First-Order Linear ODE

What is a first-order linear ODE?

An equation is first-order linear in $y$ when it can be written in the standard form $\frac{dy}{dx} + P(x) \, y = Q(x)$ where $P(x)$ and $Q(x)$ depend on $x$ only (not on $y$). "Linear" means $y$ and $y'$ appear to the first power and never multiply each other.

The given equation $\dfrac{dy}{dx} + 2y = 6$ is already in standard form with the constants $P(x) = 2$ and $Q(x) = 6$.

When $P$ and $Q$ are constants, the solution has a clean structure: $y(x) = y_{\text{equilibrium}} + \bigl(y_0 - y_{\text{equilibrium}}\bigr) e^{-Px}$ where the equilibrium (steady state) is $y_{\text{eq}} = Q/P$. Every solution relaxes to the equilibrium if $P > 0$.

Step-by-step solution (the integrating factor way)

Step 1 — Integrating factor.

Multiply both sides by $\mu(x) = e^{\int P \, dx} = e^{2x}$. The left side collapses into a single derivative: $e^{2x} \frac{dy}{dx} + 2 e^{2x} y = \frac{d}{dx}\!\left[ e^{2x} y \right]$

So the equation becomes $\frac{d}{dx}\!\left[ e^{2x} y \right] = 6 \, e^{2x}$

Step 2 — Integrate both sides. $e^{2x} y = \int 6 \, e^{2x} \, dx = 3 e^{2x} + C$

Step 3 — Solve for $y$.

Divide through by $e^{2x}$: $\boxed{\, y(x) = 3 + C e^{-2x} \,}$

Interpretation — equilibrium + decay

The solution has two pieces:

• Particular / steady-state piece: $y_{\text{eq}} = 3$. This is the equilibrium — the value where $dy/dx = 0$: $0 + 2 \cdot y = 6 \implies y = 3.$
• Transient piece: $C e^{-2x}$. This dies off exponentially (because $-2 < 0$). The time scale for decay is $\tau = 1/P = 1/2$.

Any initial condition simply picks how far above or below $3$ you start, and then the solution relaxes back to $3$.

Initial value problem

If $y(0) = 5$, then $5 = 3 + C \cdot e^{0} \implies C = 2$ $y(x) = 3 + 2 e^{-2x}$

Quick checks:

• $y(0) = 3 + 2 = 5$ ✓
• As $x \to \infty$, $e^{-2x} \to 0$, so $y \to 3$. ✓

Verification

Compute $\dfrac{dy}{dx}$ from $y = 3 + C e^{-2x}$: $\frac{dy}{dx} = -2 C e^{-2x}$

Plug into $\dfrac{dy}{dx} + 2y$: $-2 C e^{-2x} + 2 (3 + C e^{-2x}) = -2C e^{-2x} + 6 + 2C e^{-2x} = 6 \quad \checkmark$

Why the equilibrium is $Q/P$

At equilibrium $dy/dx = 0$, so the ODE reduces to $0 + P \cdot y_{\text{eq}} = Q$ → $y_{\text{eq}} = Q/P = 6/2 = 3$. Whenever $P > 0$ and $Q$ is constant, you can read off the steady state immediately.

Where this shape appears in applications

• Newton's law of cooling: $T' = -k(T - T_{\text{amb}})$ → temperature relaxes exponentially to the ambient temperature.
• RC circuits: $V' + \dfrac{1}{RC} V = \dfrac{V_{\text{source}}}{RC}$ → capacitor voltage relaxes to the source voltage.
• Drug clearance: $C' + k C = I$ (infusion rate $I$) → drug concentration stabilizes at $I/k$.
• Bank interest with steady withdrawals, terminal velocity under air drag, and many other "approach to a steady state" problems all have this form.

Common mistakes

• Forgetting to divide by the coefficient of $dy/dx$ before reading off $P(x)$. If the equation is $3 y' + 6 y = 12$, divide first: $y' + 2y = 4$.
• Using the wrong integrating factor for non-constant $P(x)$. The rule is $\mu(x) = e^{\int P(x)\, dx}$, with no extra constant.
• Forgetting the $+C$. Linear ODEs have a one-parameter family of solutions; $C$ is how you match the initial condition.
• Confusing $y_{\text{eq}}$ with the initial condition. They are different: $y_{\text{eq}}$ is determined by the ODE alone, the initial condition is extra data that picks out one curve.

Try it in the visualization

Drag the initial value $y(0)$ above and below the equilibrium line $y = 3$, and watch every curve relax back to the equilibrium with the same time constant. Increase the coefficient $P$ to see the decay become faster.