First Order Differential Equations

A first-order differential equation relates a function $y(x)$y(x) to its first derivative $\dfrac{\mathrm{d}y}{\mathrm{d}x}$dxdy​. They are the language of change: population growth, radioactive decay, cooling coffee, draining tanks and charging capacitors are all first-order ODEs. In Further Mathematics you must solve two standard types — separable equations and linear equations via an integrating factor — and, crucially, set them up from a worded modelling scenario. This lesson develops both methods rigorously, drills the integrating-factor derivation, and applies them to the classic models examiners love.

---

1 Spec mapping (AQA 7367)

First-order differential equations 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%), closing the "Further calculus" strand and leading into second-order equations. Mechanically solving a given equation is AO1 (separate-and-integrate, or build the integrating factor). The richest marks are AO3: forming the differential equation from a modelling description ("the rate of cooling is proportional to the temperature excess"), then interpreting the solution in context. Communicating the modelling assumptions and applying boundary conditions cleanly is AO2. This is a topic where a worded question can be worth 10+ marks, so a disciplined four-step routine — set up, solve, apply conditions, answer the question asked — is essential. It is also one of the most transferable topics in the whole qualification: the same two methods that solve an abstract $\tfrac{\mathrm{d}y}{\mathrm{d}x}$dxdy​ equation solve real problems in physics, chemistry, biology and economics, which is why differential equations sit at the heart of applied mathematics far beyond A-Level.

---

2 Core theory — separable equations

A first-order ODE is separable if it can be written with all the $y$y's on one side and all the $x$x's on the other:

$\frac{\mathrm{d}y}{\mathrm{d}x} = f(x)\,g(y) \;\Longrightarrow\; \frac{1}{g(y)}\,\mathrm{d}y = f(x)\,\mathrm{d}x,$dxdy​=f(x)g(y)⟹g(y)1​dy=f(x)dx,

after which you integrate both sides. The "splitting" of $\tfrac{\mathrm{d}y}{\mathrm{d}x}$dxdy​ is shorthand for the chain rule $\int \tfrac{1}{g(y)}\tfrac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x = \int f(x)\,\mathrm{d}x$∫g(y)1​dxdy​dx=∫f(x)dx; it is rigorous, not a fudge. One arbitrary constant appears (combine the two constants of integration into a single $C$C); a given boundary/initial condition then pins it down.

General versus particular solutions. Integrating once introduces exactly one arbitrary constant, so a first-order ODE has a one-parameter family of solutions — the general solution. Each value of the constant selects one curve from the family; an initial condition $y(x_0)=y_0$y(x0​)=y0​ picks out the single curve through that point, the particular solution. Geometrically the family fills the plane with non-crossing solution curves (a direction field), and the condition names which curve you are on. Always find the general solution first, with its constant, and apply the condition last — applying it mid-calculation is a frequent source of error.

Whether you can integrate the separated equation depends on the functions involved: $\int\tfrac{1}{g(y)}\,\mathrm{d}y$∫g(y)1​dy might need logarithms, partial fractions, an inverse-trig form, or a substitution. Recognising which integration technique the separated form demands is half the skill; the other half is the algebraic rearrangement at the end to make $y$y the subject (not always possible — sometimes an implicit solution $F(x,y)=C$F(x,y)=C is the best you can do).

Worked Example 1 — exponential-type

Solve $\dfrac{\mathrm{d}y}{\mathrm{d}x} = xy$dxdy​=xy, given $y(0)=2$y(0)=2. Separate and integrate:

$\int\frac{1}{y}\,\mathrm{d}y = \int x\,\mathrm{d}x \;\implies\; \ln|y| = \frac{x^2}{2} + C \;\implies\; y = A\,e^{x^2/2}, \quad A = \pm e^{C}.$∫y1​dy=∫xdx⟹ln∣y∣=2x2​+C⟹y=Aex2/2,A=±eC.

