First-Order Linear ODEs & the Integrating Factor

≈ 25 min

A first-order linear ODE has the standard form

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

Multiply through by the integrating factor $\mu(x) = e^{\int P(x)\, dx}.$ This is engineered so the left side collapses into a single derivative:

\frac{d}{dx}\big(\mu y\big) = \mu\, Q.

Then integrate both sides and divide by $\mu$ .

Worked example. Solve $\dfrac{dy}{dx} + 2y = 8.$ Here $P = 2$ , so $\mu = e^{2x}.$ Then $\dfrac{d}{dx}(e^{2x} y) = 8 e^{2x}$ , giving $e^{2x} y = 4 e^{2x} + C$ , so $y = 4 + C e^{-2x}.$ As $x \to \infty$ the transient dies and $y \to 4$ — the steady state.

For a linear ODE in standard form $\dfrac{dy}{dx} + P(x)\,y = Q(x)$ , the integrating factor is…

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

$e^{\int Q(x)\, dx}$

$\int P(x)\, dx$

$e^{P(x)\,y}$

What is the integrating factor for $\dfrac{dy}{dx} + 3y = x$ ?

$e^{3x}$

$e^{x}$

$e^{3x^2/2}$

$3x$

As $x \to \infty$ , the solution of $\dfrac{dy}{dx} + 5y = 10$ approaches a steady state. Find that steady-state value of $y$ .

The equation $\dfrac{dy}{dx} + 2y = 8$ has general solution $y = 4 + C e^{-2x}$ . If $y(0) = 1$ , find the constant $C$ .