The Integrating Factor

An equation like $\dfrac{dy}{dx} + 2y = e^x$ can’t be separated: the $y$ and the $e^x$ are added, not multiplied. But it is linear in $y$, and every linear first-order equation can be solved the same way: multiply through by a carefully chosen function, the integrating factor, and the left side becomes the derivative of a product. This is the last of the exact methods in IB Mathematics AA HL, alongside separation of variables and homogeneous equations.

Key ideas

Linear first-order equations in standard form

A first-order differential equation is linear if it can be written in the standard form

$$\frac{dy}{dx} + P(x)\,y = Q(x)$$

where $P$ and $Q$ are functions of $x$ only. The coefficient of $\dfrac{dy}{dx}$ must be $1$. If it isn’t, divide through first. For example, $x\dfrac{dy}{dx} + 2y = x^2$ becomes $\dfrac{dy}{dx} + \dfrac{2}{x}\,y = x$, so $P(x) = \dfrac{2}{x}$ and $Q(x) = x$.

Watch the sign: $\dfrac{dy}{dx} - 3y = x$ has $P(x) = -3$.

The integrating factor

The integrating factor is

$$I(x) = e^{\int P(x)\,dx}$$

You don’t need a $+ C$ in this integral: any one antiderivative works.

Why it works

The integrating factor is chosen so that $I'(x) = P(x)\,I(x)$ (by the chain rule). Multiply the standard form by $I(x)$:

$$I\frac{dy}{dx} + P\,I\,y = I\,Q \quad\Rightarrow\quad I\frac{dy}{dx} + I'\,y = I\,Q$$

The left side is exactly the product rule for $\dfrac{d}{dx}(I\,y)$. So

$$\frac{d}{dx}\big(I(x)\,y\big) = I(x)\,Q(x) \quad\Rightarrow\quad I(x)\,y = \int I(x)\,Q(x)\,dx + C$$

Divide by $I(x)$ to get $y$. Notice that the $+ C$ gets divided by $I(x)$ too.

The method

1. Rearrange into standard form and read off $P(x)$ and $Q(x)$.
2. Find $I(x) = e^{\int P\,dx}$ and simplify it. Useful facts: $e^{\ln x} = x$, $e^{2\ln x} = x^2$, $e^{-\ln x} = \dfrac{1}{x}$, $e^{-\ln \cos x} = \sec x$.
3. Multiply through, and write the left side as $\dfrac{d}{dx}\big(I\,y\big)$.
4. Integrate both sides, with $+ C$ on the right. You may need integration by parts or a substitution.
5. Solve for $y$, then use any initial condition to find $C$.

Check by substituting back: it only takes a minute and catches sign errors.

Some equations are both separable and linear (like $\dfrac{dy}{dx} = 3 - y$). Either method works and gives the same answer.

Worked examples

Example 1: A first one

Solve $\dfrac{dy}{dx} + 2y = e^x$, given that $y = 1$ when $x = 0$.

Solution. It is already in standard form, with $P(x) = 2$ and $Q(x) = e^x$. The integrating factor is $I(x) = e^{\int 2\,dx} = e^{2x}$. Multiply through:

$$\begin{aligned} e^{2x}\frac{dy}{dx} + 2e^{2x}y &= e^{3x} \\ \frac{d}{dx}\left(e^{2x}y\right) &= e^{3x} \\ e^{2x}y &= \frac{1}{3}e^{3x} + C \\ y &= \frac{1}{3}e^{x} + Ce^{-2x} \end{aligned}$$

At $x = 0$: $1 = \dfrac{1}{3} + C$, so $C = \dfrac{2}{3}$ and

$$y = \frac{1}{3}e^x + \frac{2}{3}e^{-2x}$$

Check: $\dfrac{dy}{dx} + 2y = \left(\dfrac{1}{3}e^x - \dfrac{4}{3}e^{-2x}\right) + \left(\dfrac{2}{3}e^x + \dfrac{4}{3}e^{-2x}\right) = e^x$. ✓

Solutions of $\frac{dy}{dx} + 2y = e^x$. The term $Ce^{-2x}$ dies away, so every solution ends up following $y = \frac{1}{3}e^x$.

