First-Order Linear ODEs

A first-order linear ODE can be written in standard form as:

$$y' + a(x)y = b(x).$$

The method of integrating factor reduces the problem to a simple integration. Define:

$$\mu(x) = e^{\int a(x),dx}.$$

Multiply both sides of the ODE by $\mu(x)$:

$$\mu(x) y' + \mu(x) a(x) y = \mu(x) b(x).$$

The key observation is that the left-hand side is the derivative of $\mu(x) y$:

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

Integrating both sides:

$$\mu(x) y = \int \mu(x) b(x),dx + C \quad\Rightarrow\quad y = \frac{1}{\mu(x)}\left(\int \mu(x) b(x),dx + C\right).$$

This is the general solution, valid on any interval where $a,b$ are continuous.

The Cauchy problem (initial value problem) consists of the ODE together with an initial condition $y(x_0)=y_0$. This determines the constant $C$ uniquely.

Procedure:

1. Compute the integrating factor $\mu(x)=e^{\int a(x)dx}$.
2. Write the general solution as $y = \frac{1}{\mu(x)}\left(\int \mu(x) b(x),dx + C\right)$.
3. Impose $y(x_0)=y_0$ and solve for $C$.

Existence and uniqueness (Cauchy–Lipschitz theorem): if $a(x)$ and $b(x)$ are continuous on an open interval $I$ containing $x_0$, then there exists a unique function $y=y(x)$ satisfying the ODE and $y(x_0)=y_0$ on $I$.

Example: $y' + 2xy = x$, $y(0)=0$. Here $a(x)=2x$, $b(x)=x$. $\mu = e^{\int 2x dx}=e^{x^2}$. General solution: $y=e^{-x^2}(\int x e^{x^2}dx + C) = e^{-x^2}(\frac12 e^{x^2}+C) = \frac12 + Ce^{-x^2}$. From $y(0)=0$: $0=\frac12 + C \Rightarrow C=-\frac12$. Hence $y=\frac12(1-e^{-x^2})$.

RL circuits: a resistor-inductor circuit is governed by:

$$L\frac{di}{dt} + R i = V(t),$$

where $i(t)$ is the current, $L$ inductance, $R$ resistance, $V(t)$ applied voltage. Dividing by $L$ gives a first-order linear ODE: $i' + (R/L)i = V(t)/L$. The integrating factor is $\mu = e^{Rt/L}$. For constant $V$ and zero initial current:

$$i(t) = \frac{V}{R}(1 - e^{-Rt/L}).$$

The current approaches $V/R$ with time constant $\tau = L/R$.

Logistic growth with harvesting: $y' = r y(1-y/K) - H$. For small $y$, the logistic term is negligible, giving $y' \approx r y - H$, i.e., $y' - r y = -H$ (linear).

Mixing problems: a tank contains $m(t)$ kg of solute. Inflow concentration $c_{in}$ at rate $Q$, outflow at same rate. Mass balance: $m' = Q c_{in} - (Q/V)m$, a first-order linear ODE in $m$.