FTC solves the simplest differential equations

A differential equation is an equation relating one or more unknown functions and their derivatives.
A solution of a differential equation is a specification of the unknown functions, such that the equation then holds.
For example the differential equation $\frac{dy}{dt} = 3t^2$ has a solution $y = t^3 + 100$, since if $y = t^3 + 100$, then $\frac{dy}{dt} = 3t^2$, as required.
Note that this solution is not unique: $y = t^3 - 43$ is a solution, as is $y = t^3 + C$, for any constant $C$.
However, if $y$ is any solution, then we have $\frac{d}{dt}( y - t^3) = 3t^2 - 3t^2 = 0$, so $y - t^3$ is constant.
So $y = t^3 + C$ gives all possible solutions.
This is called the general solution.
We may then select a unique solution, by giving some side condition that fixes $C$.
This can be done in various ways.
Here we may specify $y$ at some $t$.
For example, if we require that $y(3) = 7$, we get $7 = 3^3 + C = 27 + C$, so $C = - 20$ and the required solution is:

$$y = t^3 - 20.$$

The side condition or conditions are often called initial value conditions or boundary conditions.

FTC shows how to solve the some simple, but important differential equations:

• The differential equation $\frac{dy}{dx} = f(x)$ for $x$ in the interval $[a,b]$, is solved by the formula:
  $$y = \int_a^x f(t) dt.$$

• The differential equation $\frac{dy}{dx} = f(x)$ for $x$ in the interval $[a,b]$, with the initial condition $y(c) = A$ (where $c$ lies in the interval $[a,b]$) is solved uniquely by the formula:
  $$y = A + \int_c^x f(t) dt.$$

Proof:
In each case FTCI gives immediately that the function $y$ is a solution.
In the second case, putting $x = c$ gives $y = A + \int_c^c f(t) dt = A$, as required.
If another function $z$ were also a solution, so $\frac{dz}{dt} = f(t)$ and $z(c) = A$, then we have:

$$\frac{d}{dt}(y - z) = \frac{dy}{dt} - \frac{dz}{dt} = f(t) - f(t) = 0,$$

So $y - z$ is constant.
But at $x = c$, we have $y(c) = z(c) = A$, so $(y - z)(c) = 0$, so the function $y - z$ is everywhere zero, so $y = z$ and the solution is unique.