Linear equations and the integrating factor

What you will be able to do

• write a linear equation in the standard form and find its integrating factor;
• solve it, with the constant fixed from a point;
• solve the same kind of equation with $x$ as the unknown function of $y$, and read the transient and the steady state.

Pour hot water into a mug

You pour hot water into a mug in a cool room. The temperature falls towards the room temperature. The rate of fall depends on how far the water is above the room. An equation of the type $\dfrac{dy}{dx} + ay = b$ describes this, and the solution settles to a fixed value. The simulation below shows how.

An equation is linear (of the first order) when $y$ and $\dfrac{dy}{dx}$ appear only to the first power and are not multiplied together. In standard form

where $P$ and $Q$ are functions of $x$, or constants. The coefficient of $\dfrac{dy}{dx}$ must be $1$. If it is not, divide first.

Where the integrating factor comes from

The product rule says $\dfrac{d}{dx}(I\,y) = I\dfrac{dy}{dx} + \dfrac{dI}{dx}\,y$. Multiply the standard form by a function $I(x)$:

The left side becomes $\dfrac{d}{dx}(Iy)$ if $\dfrac{dI}{dx} = IP$. This is a separable equation for $I$, and its solution is $I = e^{\int P\,dx}$. We need only one such $I$, so no constant is added. This $I$ is the integrating factor (IF). Now

The steps

1. Standard form. Divide so that the equation reads $\dfrac{dy}{dx} + Py = Q$.
2. Integrating factor. $I = e^{\int P\,dx}$.
3. Integrate. $I\,y = \displaystyle\int I\,Q\,dx + C$.
4. Divide by $I$. The constant is divided too. Then use the point to fix $C$.

The same method works when $x$ is the unknown. The equation $\dfrac{dx}{dy} + P_1x = Q_1$, with $P_1$ and $Q_1$ functions of $y$, has the integrating factor $e^{\int P_1\,dy}$ and the solution $x\cdot\mathrm{IF} = \int Q_1\cdot\mathrm{IF}\,dy + C$.

Worked examples

Example 1. Solve $x\dfrac{dy}{dx} + 2y = x^2$ for $x > 0$.

1. Divide by $x$: $\dfrac{dy}{dx} + \dfrac{2}{x}y = x$. So $P = \dfrac{2}{x}$.
2. $I = e^{\int 2/x\,dx} = e^{2\ln x} = x^2$.
3. $x^2y = \displaystyle\int x^2\cdot x\,dx = \dfrac{x^4}{4} + C$.
4. Divide by $x^2$: $y = \dfrac{x^2}{4} + \dfrac{C}{x^2}$.

Example 2. Solve $\dfrac{dy}{dx} + y\tan x = \sin 2x$ for $-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}$.

1. $P = \tan x$, so $I = e^{\int\tan x\,dx} = e^{\ln\sec x} = \sec x$.
2. $y\sec x = \displaystyle\int \sin 2x\sec x\,dx$. Since $\sin 2x = 2\sin x\cos x$, the integrand is $2\sin x$.
3. $y\sec x = -2\cos x + C$, so $y = -2\cos^2 x + C\cos x$.

Example 3. Solve $y\,dx + (x - y^3)\,dy = 0$ for $y > 0$. Here $x$ is the unknown.

1. Divide by $y\,dy$: $\dfrac{dx}{dy} + \dfrac{x}{y} = y^2$. So $P_1 = \dfrac{1}{y}$ and the IF is $y$.
2. $xy = \displaystyle\int y\cdot y^2\,dy = \dfrac{y^4}{4} + C$.

Example 4. Transient and steady state. Solve $\dfrac{dy}{dx} + 2y = 6$ with $y = -2$ at $x = 0$.

1. $I = e^{2x}$. So $\dfrac{d}{dx}(ye^{2x}) = 6e^{2x}$ and $ye^{2x} = 3e^{2x} + C$.
2. $y = 3 + Ce^{-2x}$. At $x = 0$: $-2 = 3 + C$, so $C = -5$.
3. $y = 3 - 5e^{-2x}$.

The term $3$ is the steady state, because it stays when time passes. The term $-5e^{-2x}$ is the transient, which dies away. Whatever the start, every solution settles to $y = 3$.

Practise with new numbers

Try it in Python

Euler's method steps along the slope: $y_{\text{next}} = y + h\,f(x, y)$. The program uses it on $\dfrac{dy}{dx} + 2y = x$ with $y(0) = 1$, and compares the result with the exact answer from the integrating factor.

Common mistakes

• Not dividing first. In $x\dfrac{dy}{dx} + 2y = x^2$ the coefficient of $\dfrac{dy}{dx}$ is $x$, so $P = \dfrac{2}{x}$, not $2$.
• A wrong integrating factor. Check $\dfrac{dI}{dx} = IP$. For $P = \dfrac{2}{x}$ we get $I = x^2$, not $2x$.
• Not multiplying the right side by $I$. The integral is $\int IQ\,dx$, not $\int Q\,dx$.
• Forgetting to divide the constant. In $y = \dfrac{x^2}{4} + \dfrac{C}{x^2}$ the constant is divided by $I$ too.
• Using the method on a nonlinear equation. A term such as $y^2$ or $y\dfrac{dy}{dx}$ breaks linearity.

Remember these

• Standard form: $\dfrac{dy}{dx} + Py = Q$. Integrating factor: $I = e^{\int P\,dx}$.
• $I\,y = \int IQ\,dx + C$.
• For $\dfrac{dx}{dy} + P_1x = Q_1$ use $e^{\int P_1\,dy}$.
• A solution is a transient part, which fades, plus a steady part.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the integrating factor of $\dfrac{dy}{dx} + \dfrac{3}{x}y = x^2$ for $x > 0$.

2. Solve $\dfrac{dy}{dx} + y = e^{-x}$.

3. Solve $\dfrac{dy}{dx} - \dfrac{y}{x} = x$ for $x > 0$.

4. Solve $\dfrac{dy}{dx} + 2y = \sin x$.

5. Solve $\dfrac{dy}{dx} = \dfrac{1}{x + y}$ by treating $x$ as the unknown.

References

• OpenStax, Calculus Volume 2, Chapter 4, section 4.5 First-order Linear Equations.
• NCERT, Mathematics Part II for Class XII, Chapter 9, the section on linear differential equations.