Antiderivatives with Initial Conditions

An antiderivative is a whole family of functions, $F(x) + C$. One extra fact, such as a starting position, picks out a single member of the family. This is how Position, Velocity and Acceleration is recovered from acceleration.

Idea clave

To solve $f'(x) = g(x)$ with $f(a) = b$: find the general antiderivative $f(x) = G(x) + C$, then substitute $x = a$ and set it equal to $b$ to find $C$. A second-order problem ($f''$ given) needs two antiderivatives and two conditions.

Por qué funciona

All the curves $G(x) + C$ are vertical shifts of one another, and exactly one of them passes through the point $(a, b)$. Substituting the point into $G(x) + C$ finds that shift. In physics this is natural: knowing a car's velocity at every moment tells you how its position changes, but not where it started. One starting position fixes everything.

Solutions of $\frac{dy}{dx} = 6x^2 - 4$. The condition $y(1) = 3$ singles out $y = 2x^3 - 4x + 5$.

Ejemplo resuelto: One condition

Find $f$ if $f'(x) = 6x^2 - 4$ and $f(1) = 3$.

$f(x) = 2x^3 - 4x + C$.
$f(1) = 2 - 4 + C = -2 + C = 3$, so $C = 5$.

Respuesta: $f(x) = 2x^3 - 4x + 5$

Ejemplo resuelto: From gravity to height

A stone is thrown up at 48 ft/s from a 64-ft cliff; its acceleration is $-32$ ft/s². Find its height $s(t)$ and when it lands.

$v(t) = \int -32\,dt = -32t + C_1$, and $v(0) = 48$ gives $C_1 = 48$.
$$s(t) = \int(-32t + 48)\,dt = -16t^2 + 48t + C_2$$, and $s(0) = 64$ gives $C_2 = 64$.
Lands when $s = 0$: $-16t^2 + 48t + 64 = 0$, so $t^2 - 3t - 4 = (t - 4)(t + 1) = 0$, and $t = 4$.

Respuesta: $s(t) = -16t^2 + 48t + 64$; it lands after 4 seconds.

Ejemplo resuelto: Two conditions

Find $f$ if $f''(x) = 12x$, $f'(0) = 2$ and $f(0) = -1$.

$f'(x) = 6x^2 + C_1$; $f'(0) = C_1 = 2$.
$f(x) = 2x^3 + 2x + C_2$; $f(0) = C_2 = -1$.

Respuesta: $f(x) = 2x^3 + 2x - 1$

Problems like these are called initial value problems, and they are the simplest differential equations. A second-order problem needs two conditions because two antiderivatives bring two constants; in physics these are the starting position and the starting velocity. Every projectile under constant gravity follows $s(t) = -\frac{g}{2}t^2 + v_0t + s_0$, where the last two coefficients are exactly those initial values.

Error común

Assuming $C$ equals the given value, as in "$f(1) = 3$, so $C = 3$." That only works when the condition is at $x = 0$ and the other terms vanish there, and even then watch out: for $f(x) = -\cos x + C$, $f(0) = -1 + C$, not $C$. Check: substitute your final answer into the condition.

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