac{(D-a_1)x}{b_1}$ . This is legitimate because $(D-a_1)x$ just means $dx/dt - a_1x$ , an ordinary function of $t$ once $x(t)$ is known — dividing by the constant $b_1$ is nothing exotic.

Step 3. Substitute this expression for $y$ into the second equation. $-a_2x + (D-b_2)\left(\dfrac{(D-a_1)x}{b_1}\right) = 0$ . Multiply through by $b_1$ to clear the fraction: $-a_2b_1x + (D-b_2)(D-a_1)x = 0$ .

Step 4. Expand the operator product $(D-b_2)(D-a_1) = D^2 - (a_1+b_2)D + a_1b_2$ , treating $D$ exactly like a variable when multiplying two linear factors. Substituting back gives $D^2x - (a_1+b_2)Dx + a_1b_2x - a_2b_1x = 0$ , i.e.

D^2x - (a_1+b_2)Dx + (a_1b_2 - a_2b_1)x = 0.

Step 5. Read off the result. This is a single second-order constant-coefficient linear ODE in $x(t)$ alone, solved by the usual auxiliary equation $m^2 - (a_1+b_2)m + (a_1b_2-a_2b_1) = 0$ . Once $x(t)$ is found from its roots, $y(t)$ is recovered by Step 2's formula $y = \dfrac{(D-a_1)x}{b_1} = \dfrac{x' - a_1x}{b_1}$ — an original-equation back-substitution, not a fresh ODE, so no spurious constants appear.

Worked example

Solve the homogeneous system $\dfrac{dx}{dt} = 3x - 2y$ , $\dfrac{dy}{dt} = 2x - 2y$ , matching the general reduction above with $a_1=3, b_1=-2, a_2=2, b_2=-2$ .

Step 1. Write the operator equations: $(D-3)x + 2y = 0$ and $-2x + (D+2)y = 0$ .

Step 2. Eliminate $y$ . From the first equation, $y = \dfrac{(3-D)x}{2}$ . Substituting into the second and clearing the factor of $2$ gives $-4x + (D+2)(3-D)x = 0$ . Expanding $(D+2)(3-D) = -D^2+D+6$ leaves $D^2x - Dx - 2x = 0$ , matching the general formula with $a_1+b_2 = 1$ and $a_1b_2-a_2b_1 = -6-(-4) = -2$ .

Step 3. Solve the auxiliary equation $m^2-m-2=0 \implies (m-2)(m+1)=0 \implies m=2,-1$ , so

x(t) = C_1e^{2t} + C_2e^{-t}.

Step 4. Back-substitute into $\dfrac{dx}{dt}=3x-2y$ to get $y=\dfrac{3x-x'}{2}$ . With $x' = 2C_1e^{2t}-C_2e^{-t}$ ,

y(t) = \frac{3(C_1e^{2t}+C_2e^{-t}) - (2C_1e^{2t}-C_2e^{-t})}{2} = \frac{C_1}{2}e^{2t} + 2C_2e^{-t}.

Final answer: $x(t) = C_1e^{2t}+C_2e^{-t}$ , $y(t) = \frac{C_1}{2}e^{2t}+2C_2e^{-t}$ , with only two arbitrary constants total, matching the system's order.

Competitive stretch

For the second Pass-1 system, $\dfrac{dx}{dt}=y$ , $\dfrac{dy}{dt}=-x$ with $x(0)=1, y(0)=0$ , the same machinery gives $a_1=0, b_1=1, a_2=-1, b_2=0$ , so the reduced equation is $D^2x - 0\cdot Dx + (0-(-1))x = 0$ , i.e. $D^2x+x=0$ . Its auxiliary roots are $m=\pm i$ , giving $x(t)=C_1\cos t+C_2\sin t$ , and back-substitution via $y=Dx$ gives $y(t)=-C_1\sin t+C_2\cos t$ . The initial conditions fix $C_1=1, C_2=0$ , so $x(t)=\cos t$ , $y(t)=-\sin t$ — a solution that traces a unit circle in the $(x,y)$ plane, a useful geometric check that this is a coupled oscillator, not two independent equations.

Common traps

Solving for $x$ and $y$ as if they were two independent second-order ODEs. This produces four arbitrary constants when the first-order system only supports as many as its total order (here two); always find the second variable by back-substitution into an original equation.
Back-substituting into the eliminated higher-order equation instead of an original first-order one. That reintroduces spurious constants that don't actually satisfy the original coupled system.
Sign or operator-algebra slips when expanding products like $(D+2)(3-D)$ . Treat $D$ as a formal symbol and expand term by term, exactly as with any quadratic factorisation.
Skipping the final check. Always substitute both $x(t)$ and $y(t)$ back into both original equations — this is the fastest way to catch an elimination or back-substitution error before it costs marks.

Ask the tutor: if you are stuck on eliminating a variable from a trickier system, or want to check whether your reduced ODE matches the shortcut formula, ask the AI tutor to walk through your elimination step by step.

Does this elimination method work for systems with more than two equations

The elementary elimination method extends in principle to three or more coupled linear equations with constant coefficients, using the same idea of treating D as an algebraic symbol and eliminating variables pairwise. In practice the algebra grows quickly, and larger systems are normally handled with matrix and eigenvalue methods instead, which lie beyond this MA10001 unit. For the two-equation systems in this syllabus, elimination stays the fastest and most transparent route.

What happens if the system is nonhomogeneous, with f1(t) and f2(t) present

The elimination steps are identical: write the operator equations, eliminate one variable, and you get a single higher-order linear ODE in the other variable, but now with a nonzero right-hand side built from f1, f2 and their derivatives. Solve it exactly as any nonhomogeneous constant-coefficient ODE — complementary function from the auxiliary equation, plus a particular integral by undetermined coefficients or another familiar method — then back-substitute into an original equation for the second variable, just as in the homogeneous case.

How do I know I have eliminated the right variable, or that I have not made an algebra slip

Two independent checks work well. First, compare the reduced equation's coefficients to the shortcut formula, D squared x minus (a1 plus b2)Dx plus (a1b2 minus a2b1)x equals 0, derived directly from the system's coefficients — if your elimination does not match this pattern, recheck the operator expansion. Second, and always worth doing regardless, substitute the final x(t) and y(t) back into both original first-order equations; any elimination or back-substitution error shows up immediately as a mismatch.

Systems of differential equations

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads