10 min readPathway sheet1-mark MCQ on the integrating factor; a 3-mark answer when y appears awkwardly, as in y dx − (x + y³) dy = 0

Linear equations: dx/dy + P₁x = Q₁

When x is easier to treat as the unknown, write dx/dy + P₁x = Q₁ with P₁ and Q₁ in y only. Then IF = e^(∫P₁ dy).

Formdx/dy + P₁x = Q₁, where P₁ and Q₁ are functions of y only (or constants)

Integrating factorIF = e^(∫P₁ dy), a function of y

Solutionx × IF = ∫(Q₁ × IF) dy + C

When to use itWhen the DE is linear in x but not in y, e.g. y², eʸ or y³ sits with x in the denominator of dy/dx

When to switch roles

Some DEs are not linear in y but become linear in x when you turn dy/dx upside down. Take y dx − (x + y³) dy = 0. As dy/dx = y/(x + y³), it is not linear in y. But dx/dy = (x + y³)/y = x/y + y², which is dx/dy − (1/y)x = y²: linear in x. Now x is the unknown and y is the independent variable.

Two kinds of linear DE

dy/dx + Py = QUnknown is y; P and Q depend on x. IF = e^(∫P dx). Solution: y × IF = ∫ Q × IF dx + C.

dx/dy + P₁x = Q₁Unknown is x; P₁ and Q₁ depend on y. IF = e^(∫P₁ dy). Solution: x × IF = ∫ Q₁ × IF dy + C.

Ask: which variable appears only to the first power? That one is the unknown.

The method

1
Try the dy/dx form. If y appears as y², eʸ, sin y and so on, flip to dx/dy.
2
Write dx/dy + P₁x = Q₁, with P₁ and Q₁ depending on y only.
3
Find IF = e^(∫P₁ dy).
4
Write x × IF = ∫(Q₁ × IF) dy + C.
5
Integrate with respect to y, then use any given condition.

Example 1: direct

Problem

Solve y dx − (x + y³) dy = 0, y > 0.

Working

Rewrite: y dx = (x + y³) dy, so dx/dy = x/y + y².
Standard form: dx/dy − (1/y)x = y². So P₁ = −1/y, Q₁ = y².
IF = e^(∫ −(1/y) dy) = e^(−log y) = 1/y.
x × (1/y) = ∫ y² × (1/y) dy + C = ∫ y dy + C.
x/y = y²/2 + C.
Check: x = y³/2 + Cy gives dx/dy = 3y²/2 + C, and x/y + y² = y²/2 + C + y² = 3y²/2 + C.

Answer

x = y³/2 + Cy

Example 2: board-style, with a condition

Problem

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

Working

dy/dx = 1/(x + y + 1) is not linear in y. Flip it: dx/dy = x + y + 1.
Standard form: dx/dy − x = y + 1. So P₁ = −1, Q₁ = y + 1.
IF = e^(∫ −1 dy) = e⁻ʸ.
xe⁻ʸ = ∫ (y + 1)e⁻ʸ dy + C.
By parts: ∫ (y + 1)e⁻ʸ dy = −(y + 1)e⁻ʸ − e⁻ʸ = −(y + 2)e⁻ʸ.
So xe⁻ʸ = −(y + 2)e⁻ʸ + C, that is x = −(y + 2) + Ceʸ.
Put y = 0, x = 1: 1 = −2 + C, so C = 3.

Answer

x = 3eʸ − y − 2

How to spot the dx/dy type

y is paired with dx, and x appears only to the first power with dy.
Terms like y², y³, eʸ or tan⁻¹ y sit next to x in the denominator of dy/dx.
MCQ trap: the IF of a dx/dy equation is a function of y, never of x.

In real life

A delivery app may find it easier to model time taken as a function of distance than distance as a function of time. Choosing the convenient unknown is exactly the dx/dy idea.

Exam angle

1 marks: “Find the integrating factor of y(dx/dy) + 2x = y³ (y > 0).” Divide by y: dx/dy + (2/y)x = y². IF = e^(2 log y) = y².

3 marks: “Solve y dx + (x − y²eʸ) dy = 0, y > 0.” dx/dy + (1/y)x = yeʸ. IF = e^(log y) = y. xy = ∫ y²eʸ dy = eʸ(y² − 2y + 2) + C.

Remember it

y awkward? Flip it. dx/dy + P₁x = Q₁ and IF = e^(∫P₁ dy).