(M1 separate variables; A1 integrate both sides.) Apply $y(0)=2$y(0)=2: $2 = A e^{0} = A$2=Ae0=A, so

$y = 2\,e^{x^2/2}.$y=2ex2/2.

(A1 for the particular solution.)

Worked Example 2 — arctan form

Solve $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1+y^2}{1+x^2}$dxdy​=1+x21+y2​. Separate:

$\int\frac{\mathrm{d}y}{1+y^2} = \int\frac{\mathrm{d}x}{1+x^2} \;\implies\; \arctan y = \arctan x + C \;\implies\; y = \tan\big(\arctan x + C\big).$∫1+y2dy​=∫1+x2dx​⟹arctany=arctanx+C⟹y=tan(arctanx+C).

Worked Example 3 — logistic-type with partial fractions

Solve $x\dfrac{\mathrm{d}y}{\mathrm{d}x} = y(1-y)$xdxdy​=y(1−y), given $y(1)=\tfrac12$y(1)=21​. Separate, then split the left side by partial fractions $\dfrac{1}{y(1-y)} = \dfrac1y + \dfrac{1}{1-y}$y(1−y)1​=y1​+1−y1​:

$\int\left(\frac1y + \frac{1}{1-y}\right)\mathrm{d}y = \int\frac{1}{x}\,\mathrm{d}x \;\implies\; \ln|y| - \ln|1-y| = \ln|x| + C.$∫(y1​+1−y1​)dy=∫x1​dx⟹ln∣y∣−ln∣1−y∣=ln∣x∣+C.

(M1 partial fractions; A1 integrate to logs.) So $\ln\left|\dfrac{y}{1-y}\right| = \ln|x| + C$ln​1−yy​​=ln∣x∣+C, giving $\dfrac{y}{1-y} = Ax$1−yy​=Ax. Apply $y(1)=\tfrac12$y(1)=21​: $\dfrac{1/2}{1/2} = A\cdot1$1/21/2​=A⋅1, so $A=1$A=1. Then $y = x(1-y) \Rightarrow y(1+x)=x$y=x(1−y)⇒y(1+x)=x:

$y = \frac{x}{1+x}.$y=1+xx​.

(A1 for the explicit solution.)

Worked Example 3b — a homogeneous equation via substitution

Some equations are not separable as they stand but become so after a substitution. An equation expressible as $\dfrac{\mathrm{d}y}{\mathrm{d}x} = F\!\left(\dfrac{y}{x}\right)$dxdy​=F(xy​) is called homogeneous, and the substitution $v = \dfrac{y}{x}$v=xy​ (so $y = vx$y=vx and, by the product rule, $\dfrac{\mathrm{d}y}{\mathrm{d}x} = v + x\dfrac{\mathrm{d}v}{\mathrm{d}x}$dxdy​=v+xdxdv​) always reduces it to a separable equation in $v$v and $x$x. Solve $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{x+y}{x}$dxdy​=xx+y​. Write the right side as $1 + \dfrac{y}{x} = 1+v$1+xy​=1+v; then

$v + x\frac{\mathrm{d}v}{\mathrm{d}x} = 1 + v \;\implies\; x\frac{\mathrm{d}v}{\mathrm{d}x} = 1 \;\implies\; \int\mathrm{d}v = \int\frac{1}{x}\,\mathrm{d}x \;\implies\; v = \ln|x| + C.$v+xdxdv​=1+v⟹xdxdv​=1⟹∫dv=∫x1​dx⟹v=ln∣x∣+C.

(M1 substitution $y=vx$y=vx; M1 separate in $v$v; A1 integrate.) Reverting via $v=\tfrac{y}{x}$v=xy​,

$y = x\big(\ln|x| + C\big).$y=x(ln∣x∣+C).

