# 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 $$ \begin{array}{ll} \frac{dy}{dx} + 2xy = 0 &\implies \int \frac{1}{y} dy = -2\int xdx \end{array} $$