Example 2: Rearranging into standard form first

Solve $x\dfrac{dy}{dx} + 2y = x^2$ for $x \gt 0$, given that $y = 1$ when $x = 1$.

Solution. Divide by $x$: $\dfrac{dy}{dx} + \dfrac{2}{x}\,y = x$. Then

$$I(x) = e^{\int \frac{2}{x}\,dx} = e^{2\ln x} = e^{\ln x^2} = x^2$$

(For $x \gt 0$ we can drop the absolute value in $\ln|x|$.) Multiply through by $x^2$:

$$\begin{aligned} x^2\frac{dy}{dx} + 2xy &= x^3 \\ \frac{d}{dx}\left(x^2y\right) &= x^3 \\ x^2y &= \frac{x^4}{4} + C \\ y &= \frac{x^2}{4} + \frac{C}{x^2} \end{aligned}$$

At $(1, 1)$: $1 = \dfrac{1}{4} + C$, so $C = \dfrac{3}{4}$ and $y = \dfrac{x^2}{4} + \dfrac{3}{4x^2}$.

Example 3: A trigonometric integrating factor

Solve $\dfrac{dy}{dx} + y\tan x = \cos x$ for $-\dfrac{\pi}{2} \lt x \lt \dfrac{\pi}{2}$, given that $y(0) = 2$. (Radians.)

Solution. $P(x) = \tan x$. Since $\displaystyle\int \tan x\,dx = -\ln|\cos x|$ and $\cos x \gt 0$ on this interval,

$$I(x) = e^{-\ln \cos x} = \frac{1}{\cos x} = \sec x$$

Multiply through by $\sec x$. On the right, $\cos x \sec x = 1$:

$$\frac{d}{dx}\left(y\sec x\right) = 1 \quad\Rightarrow\quad y\sec x = x + C \quad\Rightarrow\quad y = (x + C)\cos x$$

At $x = 0$: $2 = (0 + C)(1)$, so $C = 2$ and $y = (x + 2)\cos x$.

Example 4: A mixing tank

A tank holds $50$ L of pure water. Brine containing $2$ g of salt per litre flows in at $3$ L/min, and the well-mixed solution flows out at $2$ L/min. Let $S$ be the mass of salt (in grams) in the tank after $t$ minutes.

(a) Explain why $\dfrac{dS}{dt} = 6 - \dfrac{2S}{50 + t}$.

(b) Solve the differential equation, and find the concentration of salt after $50$ minutes.

Solution.

(a) Salt flows in at $(2 \text{ g/L})(3 \text{ L/min}) = 6$ g/min. The volume grows by $1$ L each minute, so it is $50 + t$ litres, and the concentration is $\dfrac{S}{50 + t}$ g/L. Salt flows out at $\dfrac{S}{50 + t} \times 2$ g/min. Rate of change = rate in − rate out.

(b) Standard form: $\dfrac{dS}{dt} + \dfrac{2}{50 + t}\,S = 6$. The integrating factor is

$$I(t) = e^{\int \frac{2}{50 + t}\,dt} = e^{2\ln(50 + t)} = (50 + t)^2$$

So

$$\begin{aligned} \frac{d}{dt}\Big((50 + t)^2 S\Big) &= 6(50 + t)^2 \\ (50 + t)^2 S &= 2(50 + t)^3 + C \end{aligned}$$

At $t = 0$ the water is pure, so $S = 0$: $0 = 2(50)^3 + C$, giving $C = -250\,000$. Then

$$S = 2(50 + t) - \frac{250\,000}{(50 + t)^2}$$

At $t = 50$: $S = 2(100) - \dfrac{250\,000}{10\,000} = 200 - 25 = 175$ g, in $100$ L of solution. The concentration is $1.75$ g/L. (In the long run it approaches the inflow concentration of $2$ g/L, which makes sense.)

Common mistakes

Not dividing by the coefficient of dy/dx first. The integrating factor formula only works when the equation starts $\dfrac{dy}{dx} + \ldots$. For $x\dfrac{dy}{dx} + 2y = x^2$, the $P(x)$ is $\dfrac{2}{x}$, not $2$.

