Solution to First Order Initial Value Problem

Theorem

Let $\map y x$ be a solution to the first order ordinary differential equation:

$\dfrac {\d y} {\d x} = \map f {x, y}$

which is subject to an initial condition: $\tuple {a, b}$.

Then this problem is equivalent to the integral equation:

$\ds y = b + \int_a^x \map f {t, \map y t} \rd t$

Proof

From Solution to First Order ODE, the general solution of:

$\dfrac {\d y} {\d x} = \map f {x, y}$

is:

$\ds y = \int \map f {x, \map y x} \rd x + C$

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

Thus:

$\ds b = \valueat {\int \map f {x, \map y x} \rd x + C} a$

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which gives:

$\ds C = b - \valueat {\int \map f {x, \map y x} \rd x} a$

and so:

$\ds y = b + \int \map f {x, \map y x} \rd x - \valueat {\int \map f {x, \map y x} \rd x} a$

whence the result, by the Fundamental Theorem of Calculus.

$\blacksquare$