(A1 for the explicit solution.) The homogeneous substitution is a standard tool: whenever every term in $\dfrac{\mathrm{d}y}{\mathrm{d}x}$dxdy​ can be written as a function of $\dfrac{y}{x}$xy​ alone, reach for $v=\tfrac{y}{x}$v=xy​. The same change-of-variable philosophy — convert an awkward equation into a familiar separable one — recurs throughout the study of differential equations.

---

3 Linear equations and the integrating factor

A linear first-order ODE has the form

$\frac{\mathrm{d}y}{\mathrm{d}x} + P(x)\,y = Q(x).$dxdy​+P(x)y=Q(x).

Multiply through by the integrating factor $\mu(x) = e^{\int P(x)\,\mathrm{d}x}$μ(x)=e∫P(x)dx. The point of $\mu$μ is that the left side becomes an exact derivative: since $\mu' = \mu P$μ′=μP (by construction),

$\mu\frac{\mathrm{d}y}{\mathrm{d}x} + \mu P\,y = \mu\frac{\mathrm{d}y}{\mathrm{d}x} + \mu' y = \frac{\mathrm{d}}{\mathrm{d}x}\big[\mu y\big].$μdxdy​+μPy=μdxdy​+μ′y=dxd​[μy].

So the equation collapses to $\dfrac{\mathrm{d}}{\mathrm{d}x}[\mu y] = \mu Q$dxd​[μy]=μQ, and integrating once gives $\mu y = \int \mu Q\,\mathrm{d}x$μy=∫μQdx. This is the whole method; the only skill is computing $\mu$μ correctly and recognising the left side as a product-rule derivative.

Why does $\mu = e^{\int P\,\mathrm{d}x}$μ=e∫Pdx work? It is engineered to make $\mu' = \mu P$μ′=μP: differentiating $e^{\int P\,\mathrm{d}x}$e∫Pdx by the chain rule gives $e^{\int P\,\mathrm{d}x}\cdot P = \mu P$e∫Pdx⋅P=μP. With that property, the left side $\mu y' + \mu P y$μy′+μPy is exactly the product-rule expansion of $\dfrac{\mathrm{d}}{\mathrm{d}x}[\mu y]$dxd​[μy], so the equation telescopes into a single derivative that can be undone by one integration. The construction is the same trick — manufacture a factor that turns a sum into an exact derivative — that recurs throughout differential equations. Note that the constant of integration in $\int P\,\mathrm{d}x$∫Pdx can be dropped: it merely multiplies $\mu$μ (and hence both sides) by a constant, which cancels. One subtlety: the equation must be in standard form $y' + P y = Q$y′+Py=Q (coefficient of $y'$y′ equal to $1$1) before you read off $P$P; if it begins as $a(x)y' + b(x)y = c(x)$a(x)y′+b(x)y=c(x), divide through by $a(x)$a(x) first.

Worked Example 4 — $\mu = x$μ=x

Solve $\dfrac{\mathrm{d}y}{\mathrm{d}x} + \dfrac{y}{x} = x$dxdy​+xy​=x for $x>0$x>0. Here $P=\tfrac1x$P=x1​, so $\mu = e^{\int (1/x)\,\mathrm{d}x} = e^{\ln x} = x$μ=e∫(1/x)dx=elnx=x. (M1 for $\mu = e^{\int P}$μ=e∫P; A1 for $\mu = x$μ=x.) Multiply through:

$x\frac{\mathrm{d}y}{\mathrm{d}x} + y = x^2 \;\implies\; \frac{\mathrm{d}}{\mathrm{d}x}[xy] = x^2 \;\implies\; xy = \frac{x^3}{3} + C.$xdxdy​+y=x2⟹dxd​[xy]=x2⟹xy=3x3​+C.

(M1 recognise the exact derivative and integrate.) Hence

$y = \frac{x^2}{3} + \frac{C}{x}.$y=3x2​+xC​.

(A1 general solution.)

