Multiply a linear equation by the right function of the input and its left-hand side becomes the derivative of a single product. After that one integration finishes the job. The right function is the exponential of the integral of the coefficient of the unknown, and it is called the integrating factor; this lesson is about using it, and the next is about why it works.

A lock whose key is cut from the lock itself. The coefficient p is already in the equation, and exponentiating its integral makes the one multiplier that folds a sum of two terms into a single derivative.

Solutions of $y'+2y=4$. The factor $e^{2x}$ gives $y=2+Ce^{-2x}$, and whatever the starting value the term in $C$ dies away: every solution settles at two.

For $y'+p(x)y=q(x)$, let $\mu(x)=e^{\int p(x)\,dx}$. Multiplying by turns the left-hand side into . Integrating gives , and dividing by gives the general solution. Any antiderivative of will do: adding a constant to it only multiplies by a constant, which divides out.

• For $y'+2y=4$: $\mu=e^{2x}$, , so and .

The integrating factor is built from $p$ alone, once the equation is in standard form.

The constant of integration appears when $(\mu y)'=\mu q$ is integrated, and it is divided by $\mu$ along with everything else.

Great work! You now know how to:

• compute the integrating factor from p, with the right sign
• turn the left-hand side into the derivative of a product and integrate
• put the constant in before dividing, so the whole family survives
• find the level every solution settles at when the coefficients are constant