is an example of an initial-value problem. Since the solutions of the differential equation are y = 2x3 +C y = 2 x 3 + C, to find a function y y that also satisfies the initial condition, we need to find C C such that y(1) = 2(1)3 +C =5 y ( 1) = 2 ( 1) 3 + C = 5.
that has a derivative in it is called a differential equation. Differential equations are an important topic in calculus, engineering, and the sciences.
22.04.2020 · What Is Initial Value? Initial value in calculus is a type of problem involving the use of an initial condition. This type of problem produces an …
An initial condition is a starting point; Specifically, it gives dependent variable values (or one of its derivatives) for a certain independent variable. It ...
Initial Value Problems : Example Question #2 ... Explanation: So this is a separable differential equation, but it is also subject to an initial condition. This ...
A differential equation with an initial condition is called an initial value problem. The most general antiderivative of the function gives the general ...
Initial Value Problems – Informal Calculus 49 Initial Value Problems Often a differential equation has many solutions. Consider the population equation P ′(t) = 0.03P (t) P ′ ( t) = 0.03 P ( t) We saw in the last section that P (t) = e0.03t P ( t) = e 0.03 t solves this differential equation.