2.6 - First-order linear differential equations | Cambridge A-Level Further Mathematics

Some differential equations cannot be separated, yet become a single product derivative after multiplication by a suitable function. You will identify these linear equations, derive an integrating factor and use an initial condition to select a solution on a valid interval.

Identifying a linear equation

How can you solve $xy'+2y=x^3$ when the terms involving $x$ and $y$ will not separate? The first step is to make the coefficient of the derivative equal to one.

Here and below, $y'=\mathrm{d}y/\mathrm{d}x$ . Dividing by $x$ , on an interval where $x\ne0$ , gives $y'+(2/x)y=x^2$ . The unknown function and its derivative occur to the first power, with coefficients depending only on $x$ .

First-order linear equation

A first-order linear differential equation can be written

y'+P(x)y=Q(x).

The functions $P$ and $Q$ depend only on the independent variable. For example, $y'+xy=\sin x$ is linear; $y'+xy^2=\sin x$ is not linear in $y$ .

“Linear” describes the dependence on the unknown function. The coefficient $P(x)$ itself need not be a linear function of $x$ .

Worked example 1: making the derivative coefficient one

Question. Write $xy'+2y=x^3$ , for $x>0$ , in standard linear form and identify $P(x)$ and $Q(x)$ .

Plan. Divide every term by the nonzero coefficient of $y'$ .

y'+\frac2x y=x^2.

The coefficient of $y'$ is now one, so the coefficient of $y$ is the required $P$ .

Answer. $P(x)=2/x$ and $Q(x)=x^2$ , on $x>0$ .

Check. Multiplication by $x$ restores all three original terms, including the right-hand side $x^3$ .

Change only the sign of the term in $y$ . That sign will change the integrating factor in the next part.

Worked example 2: retaining the signed coefficient

Question. Write $xy'-2y=x^3$ , for $x>0$ , in standard linear form and identify $P(x)$ and $Q(x)$ .

Plan. Divide through by $x$ , retaining the minus sign.

y'-\frac2x y=x^2.

Comparing with $y'+Py=Q$ means that the sign belongs to $P$ .

Answer. $P(x)=-2/x$ and $Q(x)=x^2$ .

Check. At $x=2$ , the normalised equation is $y'-y=4$ . Doubling it gives $2y'-2y=8$ , exactly the original equation at that value of $x$ .

The interval is part of the problem. Dividing by $x$ cannot establish a solution across $x=0$ ; a solution on a negative interval can be treated separately.

Watch out

In $a(x)y'+b(x)y=c(x)$ , the standard-form coefficient is $P=b/a$ , not $b$ . Divide the right-hand side by $a$ as well.

Identify both coefficients in this negative-domain example.

The integrating factor

The product rule explains the multiplier we need: $(x^2y)'=x^2y'+2xy$ , so multiplying $y'+(2/x)y=x^2$ by $x^2$ creates a product derivative.

For a general multiplier $\mu(x)$ , the product rule gives

(\mu y)'=\mu y'+\mu' y.

After multiplying the standard equation by $\mu$ , we need $\mu'=P\mu$ . Separating this auxiliary equation gives a convenient choice $\mu=\exp(\int P\,\mathrm{d}x)$ .

Integrating factor

For $y'+Py=Q$ on an interval where $P$ and $Q$ are continuous, choose

\mu=e^{\int P\,\mathrm{d}x},\qquad (\mu y)'=\mu Q.

Then

y=\frac{\int\mu Q\,\mathrm{d}x+C}{\mu}.

A nonzero constant multiple of $\mu$ works equally well, so no arbitrary constant is needed in the chosen integrating factor. Not in the formula booklet: learn it.

Multiply the whole equation, integrate the right-hand side, then divide the whole result by the integrating factor. This final division also applies to the constant.

Worked example 3: a positive power as integrating factor

Question. Find the general solution of $xy'+2y=x^3$ , for $x>0$ .

Plan. Normalise, multiply by $x^2$ , then integrate the product derivative.

\begin{aligned} y'+\frac2x y&=x^2 \\ \mu&=e^{2\ln x}=x^2 \\ (x^2y)'&=x^4 \\ x^2y&=\frac{x^5}{5}+C. \end{aligned}

The last line is an equation for $x^2y$ , so divide both terms on the right by $x^2$ .

Answer. $y=x^3/5+C/x^2$ , for $x>0$ .

Check. Differentiating gives $y'=3x^2/5-2C/x^3$ . Thus $xy'+2y=(3x^3/5-2C/x^2)+(2x^3/5+2C/x^2)=x^3$ .

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2.6 - First-order linear differential equations

Identifying a linear equation

The integrating factor

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