IB Analysis and Approaches HL · Integral calculus · Section 7 of 13

Linear first-order differential equations

3 worked examples · By Álvaro López

Key idea

A first-order linear equation, $\frac{d y}{d x} + P (x) y = Q (x)$ , cannot be separated, but if you multiply it by the integrating factor the left-hand side becomes the derivative of a product, and it is enough to integrate. This is HL content and appears in extended-response questions, often with an initial condition.

The standard form

The integrating factor

The idea is to look for a function $μ (x)$ such that $μ y^{'} + μ P y$ is the derivative of $μ y$ . As $(μ y)^{'} = μ^{'} y + μ y^{'}$ , we need $μ^{'} = μ P$ , and this is satisfied by $μ = e^{\int P d x}$ .

To find $μ$ you need only one antiderivative of $P$ : no constant is needed, because any multiplicative constant in $μ$ cancels in the formula.

A very useful trick: $e^{l n g (x)} = g (x)$ , so if $\int P = ln ∣ g ∣$ the integrating factor is directly $g$ (or $∣ g ∣$ , with the appropriate sign).

Worked examples

Example 1 Basic

Constant coefficients

Solve $\frac{d y}{d x} + 2 y = 6$ given that $y = 1$ when $x = 0$ .

Solution

Here $P = 2$ and $Q = 6$ . The integrating factor is $μ = e^{\int 2 d x} = e^{2 x}$ . Multiply: $\frac{d}{d x} (y e^{2 x}) = 6 e^{2 x} \Rightarrow y e^{2 x} = 3 e^{2 x} + C .$ Divide by $e^{2 x}$ : $y = 3 + C e^{- 2 x}$ . With $y = 1$ at $x = 0$ : $1 = 3 + C$ , so $C = - 2$ . $y = 3 - 2 e^{- 2 x} .$

Example 2 Medium

Writing in standard form first

Solve $x \frac{d y}{d x} + 2 y = 4 x$ , for $x > 0$ , given that $y = \frac{5}{3}$ when $x = 1$ .

Solution

Divide by $x$ to have coefficient $1$ on $y^{'}$ : $\frac{d y}{d x} + \frac{2}{x} y = 4$ . Then $P = \frac{2}{x}$ and $Q = 4$ .

Integrating factor: $μ = e^{\int \frac{2}{x} d x} = e^{2 l n x} = x^{2}$ . Multiply: $\frac{d}{d x} (x^{2} y) = 4 x^{2} \Rightarrow x^{2} y = \frac{4 x ^{3}}{3} + C \Rightarrow y = \frac{4 x}{3} + \frac{C}{x ^{2}} .$ With $y = \frac{5}{3}$ at $x = 1$ : $\frac{5}{3} = \frac{4}{3} + C$ , so $C = \frac{1}{3}$ . $y = \frac{4 x}{3} + \frac{1}{3 x ^{2}} .$

Example 3 Hard

With a trigonometric function

Consider the differential equation $\frac{d y}{d x} + y tan x = cos^{2} x, - \frac{π}{2} < x < \frac{π}{2},$ with $y = 2$ when $x = 0$ .

a) Find the integrating factor.

b) Find the general solution.

c) Find the particular solution and calculate $y (\frac{π}{3})$ .

Solution

a) Here $P = tan x$ , and $\int tan x d x = - ln ∣ cos x ∣$ . As $cos x > 0$ on the interval, $μ = e^{- l n c o s x} = \frac{1}{cos x} .$

b) Multiply by $μ$ : $\frac{d}{d x} (\frac{y}{cos x}) = \frac{cos ^{2} x}{cos x} = cos x$ . Integrate: $\frac{y}{cos x} = sin x + C \Rightarrow y = sin x cos x + C cos x .$

c) With $y = 2$ at $x = 0$ : $2 = 0 + C \cdot 1$ , so $C = 2$ and $y = sin x cos x + 2 cos x .$ For $x = \frac{π}{3}$ : $y = \frac{3}{2} \cdot \frac{1}{2} + 2 \cdot \frac{1}{2} = \frac{3}{4} + 1$ .

In the exam: marks are given for writing down $μ$ and recognising that $μ y$ has derivative $μ Q$ . Do not forget to divide the constant by $μ$ when solving for $y$ : here $C$ ends up multiplied by $cos x$ .

Common mistakes

Summary

Standard form: $\frac{d y}{d x} + P (x) y = Q (x)$ (coefficient $1$ on $y^{'}$ ).
Integrating factor: $μ = e^{\int P d x}$ (one antiderivative is enough, with no constant).
Multiplying: $(μ y)^{'} = μ Q$ , and therefore $y = \frac{1}{μ} (\int μ Q d x + C)$ .
Divide the constant by $μ$ too, and apply the initial condition at the end.
It is HL content.

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