Getting the sign of P wrong. For $\dfrac{dy}{dx} - \dfrac{2}{x}y = \ldots$, $P(x) = -\dfrac{2}{x}$ and $I(x) = e^{-2\ln x} = x^{-2}$. Move every $y$ term to the left before reading off $P$.

Forgetting to multiply the right side by I(x). Both sides get multiplied. The left side becomes $\dfrac{d}{dx}(Iy)$ automatically; the right side becomes $I(x)Q(x)$, which is what you integrate.

Dividing only part of the answer by I(x). From $e^{2x}y = \dfrac{1}{3}e^{3x} + C$, the solution is $y = \dfrac{1}{3}e^x + Ce^{-2x}$, not $\dfrac{1}{3}e^x + C$. The constant gets divided too, and that’s what makes the solution curves different shapes.

Not simplifying the integrating factor. Leaving $I(x) = e^{2\ln x}$ makes the next step messy. Use log laws: $e^{2\ln x} = x^2$.

Adding the constant too late. The $+ C$ belongs on the right as soon as you integrate. Adding it after dividing by $I(x)$ gives the wrong family of solutions.

Practice

1. (Warm-up) For each equation, state $P(x)$ and $Q(x)$ and find the integrating factor, simplified.

• (a) $\dfrac{dy}{dx} + 3y = x$
• (b) $\dfrac{dy}{dx} - \dfrac{1}{x}\,y = x^2$, $x \gt 0$
• (c) $\dfrac{dy}{dx} + 2xy = x$

Solution

(a) $P(x) = 3$, $Q(x) = x$, $I(x) = e^{3x}$.

(b) $P(x) = -\dfrac{1}{x}$, $Q(x) = x^2$, $I(x) = e^{-\ln x} = \dfrac{1}{x}$.

(c) $P(x) = 2x$, $Q(x) = x$, $I(x) = e^{x^2}$.

2. (Warm-up) Find the general solution of $\dfrac{dy}{dx} - y = e^{2x}$.

Solution

$P(x) = -1$, so $I(x) = e^{-x}$.

$$\frac{d}{dx}\left(e^{-x}y\right) = e^{-x}e^{2x} = e^x \quad\Rightarrow\quad e^{-x}y = e^x + C \quad\Rightarrow\quad y = e^{2x} + Ce^x$$

3. (Warm-up) Find the general solution of $\dfrac{dy}{dx} + 2xy = 2x$.

Solution

$I(x) = e^{x^2}$.

$$\frac{d}{dx}\left(e^{x^2}y\right) = 2xe^{x^2} \quad\Rightarrow\quad e^{x^2}y = e^{x^2} + C \quad\Rightarrow\quad y = 1 + Ce^{-x^2}$$

(The integral $\displaystyle\int 2xe^{x^2}\,dx = e^{x^2} + C$ uses the substitution $u = x^2$. This equation is also separable; try it that way too.)

4. (Core) Solve $x\dfrac{dy}{dx} - 2y = x^3e^x$, $x \gt 0$, given that $y = 0$ when $x = 1$.

Solution

Divide by $x$: $\dfrac{dy}{dx} - \dfrac{2}{x}\,y = x^2e^x$. Then $I(x) = e^{-2\ln x} = x^{-2}$.

$$\frac{d}{dx}\left(\frac{y}{x^2}\right) = x^{-2}\cdot x^2e^x = e^x \quad\Rightarrow\quad \frac{y}{x^2} = e^x + C$$

At $(1, 0)$: $0 = e + C$, so $C = -e$ and

$$y = x^2\left(e^x - e\right)$$

5. (Core) Solve $\dfrac{dy}{dx} = x - y$ with $y(0) = 2$, and find $y(1)$ to 3 s.f.

Solution

Rearrange: $\dfrac{dy}{dx} + y = x$, so $I(x) = e^x$.

$$\frac{d}{dx}\left(e^xy\right) = xe^x$$

By parts, $\displaystyle\int xe^x\,dx = xe^x - e^x + C$. So $e^xy = (x - 1)e^x + C$ and $y = x - 1 + Ce^{-x}$.

At $x = 0$: $2 = -1 + C$, so $C = 3$ and $y = x - 1 + 3e^{-x}$. Then $y(1) = 3e^{-1} \approx 1.10$ (3 s.f.).

(Compare with the rough estimate of $0.75$ from two steps of Euler’s method for the same equation.)

