Solving Initial Value Problems

Calculus 1 · Axiom Academy

EXAMPLE Solving Initial Value Problems Recover a function from its derivative, then pin down the constant of integration with an initial condition. Solve the initial value problem with the initial condition y(1) = 5 . That is, find the function y(x) whose derivative is 3x^2 and that passes through the point (1, 5) . Nice work. You solved an initial value problem by reversing differentiation and then using a single data point to lock in the constant. Antiderivative first: integrating gives the general solution y = x^3 + C , a whole family of curves. The condition pins down C: the initial condition y(1) = 5 selects exactly one curve — substitute it and solve, giving C = 4 . Result: the particular solution is y = x^3 + 4 , and you can check it: and y(1) = 1 + 4 = 5 . Every initial value problem follows this rhythm: integrate to get the general solution, then apply the condition to find the one constant the general solution leaves open.

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