Solution to First Order Initial Value Problem

Theorem
Let $$y \left({x}\right)$$ be a solution to the first order ordinary differential equation:
 * $$\frac {dy} {dx} = f \left({x, y}\right)$$

which is subject to an initial condition: $$\left({a, b}\right)$$.

Then this problem is equivalent to the integral equation:


 * $$y = b + \int_a^x f \left({t, y \left({t}\right)}\right) dt$$

Proof
From Solution to First Order ODE, the general solution of:
 * $$\frac {dy} {dx} = f \left({x, y}\right)$$

is:
 * $$y = \int f \left({x, y \left({x}\right)}\right) dx + C$$

When $$x = a$$, we have $$y = b$$.

Thus:
 * $$b = \left[{\int f \left({x}\right) dx + C}\right]_a$$

which gives:
 * $$C = b - \left[{\int f \left({x}\right) dx}\right]_a$$

and so:

$$y = \int f \left({x}\right) dx + b - \left[{\int f \left({x}\right) dx}\right]_a$$

... whence the result.