Initial Value Problems - Finding the Constant

Calculus 1 · Axiom Academy

LESSON Pinning Down the Constant An antiderivative is a whole family of curves. One extra fact — an initial condition — picks out the single one you want. Suppose all you know is the derivative: F'(x) = 2x . Integrating gives an antiderivative, but the constant term is invisible to the derivative — and are identical . So integrating returns not one function but a whole family , one curve for every value of C . Every value of C is a valid antiderivative 2. One Condition Selects One Curve An initial value problem adds a constraint: the solution must pass through a specific point. Out of the infinite stack, only one parabola actually goes through (1, 5) — and that point acts like a filter, discarding every other member of the family. The faded curves are the rejected candidates; the highlighted one is the unique solution. We still have to compute which C that is — that's the next step. Finding C is the same move every time: write the general antiderivative, then substitute the initial condition and solve. The animation tunes C until the curve's height at x = 1 lands exactly on the target value 5 . Solve the initial value problem F'(x) = 2x with F(1) = 5 . Velocity is the derivative of position, so position is an antiderivative of velocity — and the constant is the starting point. A ball thrown upward has velocity v(t) = 20 - 10t (metres per second) and starts at ground level, s(0) = 0 : Infinitely many trajectories — same velocity pattern, every possible starting height.

This is the written version of the interactive lesson above. See the full Calculus 1 course.