Linear First Order ODE/dy = f(x) dx/Initial Condition

Theorem
Let $f: \R \to \R$ be an integrable real function.

Consider the linear first order ODE:
 * $(1): \quad \dfrac {\d y} {\d x} = \map f x$

subject to the initial condition:
 * $y = y_0$ when $x = x_0$

$(1)$ has the particular solution:
 * $y = y_0 + \ds \int_{x_0}^x \map f \xi \rd \xi$

where $\ds \int \map f x \rd x$ denotes the primitive of $f$.

Proof
It is seen that $(1)$ is an instance of the first order ordinary differential equation:
 * $\dfrac {\d y} {\d x} = \map f {x, y}$

which is:
 * subject to an initial condition: $\tuple {x_0, y_0}$

where:
 * $\map f {x, y}$ is actually $\map f x$

From Solution to First Order Initial Value Problem, this problem is equivalent to the integral equation:


 * $\ds y = y_0 + \int_{x_0}^x \map f {\xi, \map y \xi} \rd \xi$

As $\map y \xi$ does not contribute towards $\map f x$, it can be ignored.

Hence we have:
 * $\ds y = y_0 + \int_{x_0}^x \map f \xi \rd \xi$