Not every first-order ODE is
separable
— the variables often refuse to come apart. The next class we can always crack is the
first-order linear equation, recognisable because
y and y' appear only to the first power and
are never multiplied together:
\frac{dy}{dx} + P(x)\,y = Q(x).
"Linear" means linear in the unknown y and its derivative; the
coefficients P and Q may be any functions of
x. The trouble is that the left side,
y' + Py, is almost but not quite the derivative of a product. The
whole method is one clever multiplication that fixes that.
The integrating factor, line by line
The idea: multiply the whole equation by a cunning function
\mu(x) chosen so the left side collapses into a single
product derivative (\mu y)'. Then one integration finishes the job.
Step 1 — multiply through by an unknown factor
\mu(x):
\mu\,y' + \mu P\,y = \mu Q.
Step 2 — demand the left side be a product derivative. By the product rule,
(\mu y)' = \mu y' + \mu' y. Comparing with the left side above, the two
match provided
\mu' = \mu P.
Step 3 — solve that little separable ODE for \mu.
The condition \mu' = P\mu is itself separable
(d\mu/\mu = P\,dx), giving \ln\mu = \int P\,dx,
hence the integrating factor
\mu(x) = e^{\int P(x)\,dx}.
Step 4 — rewrite the equation as a single derivative. With this
\mu the left side is exactly (\mu y)':
\big(\mu(x)\,y\big)' = \mu(x)\,Q(x).
Step 5 — integrate both sides in x. The left side
integrates trivially (it is already a derivative); the right side carries the constant:
\mu(x)\,y = \int \mu(x)\,Q(x)\,dx + C.
Step 6 — divide by \mu to isolate
y:
\boxed{\,y = \frac{1}{\mu(x)}\left(\int \mu(x)\,Q(x)\,dx + C\right).}
A worked example: y' + y = x
Here P(x) = 1 and Q(x) = x. Run the recipe.
Step 1 — compute the integrating factor.
\mu = e^{\int 1\,dx} = e^{x}.
Step 2 — multiply through and recognise the product derivative.
e^{x}y' + e^{x}y = x\,e^{x} \quad\Longrightarrow\quad \big(e^{x}y\big)' = x\,e^{x}.
Step 3 — integrate both sides. The right side needs integration by parts
(\int x e^x\,dx = (x-1)e^x):
e^{x}y = (x - 1)e^{x} + C.
Step 4 — divide by e^{x}:
\boxed{\,y = x - 1 + C\,e^{-x}.}
Step 5 — sanity check. Then y' = 1 - Ce^{-x}, and
y' + y = (1 - Ce^{-x}) + (x - 1 + Ce^{-x}) = x. \quad\checkmark
Notice the structure of the answer: a particular solution
x - 1 (which alone satisfies y' + y = x) plus
C e^{-x}, the general solution of the homogeneous equation
y' + y = 0. The transient Ce^{-x} fades as
x \to \infty, leaving the steady line x - 1 —
a pattern that recurs everywhere linear systems appear.
The first-order linear ODE y' + P(x)\,y = Q(x) is solved as follows.
-
Form the integrating factor
\mu(x) = e^{\int P(x)\,dx} (any one antiderivative of
P will do — the constant cancels).
-
Multiplying by \mu turns the left side into a single product
derivative: (\mu y)' = \mu Q.
-
Integrating gives the general solution
y = \frac{1}{\mu(x)}\left(\int \mu(x)\,Q(x)\,dx + C\right).
-
It is the sum of one particular solution and the
homogeneous solution C/\mu(x) = C\,e^{-\int P}; an
initial condition fixes C.
First-order linear ODEs are the equations of things filling and draining. Pour brine into
a tank holding V litres at r L/min with
concentration c_{\text{in}}, draining at the same rate. The amount of
salt y(t) obeys "rate in minus rate out":
\frac{dy}{dt} = r\,c_{\text{in}} - \frac{r}{V}\,y \quad\Longleftrightarrow\quad y' + \frac{r}{V}\,y = r\,c_{\text{in}},
textbook first-order linear with P = r/V constant. The solution
relaxes exponentially to the equilibrium salt content V c_{\text{in}} —
the Ce^{-rt/V} transient dies, the steady state survives.
The same equation runs an RC circuit. A resistor and capacitor in series with a
source voltage V(t) give, by Kirchhoff's law,
R\,q' + q/C = V(t), i.e.
q' + \tfrac{1}{RC}q = V(t)/R. The capacitor charges toward
CV with the famous time constant \tau = RC:
one integrating factor, two textbooks' worth of applications.