Integrating Factor Theorem

The integrating factor method transforms non-exact equations into exact ones by multiplication with a suitable function. This theorem formalizes when and how such factors can be found.

TheoremIntegrating Factor for Linear Equations

Every linear first-order differential equation:

$\frac{dy}{dx} + p(x)y = q(x)$

where $p(x)$ is continuous, has an integrating factor:

$\mu(x) = e^{\int p(x)dx}$

Multiplying the equation by $\mu(x)$ yields:

$\frac{d}{dx}[\mu(x)y] = \mu(x)q(x)$

which can be integrated directly to obtain the general solution.

This theorem guarantees that every linear first-order ODE can be solved by quadratures (integration), making the class of linear equations completely solvable in principle.

Proof Sketch

Given $\frac{dy}{dx} + p(x)y = q(x)$ , multiply by $\mu(x)$ :

$\mu(x)\frac{dy}{dx} + \mu(x)p(x)y = \mu(x)q(x)$

For the left side to equal $\frac{d}{dx}[\mu(x)y]$ , we need:

$\mu(x)\frac{dy}{dx} + \mu'(x)y = \mu(x)\frac{dy}{dx} + \mu(x)p(x)y$

This requires $\mu'(x) = \mu(x)p(x)$ , a separable equation:

$\frac{d\mu}{\mu} = p(x)dx$

Integrating: $\ln|\mu| = \int p(x)dx$ , giving $\mu(x) = e^{\int p(x)dx}$ .

ExampleVariable Coefficient Equation

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

Here $p(x) = 2/x$ , so:

$\mu(x) = e^{\int (2/x)dx} = e^{2\ln x} = x^2$

Multiply through:

$x^2\frac{dy}{dx} + 2xy = x^4$

$\frac{d}{dx}[x^2 y] = x^4$

Integrate:

$x^2 y = \frac{x^5}{5} + C$

$y = \frac{x^3}{5} + \frac{C}{x^2}$

TheoremIntegrating Factors for Non-Exact Equations

Consider the equation $M(x,y)dx + N(x,y)dy = 0$ which is not exact.

(a) If $\frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\right) = h(x)$ depends only on $x$ , then:

$\mu(x) = e^{\int h(x)dx}$

is an integrating factor.

(b) If $\frac{1}{M}\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) = k(y)$ depends only on $y$ , then:

$\mu(y) = e^{\int k(y)dy}$

is an integrating factor.

ExampleFinding an Integrating Factor

Consider $(3xy + y^2)dx + (x^2 + xy)dy = 0$ .

Check exactness: $M = 3xy + y^2$ , $N = x^2 + xy$

$\frac{\partial M}{\partial y} = 3x + 2y, \quad \frac{\partial N}{\partial x} = 2x + y$

Not exact. Try case (a):

$\frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\right) = \frac{(3x+2y)-(2x+y)}{x^2+xy} = \frac{x+y}{x(x+y)} = \frac{1}{x}$

This depends only on $x$ ! So $\mu(x) = e^{\int (1/x)dx} = x$ .

Multiply the original equation by $x$ :

$(3x^2y + xy^2)dx + (x^3 + x^2y)dy = 0$

Now $\frac{\partial M}{\partial y} = 3x^2 + 2xy = \frac{\partial N}{\partial x}$ . Exact!

Remark

While the theorem provides conditions for finding integrating factors depending on one variable, finding general integrating factors (depending on both $x$ and $y$ ) is usually very difficult. The cases where $\mu = \mu(x)$ or $\mu = \mu(y)$ are the most practically useful.

The integrating factor theorem bridges the gap between exact and non-exact equations, showing that many seemingly different equation types can be unified under the exact equation framework.