We look at techniques for integrating a large variety of functions involving products, quotients, and compositions later in the text. Here we turn to one common use for antiderivatives that arises often in many applications: solving differential equations.
A
differential equation is an equation that relates an unknown function and one or more of its derivatives. The equation
$\frac{dy}{dx}=f\left(x\right)$
is a simple example of a differential equation. Solving this equation means finding a function
$y$ with a derivative
$f.$ Therefore, the solutions of
[link] are the antiderivatives of
$f.$ If
$F$ is one antiderivative of
$f,$ every function of the form
$y=F\left(x\right)+C$ is a solution of that differential equation. For example, the solutions of
$\frac{dy}{dx}=6{x}^{2}$
are given by
$y={\displaystyle \int 6{x}^{2}dx=2{x}^{3}+C}.$
Sometimes we are interested in determining whether a particular solution curve passes through a certain point
$\left({x}_{0},{y}_{0}\right)$ —that is,
$y\left({x}_{0}\right)={y}_{0}.$ The problem of finding a function
$y$ that satisfies a differential equation
$\frac{dy}{dx}=f\left(x\right)$
with the additional condition
$y\left({x}_{0}\right)={y}_{0}$
is an example of an
initial-value problem . The condition
$y\left({x}_{0}\right)={y}_{0}$ is known as an
initial condition . For example, looking for a function
$y$ that satisfies the differential equation
$\frac{dy}{dx}=6{x}^{2}$
and the initial condition
$y\left(1\right)=5$
is an example of an initial-value problem. Since the solutions of the differential equation are
$y=2{x}^{3}+C,$ to find a function
$y$ that also satisfies the initial condition, we need to find
$C$ such that
$y\left(1\right)=2{\left(1\right)}^{3}+C=5.$ From this equation, we see that
$C=3,$ and we conclude that
$y=2{x}^{3}+3$ is the solution of this initial-value problem as shown in the following graph.
Next we need to look for a solution
$y$ that satisfies the initial condition. The initial condition
$y\left(0\right)=5$ means we need a constant
$C$ such that
$\text{\u2212}\text{cos}\phantom{\rule{0.1em}{0ex}}x+C=5.$ Therefore,
$C=5+\text{cos}\left(0\right)=6.$
The solution of the initial-value problem is
$y=\text{\u2212}\text{cos}\phantom{\rule{0.1em}{0ex}}x+6.$
(mathematics) For a complex number a+bi, the principal square root of the sum of the squares of its real and imaginary parts, √a2+b2 . Denoted by | |.
The absolute value |x| of a real number x is √x2 , which is equal to x if x is non-negative, and −x if x is negative.