12 min readPathway sheet2-mark and 3-mark questions, often with an initial condition; growth and decay problems in case-based questions

Solving by separating the variables

If dy/dx = g(x)·h(y), put all the y terms with dy and all the x terms with dx, then integrate both sides and add one constant.

Formdy/dx = g(x)·h(y): a function of x only times a function of y only

Separatedy/h(y) = g(x) dx, for h(y) ≠ 0

Integrate∫ dy/h(y) = ∫ g(x) dx + C (one arbitrary constant is enough)

Growth and decaydy/dt = ky gives y = y₀e^(kt), where y₀ is the value at t = 0

The method

1
Write the DE as dy/dx = g(x)·h(y). Factorise first if needed, e.g. e^(x + y) = eˣ·eʸ.
2
Separate: dy/h(y) = g(x) dx. Every y goes with dy, every x goes with dx.
3
Integrate both sides: ∫ dy/h(y) = ∫ g(x) dx.
4
Add one arbitrary constant C on one side. This is the general solution.
5
If a condition such as y(0) = 1 is given, substitute it to find C. This gives the particular solution.

Example 1: direct

Problem

Find the general solution of dy/dx = e^(3x + 2y).

Working

Use the law of indices: e^(3x + 2y) = e³ˣ·e²ʸ.
So dy/dx = e³ˣ·e²ʸ. Separate: e⁻²ʸ dy = e³ˣ dx.
Integrate both sides: ∫ e⁻²ʸ dy = ∫ e³ˣ dx.
−(1/2)e⁻²ʸ = (1/3)e³ˣ + C
Multiply by 6: −3e⁻²ʸ = 2e³ˣ + 6C.
Write K = −6C: 2e³ˣ + 3e⁻²ʸ = K.

Answer

2e³ˣ + 3e⁻²ʸ = K, where K is an arbitrary constant.

Example 2: with an initial condition

Problem

Find the particular solution of (1 + x²) dy/dx = 2x(y + 1), given that y = 2 when x = 0.

Working

Separate: dy/(y + 1) = 2x/(1 + x²) dx.
Integrate: ∫ dy/(y + 1) = ∫ 2x/(1 + x²) dx.
log|y + 1| = log(1 + x²) + log|C| (writing the constant as log|C| makes the next step easy).
So y + 1 = C(1 + x²). This is the general solution.
Use y = 2 when x = 0: 2 + 1 = C(1 + 0), so C = 3.
y + 1 = 3(1 + x²), so y = 3x² + 2.
Check: dy/dx = 6x. LHS = (1 + x²)(6x) = 6x + 6x³. RHS = 2x(3x² + 3) = 6x³ + 6x.

Answer

y = 3x² + 2

Example 3: money that grows continuously

Problem

₹ 5000 is kept in a bank where it grows continuously at 4% per year. In how many years will it double? (Take logₑ 2 = 0.6931.)

Working

Let A be the amount after t years. dA/dt = (4/100)A = 0.04A.
Separate: dA/A = 0.04 dt.
Integrate: logₑ A = 0.04t + C.
At t = 0, A = 5000, so C = logₑ 5000.
logₑ(A/5000) = 0.04t, that is A = 5000e^(0.04t).
To double, A = 10000: logₑ 2 = 0.04t.
t = 0.6931/0.04 = 17.3275

Answer

About 17.33 years (rounded to 2 decimal places).

Mistakes that cost marks

Adding a constant on both sides. One C is enough.
Leaving an x on the dy side. Separate completely before you integrate.
Writing log y instead of log|y| when y may be negative.
Forgetting to use the given condition. A question that gives y(0) wants a particular solution.

In real life

The rule dN/dt = kN predicts how fast bacteria multiply in milk or cooked rice left out on a hot day. That is why food safety advice says to refrigerate leftovers quickly.

Exam angle

1 marks: “Find the general solution of dy/dx = y/x (x, y ≠ 0).” dy/y = dx/x, so log|y| = log|x| + log|C|. Hence y = Cx.

2 marks: “Solve dy/dx = x³e⁻ʸ.” Separate: eʸ dy = x³ dx. Integrate: eʸ = x⁴/4 + C.

3 marks: “Find the particular solution of dy/dx = 3x²(1 + y²), given y = 1 when x = 0.” dy/(1 + y²) = 3x² dx, so tan⁻¹ y = x³ + C. At x = 0, y = 1: C = tan⁻¹ 1 = π/4. So y = tan(x³ + π/4).

Remember it

Separate, integrate, add one C, then use the condition.