Solution to First Order Initial Value Problem

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

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

Then this problem is equivalent to the integral equation:


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

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

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

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

Thus:
 * $\displaystyle b = \left[{\int f \left({x, y \left({x}\right)}\right) \, \mathrm d x + C}\right]_a$

which gives:
 * $\displaystyle C = b - \left[{\int f \left({x, y \left({x}\right)}\right) \, \mathrm d x}\right]_a$

and so:

$\displaystyle y = b + \int f \left({x, y \left({x}\right)}\right) \, \mathrm d x - \left[{\int f \left({x, y \left({x}\right)}\right) \, \mathrm d x}\right]_a$

whence the result, by the Fundamental Theorem of Calculus.