Worked Example 5 — exponential $\mu$μ, with a condition

Solve $\dfrac{\mathrm{d}y}{\mathrm{d}x} + 2y = e^{-x}$dxdy​+2y=e−x, given $y(0)=3$y(0)=3. $P=2\Rightarrow \mu = e^{2x}$P=2⇒μ=e2x. Multiply:

$\frac{\mathrm{d}}{\mathrm{d}x}\big[e^{2x}y\big] = e^{2x}e^{-x} = e^{x} \;\implies\; e^{2x}y = e^{x} + C \;\implies\; y = e^{-x} + Ce^{-2x}.$dxd​[e2xy]=e2xe−x=ex⟹e2xy=ex+C⟹y=e−x+Ce−2x.

Apply $y(0)=3$y(0)=3: $3 = 1 + C$3=1+C, so $C=2$C=2 and $y = e^{-x} + 2e^{-2x}$y=e−x+2e−2x. (A1 for $C$C; A1 final.)

Worked Example 6 — trig right-hand side

Solve $\dfrac{\mathrm{d}y}{\mathrm{d}x} - 3y = \sin x$dxdy​−3y=sinx. $P=-3\Rightarrow \mu = e^{-3x}$P=−3⇒μ=e−3x, giving $\dfrac{\mathrm{d}}{\mathrm{d}x}[e^{-3x}y] = e^{-3x}\sin x$dxd​[e−3xy]=e−3xsinx. Using $\int e^{ax}\sin bx\,\mathrm{d}x = \dfrac{e^{ax}(a\sin bx - b\cos bx)}{a^2+b^2}$∫eaxsinbxdx=a2+b2eax(asinbx−bcosbx)​ with $a=-3,\ b=1$a=−3, b=1:

$e^{-3x}y = \frac{e^{-3x}(-3\sin x - \cos x)}{10} + C \;\implies\; y = \frac{-3\sin x - \cos x}{10} + Ce^{3x}.$e−3xy=10e−3x(−3sinx−cosx)​+C⟹y=10−3sinx−cosx​+Ce3x.

---

4 Specimen-style exam question

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

Water leaks from a tank so that the depth $h$h (cm) at time $t$t (minutes) satisfies $\dfrac{\mathrm{d}h}{\mathrm{d}t} = -k\sqrt{h}$dtdh​=−kh​, where $k>0$k>0 is constant. Initially $h=16$h=16, and after $2$2 minutes $h=9$h=9. (a) Solve the differential equation, expressing $\sqrt{h}$h​ as a function of $t$t. (b) Find the time at which the tank is empty.

(a) Separate: $\displaystyle\int h^{-1/2}\,\mathrm{d}h = \int -k\,\mathrm{d}t$∫h−1/2dh=∫−kdt, so $2\sqrt{h} = -kt + C$2h​=−kt+C. At $t=0,\ h=16$t=0, h=16: $2\cdot4 = C$2⋅4=C, so $C=8$C=8. At $t=2,\ h=9$t=2, h=9: $2\cdot3 = -2k + 8\Rightarrow 6 = 8-2k\Rightarrow k=1$2⋅3=−2k+8⇒6=8−2k⇒k=1. Hence

$2\sqrt{h} = 8 - t \;\implies\; \sqrt{h} = 4 - \tfrac{t}{2}, \quad\text{i.e. } h = \left(4 - \tfrac{t}{2}\right)^2.$2h​=8−t⟹h​=4−2t​,i.e. h=(4−2t​)2.

(b) The tank is empty when $h=0$h=0, i.e. $\sqrt{h}=0$h​=0: $4 - \tfrac{t}{2} = 0 \Rightarrow t = 8$4−2t​=0⇒t=8 minutes. (This is Torricelli's law of draining — the depth falls as a downward parabola, reaching zero in finite time.)

