H2 Maths Differential Equations: Connect the Rate to the Initial Condition

Begin with what is changing

In a public SGExams H2 maths-help discussion, students asked how to study the subject and approach integration questions. The discussion also raised the value of defining variables and checking contextual answers. Differential equations bring those ideas together: a successful solution needs both correct integration and a clear account of what the rate describes.

The official 2026 H2 Mathematics syllabus, section 5.5, includes separable differential equations, general and particular solutions, forming equations from situations, and interpreting solutions. This guide uses an original hypothetical model.

Translate the words before separating variables

Suppose the amount of material collected in a process is y grams after t minutes. The model states that its rate of increase is proportional to the difference between 100 grams and the amount already collected. Initially, 20 grams have been collected. Take the proportionality constant as 0.2 per minute.

The rate is dy/dt. The remaining difference is 100 − y, so the equation is dy/dt = 0.2(100 − y), with y(0) = 20. The constant has units of inverse minutes: multiplying it by a quantity in grams gives a rate in grams per minute.

Check the sign before solving. At y = 20, the rate is positive. As y approaches 100 from below, the rate gets smaller. A sketch should therefore start at 20, rise, and flatten towards 100. This forecast gives you something independent against which to check the algebra.

Separate and integrate carefully

For a solution below 100, separate the variables to obtain dy/(100 − y) = 0.2 dt. Integrating gives −ln|100 − y| = 0.2t + C. The minus sign comes from the derivative of 100 − y with respect to y, which is −1.

Because the branch beginning at 20 remains below 100, write ln(100 − y) = −0.2t + C₁. Exponentiating gives 100 − y = Ae−0.2t. The initial condition gives 100 − 20 = A, so A = 80. The particular solution is y = 100 − 80e−0.2t.

There is also a constant solution y = 100 to the differential equation. Dividing by 100 − y during separation excludes it from that calculation, so recognise it separately. It does not satisfy the initial condition y(0) = 20 and therefore is not the solution to this initial-value problem.

Use the function to answer the actual question

When does the model reach 80 grams? Substitute y = 80 into the particular solution: 20 = 80e−0.2t, so e−0.2t = 1/4. Taking logarithms gives t = ln 4 ÷ 0.2, approximately 6.93 minutes.

The time is positive, as expected. Also distinguish reaching 80 grams from reaching the limiting value of 100 grams. In this model, the exponential term stays positive for every finite t, so the amount approaches 100 without reaching it at a finite time. That is a consequence of the stated mathematical model, not evidence about a real collection process.

Check three different things

First, test the initial condition: at t = 0, the expression gives 100 − 80 = 20. Second, differentiate the answer. Its derivative is 16e−0.2t, which equals 0.2(100 − y). Third, compare its shape with the earlier forecast: it increases towards 100 and its rate decreases.

These checks catch different mistakes. An incorrect constant may fail the initial value while still producing a plausible shape. A missing minus sign may produce growth in the wrong direction. A result without units may conceal a mismatch between the time variable and the proportionality constant.

Make revision specific

Choose a suitable question from the TestPapersFree JC H2 Maths collection. Before solving, write the variable definitions, rate equation and initial condition. After solving, add one derivative check and one sentence interpreting the answer. If you cannot form the equation, practise translating the words; if the equation is correct but integration fails, work on that precise integration step.