6. (Core) Solve $\cos x\,\dfrac{dy}{dx} + y\sin x = 1$ for $0 \le x \lt \dfrac{\pi}{2}$, given that $y(0) = 3$.

Solution

Divide by $\cos x$: $\dfrac{dy}{dx} + y\tan x = \sec x$. As in Example 3, $I(x) = \sec x$.

$$\frac{d}{dx}\left(y\sec x\right) = \sec^2 x \quad\Rightarrow\quad y\sec x = \tan x + C \quad\Rightarrow\quad y = \sin x + C\cos x$$

At $x = 0$: $3 = 0 + C$. So $y = \sin x + 3\cos x$.

7. (Core) A pollutant enters a lake, and the mass $P$ kg of pollutant in the lake after $t$ days satisfies

$$\frac{dP}{dt} + 0.1P = 5e^{-0.05t}, \qquad P(0) = 0$$

• (a) Find $P$ in terms of $t$.
• (b) Find the maximum mass of pollutant in the lake, and when it occurs (to 3 s.f.).

Solution

(a) $I(t) = e^{0.1t}$.

$$\frac{d}{dt}\left(e^{0.1t}P\right) = 5e^{0.05t} \quad\Rightarrow\quad e^{0.1t}P = 100e^{0.05t} + C$$

At $t = 0$: $0 = 100 + C$, so $C = -100$ and

$$P = 100\left(e^{-0.05t} - e^{-0.1t}\right)$$

(b) $\dfrac{dP}{dt} = 100\left(-0.05e^{-0.05t} + 0.1e^{-0.1t}\right) = 0$ gives $e^{-0.05t} = 2e^{-0.1t}$, so $e^{0.05t} = 2$ and $t = 20\ln 2 \approx 13.9$ days. Then $e^{-0.05t} = \dfrac{1}{2}$ and $e^{-0.1t} = \dfrac{1}{4}$, so $P = 100\left(\dfrac{1}{2} - \dfrac{1}{4}\right) = 25$ kg. (It is a maximum: $P$ starts at $0$, rises, and tends to $0$ again.)

8. (Challenge) Solve $\dfrac{dy}{dx} + \dfrac{y}{x} = \sin x$, $x \gt 0$, given that $y = 1$ when $x = \pi$.

Solution

$I(x) = e^{\ln x} = x$.

$$\frac{d}{dx}(xy) = x\sin x$$

By parts, with $u = x$ and $dv = \sin x\,dx$: $\displaystyle\int x\sin x\,dx = -x\cos x + \int \cos x\,dx = \sin x - x\cos x + C$. So

$$xy = \sin x - x\cos x + C$$

At $(\pi, 1)$: $\pi = 0 - \pi(-1) + C = \pi + C$, so $C = 0$ and

$$y = \frac{\sin x}{x} - \cos x$$

9. (Challenge) Solve $(x^2 + 1)\dfrac{dy}{dx} + 4xy = x$, given that $y(0) = 0$, and describe what happens to $y$ as $x \to \infty$.

Solution

Divide by $x^2 + 1$: $\dfrac{dy}{dx} + \dfrac{4x}{x^2 + 1}\,y = \dfrac{x}{x^2 + 1}$. Then

$$I(x) = e^{\int \frac{4x}{x^2 + 1}\,dx} = e^{2\ln(x^2 + 1)} = (x^2 + 1)^2$$

Multiply through:

$$\frac{d}{dx}\Big((x^2 + 1)^2 y\Big) = x(x^2 + 1)$$

With $u = x^2 + 1$, $\displaystyle\int x(x^2 + 1)\,dx = \frac{(x^2 + 1)^2}{4} + C$. So $(x^2 + 1)^2 y = \dfrac{(x^2 + 1)^2}{4} + C$, and

$$y = \frac{1}{4} + \frac{C}{(x^2 + 1)^2}$$

At $x = 0$: $0 = \dfrac{1}{4} + C$, so $C = -\dfrac{1}{4}$ and $y = \dfrac{1}{4} - \dfrac{1}{4(x^2 + 1)^2}$.

As $x \to \infty$, the second term tends to $0$, so $y \to \dfrac{1}{4}$ (from below).