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3.8 Differential equations

Syllabus
9709–2028–2029
Topic
3.8
Level
A2

Learning objectives

3.8.1Differential equations• formulate a simple statement involving a rate of change as a differential equation The introduction and evaluation of a constant of proportionality, where necessary, is included.3.8.2Differential equations• find by integration a general form of solution for a first order differential equation in which the variables are separable Including any of the integration techniques from topic 3.5 above.3.8.3An initial condition to find a• use an initial condition to find a particular solution3.8.4Differential equations• interpret the solution of a differential equation in the context of a problem being modelled by the equation. Where a differential equation is used to model a 'real-life' situation, no specialised knowledge of the context will be required.

Translate rate language into a signed differential equation

Statement about y(t)y(t) Differential equation
rate proportional to yy dy/dt=kydy/dt=ky
rate of decrease proportional to yy dy/dt=−kydy/dt=-ky, k>0k>0
rate proportional to A−yA-y dy/dt=k(A−y)dy/dt=k(A-y)

Name dependent/independent variables, convert “rate of change” to a derivative, make the right side depend only on stated quantities, introduce kk with sign/units, then use any given value and rate at the same instant to evaluate kk.

If $M$ decreases at a rate proportional to $M$ and when $M=50$ kg its rate is $-3$ kg h$^{-1}$, then\frac{dM}{dt}=-kM,\quad -3=-50k,\quad k=0.06\ \text{h}^{-1}.

Check dimensions: in dy/dt=kydy/dt=ky, kk has units inverse to the independent variable. State an initial relation separately if supplied.

This objective is formulation, not solution. Integrating factors are not part of the syllabus; do not introduce them.

Separate, integrate and state the general solution

ForFor\frac{dy}{dx}=f(x)g(y),movevariablestomove variables to\frac{1}{g(y)},dy=f(x),dxwheredivisionislegal,thenintegratebothsidesandcombineconstants.where division is legal, then integrate both sides and combine constants.

Check constant solutions from g(y)=0g(y)=0 before dividing. Separate completely, use any required technique from 3.5, include one arbitrary constant, then rearrange explicitly or leave a valid implicit relation as appropriate.

\frac{dy}{dx}=xy\Rightarrow\frac{dy}{y}=x,dx\Rightarrow\ln|y|=\frac{x^2}{2}+C,so $y=Ae^{x^2/2}$; $A=0$ includes the constant solution $y=0$.

Differentiate the general solution and substitute it into the original equation. Record intervals/domain restrictions created by logarithms or division.

Only separable first-order equations are required. Do not use a first-order linear integrating factor method.

Use one initial point to select a particular solution

A general first-order solution contains one arbitrary constant. Substitute the stated condition y(x0)=y0y(x_0)=y_0 into that integrated family to determine it, then state the resulting particular solution.

Solve generally first, apply the actual initial point (not automatically x=0x=0), solve the constant exactly, and check both the differential equation and the initial condition.

If $dy/dx=3x^2-2$ and $y(1)=4$, theny=x^3-2x+C,\quad4=1-2+C,so $C=5$ and $y=x^3-2x+5$.

Choose the branch/interval containing the initial point when logs, square roots or implicit relations create multiple possibilities.

This syllabus objective concerns first-order equations; do not add second-order condition counting or unrelated initial-velocity theory.

Translate a solution back into model meaning and limits

Feature of solution Context question
variable and derivative what quantity/rate, with what units?
sign/monotonicity increasing or decreasing when?
constants/initial value what do they represent?
limiting value or growth what happens as time increases?
domain when are time and predicted values physically meaningful?

If M(t)=50e−0.06tM(t)=50e^{-0.06t} kg for t≥0t\ge0, then M(0)=50M(0)=50 kg, the model predicts continuous decay, M(t)>0M(t)>0, and M(t)→0M(t)\to0 as t→∞t\to\infty. The constant 0.060.06 has units h−1^{-1} if tt is hours.

To find when $M$ reaches $20$ kg, solve50e^{-0.06t}=20\Rightarrow t=-\frac{\ln(0.4)}{0.06}.Interpretthepositiveresultinhours,notasanewmodelconstant.Interpret the positive result in hours, not as a new model constant.

Also verify the formula satisfies the ODE and initial data, but algebraic validity is only the start: state the contextual conclusion in words and sensible accuracy.

A mathematically valid formula may be unrealistic outside the stated time/value range. Do not claim negative time, negative populations/masses or indefinite model validity without support.