Linear First-Order Equations

Linear First-Order Differential Equations

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

\[\frac{dy}{dx} + p(x)y = g(x), \quad y(x_0) = y_0\]

To solve such equations, we need to find a function $\mu = \mu(x)$ to obtain:

\[\mu(x) (\frac{dy}{dx} + p(x)y) = \mu(x) g(x)\]

This $\mu$ should satisfy:

\[\mu(x) (\frac{dy}{dx} + p(x)y) = \frac{d}{dx}(\mu(x) y)\]

Thus:

\[\begin{aligned} \mu(x) (\frac{dy}{dx} + p(x)y) &= \frac{d}{dx}(\mu(x) y) \\ \mu \frac{dy}{dx} + \mu py &= \mu \frac{dy}{dx} + \frac{d \mu}{dx} y \\ \frac{d \mu}{dx} &= p(x) \mu \end{aligned}\]

Here, $\frac{d \mu}{dx} = p(x) \mu$ is another separable differential equation that can written in $\frac{1}{\mu} d \mu = p(x) dx$.

Given $y(x_0) = y_0$, we can choose another arbiturary $\mu (x_0) = 1$. Then, solve the differential equation.

\[\begin{aligned} \int_{1}^{\mu} \frac{1}{\mu} d\mu &= \int_{x_0}^{x} p(x) dx \\ ln(\mu) &= \int_{x_0}^{x} p(x) dx \\ \mu(x) &= e^{\int_{x_0}^{x} p(x) dx} \end{aligned}\]

Since $\mu$ satisfies $\mu(x) (\frac{dy}{dx} + p(x)y) = \frac{d}{dx}(\mu(x) y)$

\[\begin{aligned} \mu(x) (\frac{dy}{dx} + p(x)y) = \frac{d}{dx}(\mu(x) y) &= \mu(x) g(x) \\ \int_{x_0}^{x} \frac{d}{dx}(\mu(x) y) dx &= \int_{x_0}^{x} \mu(x) g(x) dx \\ \end{aligned}\]

According to the Fundamental Theorem of Calculus,

\[\begin{aligned} \int_{x_0}^{x} \frac{d}{dx}(\mu(x) y) dx &= (\mu(x)y) \vert_{x_0}^{x} \\ &= \mu(x) y - \mu(x_0) y_0 = \mu(x) y - y_0 \\ \end{aligned}\]

Therefore,

\[\begin{aligned} \mu(x) y - y_0 &= \int_{x_0}^{x} \mu(x) g(x) dx \\ y(x) &= \frac{1}{\mu(x)}(y_0 + (\int_{x_0}^{x} \mu(x) g(x) dx)) \end{aligned}\]

Examples

Solve linear ODE: $\frac{dy}{dx} = x - y, y(0) = -1$

Convert to the standard form:

\[\begin{aligned} \frac{dy}{dx} + y &= x, \quad p(x) = 1, g(x) = x \\ \mu(x) = e^{\int_{0}^{x} 1 dx} &= e^x \\ y(x) &= e^{-x} (-1 + \int_{0}^{x} (e^x x)dx) \end{aligned}\]

Integrate $\int_{0}^{x} (e^x x)dx$ by parts:

\[\begin{aligned} \int_{0}^{x} & (e^x x)dx \\ u = x &\quad dv = e^x dx \\ du = dx &\quad v = e^x \\ \int_{0}^{x} (e^x x)dx &= (xe^x) \vert_{0}^{x} + \int_{0}^{x}e^x dx \end{aligned}\]

Thus,

\[\begin{aligned} y(x) &= e^{-x} (-1 + \int_{0}^{x} (e^x x)dx) \\ &= e^{-x} (-1 + (xe^x) \vert_{0}^{x} + \int_{0}^{x}e^x dx) \\ &= e^{-x} (-1 + xe^x - e^x + 1) \\ &= x - 1 \end{aligned}\]