First-Order Linear Equations

The hook: a bank account that both earns and spends

Picture a savings account with balance y(t) that earns continuous interest at rate r — so left alone it would grow like y' = ry — while you simultaneously deposit or withdraw money at some outside rate d(t) (positive for a deposit, negative for a withdrawal). The balance obeys both effects at once:

\frac{dy}{dt} = r\,y + d(t) \quad\Longleftrightarrow\quad y' - r\,y = d(t).

The two influences — the account's own exponential growth and your outside interference — are hopelessly tangled together on the right; there is no way to shuffle y and t onto separate sides. This exact shape, "the unknown's own rate of change plus a multiple of itself, driven by an outside term," is the first-order linear equation, and it turns out to have a clean trick of its own.

Not every first-order ODE is separable — the variables often refuse to come apart, exactly as in the bank-account equation above. 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.

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.

A second worked example: y' + 2y = e^{-x}

Here P(x) = 2 (a constant) and Q(x) = e^{-x}.

Step 1 — the integrating factor. \mu = e^{\int 2\,dx} = e^{2x}.

Step 2 — multiply through by \mu — every term, including Q:

e^{2x}y' + 2e^{2x}y = e^{2x}\cdot e^{-x} \quad\Longrightarrow\quad \big(e^{2x}y\big)' = e^{x}.

Step 3 — integrate both sides: e^{2x}y = e^{x} + C.

Step 4 — divide by e^{2x}:

\boxed{\,y = e^{-x} + C\,e^{-2x}.}

Check: y' = -e^{-x} - 2Ce^{-2x}, so y' + 2y = (-e^{-x} - 2Ce^{-2x}) + 2(e^{-x} + Ce^{-2x}) = e^{-x}. ✓ Once again the answer splits into a forced piece e^{-x} and a homogeneous transient Ce^{-2x} that dies out.

A third worked example: a mixing tank, with real numbers

A tank holds V = 200 litres of brine; it starts with y(0) = 20 kg of dissolved salt. Fresh brine flows in at r = 5 L/min with concentration c_{\text{in}} = 0.4 kg/L (so salt arrives at r c_{\text{in}} = 2 kg/min), and the well-stirred mixture drains out at the same rate, carrying salt away at \tfrac{r}{V}y = \tfrac{5}{200}y = 0.025\,y kg/min. "Rate in minus rate out" gives

y' = 2 - 0.025\,y \quad\Longleftrightarrow\quad y' + 0.025\,y = 2.

Integrating factor: \mu = e^{0.025t}. Multiplying through and collapsing the left side:

\big(e^{0.025t}y\big)' = 2\,e^{0.025t} \;\Longrightarrow\; e^{0.025t}y = 80\,e^{0.025t} + C \;\Longrightarrow\; y = 80 + C\,e^{-0.025t}.

Apply y(0) = 20: 20 = 80 + C, so C = -60, giving

\boxed{\,y(t) = 80 - 60\,e^{-0.025t}\ \text{kg}.}

As t \to \infty the salt content creeps up to 80 kg and stays there — exactly V c_{\text{in}} = 200 \times 0.4, the equilibrium the vignette above predicted, reached by decaying away from the starting imbalance of 60 kg.

These two mistakes are the most common way to wreck an otherwise correct calculation:

The integrating-factor trick is old. Gottfried Leibniz worked out how to solve exactly this class of equation in 1694, trading letters with Johann Bernoulli about the general "linear" differential equation — barely a decade after he and Newton had invented calculus itself.

Three centuries later the same equation quietly runs the electronics in your pocket. Charging a capacitor C through a resistor R toward a supply voltage V obeys q' + \tfrac{1}{RC}q = V/R — constant P = 1/RC, exactly like the mixing tank's constant P = r/V above. Swap "salt" for "charge" and "litres" for "coulombs" and it is the identical differential equation, right down to the exponential approach to equilibrium.

The transient fading, live

Below is the solution y = x - 1 + Ce^{-x} of y' + y = x. Slide the constant C (set by the initial value): no matter where the curve starts, the Ce^{-x} transient decays and every solution is drawn toward the same particular solution, the dashed line y = x - 1.