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docs/mathematics/ordinary-differential-equations/first-order-ode.md

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+# First-order differential equations
+
+## First-order linear differential equations
+
+A first-order **linear** differential equation is one of the type
+
+$$
+\frac{dy}{dx} + p(x) y = q(x).
+$$
+
+where $p(x)$ and $q(x)$ are given functions, which may be assumed to be continuous. The equation is called **nonhomogeneous** unless $q(x)$ is dentically zero. The corresponding **homogeneous** equation
+
+$$
+\frac{dy}{dx} + p(x)y = 0,
+$$
+
+is separable and so is easily solved. By seperation of variables
+
+$$
+\frac{dy}{dx} = -p(x)y \implies \int \frac{1}{y}dy = \int p(x)dx.
+$$
+
+Though, pay attention to absolute values.
+
+There are two methods for solving nonhomogeneous equations.
+
+### Integration factor
+
+The first method is by using an integrating factor. Let $\mu(x)$ be an antiderivative of $p(x)$ and multiply the equation by $e^{\mu(x)}$.
+
+$$
+\begin{array}{ll}
+e^{\mu(x)} \frac{dy}{dx} + e^{\mu(x)} p(x) y = e^{\mu(x)} q(x) &\implies \frac{d}{dx}(e^{\mu(x)} y) = q(x) e^{\mu(x)}, \\
+&\implies e^{\mu(x)} y = \int q(x) e^{\mu(x)}dx, \\
+&\implies y(x) = e^{-\mu(x)} \int q(x) e^{\mu(x)}dx.
+\end{array}
+$$
+
+### Variation of the constant
+
+The second method is by using a variation of a constant. Let $\mu(x)$ be an antiderivative of $p(x)$ and solve the homegeneous equation
+
+$$
+\frac{dy}{dx} + p(x)y = 0 \implies y(x) = k e^{-\mu(x)}.
+$$
+
+Try:
+
+$$
+y'(x) + p(x) y(x) = q(x) = k'(x) e^{\mu(x)},
+$$
+
+thus $k'(x) = q(x) e^{\mu(x)}$.
+
+#### Example
+
+Solve $\frac{dy}{dx} + 2xy = x$ with $y(0) = 3$.
+
+
+First solving the homogeneous equation
+
+$$
+\begin{array}{ll}
+\frac{dy}{dx} + 2xy = 0 &\implies \int \frac{1}{y} dy = -2\int xdx
+\end{array}
+$$