---

5 Modelling with first-order ODEs

The exam payoff is forming the equation. Three classics recur:

Population growth / decay. "Rate of change proportional to amount" $\Rightarrow \dfrac{\mathrm{d}P}{\mathrm{d}t} = kP$⇒dtdP​=kP, giving $P = P_0 e^{kt}$P=P0​ekt (growth $k>0$k>0, decay $k<0$k<0).

Newton's law of cooling. "Rate of cooling proportional to the temperature excess over the surroundings $T_s$Ts​" $\Rightarrow \dfrac{\mathrm{d}T}{\mathrm{d}t} = -k(T-T_s)$⇒dtdT​=−k(T−Ts​). Separating, $\int\dfrac{\mathrm{d}T}{T-T_s} = \int -k\,\mathrm{d}t \Rightarrow \ln|T-T_s| = -kt + C$∫T−Ts​dT​=∫−kdt⇒ln∣T−Ts​∣=−kt+C, so $T = T_s + Ae^{-kt}$T=Ts​+Ae−kt.

Mixing problems. Track an amount $S(t)$S(t): $\dfrac{\mathrm{d}S}{\mathrm{d}t} = (\text{rate in}) - (\text{rate out})$dtdS​=(rate in)−(rate out), where each rate is concentration $\times$× flow. If the inflow carries no solute the equation is separable (pure dilution); if the inflow is itself a solution the equation is linear and needs an integrating factor.

The skill examiners test is the translation from words to equation. Phrases map directly: "the rate of increase of $N$N is proportional to $N$N" $\to \dfrac{\mathrm{d}N}{\mathrm{d}t} = kN$→dtdN​=kN; "decreases at a rate proportional to its square" $\to \dfrac{\mathrm{d}N}{\mathrm{d}t} = -kN^2$→dtdN​=−kN2; "the excess over the surroundings" $\to (T-T_s)$→(T−Ts​). Read carefully whether a rate is of increase (positive $k$k) or of decrease (negative, or an explicit minus sign), and whether the proportionality is to the quantity, its square, or some other function — getting this set-up right is worth the first cluster of marks before any calculus begins.

Worked Example 7 — cooling tea

A cup of tea at $90^\circ$90∘C is placed in a $20^\circ$20∘C room; after $5$5 min it is $70^\circ$70∘C. Find its temperature after $10$10 min. From $T = 20 + Ae^{-kt}$T=20+Ae−kt: at $t=0$t=0, $90 = 20 + A\Rightarrow A=70$90=20+A⇒A=70. At $t=5$t=5, $70 = 20 + 70e^{-5k}\Rightarrow e^{-5k} = \tfrac{5}{7}$70=20+70e−5k⇒e−5k=75​. Then at $t=10$t=10,

$T = 20 + 70e^{-10k} = 20 + 70\big(e^{-5k}\big)^2 = 20 + 70\cdot\left(\frac{5}{7}\right)^2 = 20 + 70\cdot\frac{25}{49} = 20 + \frac{250}{7} = \frac{390}{7} \approx 55.7^\circ\text{C}.$T=20+70e−10k=20+70(e−5k)2=20+70⋅(75​)2=20+70⋅4925​=20+7250​=7390​≈55.7∘C.

Note the neat trick: rather than solving for $k$k (which would force a logarithm and a rounded decimal), use $e^{-10k} = (e^{-5k})^2$e−10k=(e−5k)2 — faster, exact, and far less error-prone. Recognising that $t=10$t=10 is twice $t=5$t=5 lets you reuse the value of $e^{-5k}$e−5k directly.

Worked Example 8 — a mixing tank

A tank holds $100$100 L of water with $10$10 kg of salt dissolved. Fresh water enters at $5$5 L/min and the well-mixed solution leaves at $5$5 L/min. Find the salt $S(t)$S(t). The volume stays $100$100 L, so the outflow concentration is $\tfrac{S}{100}$100S​ kg/L; rate in $=0$=0, rate out $= \tfrac{S}{100}\cdot5 = \tfrac{S}{20}$=100S​⋅5=20S​: