Section 8.1 Initial Value Problems
The differential equation, for example, \(\dfdx{y}{x}=2x\text{.}\)
The initial value. So called because when the independent variable is time we usually know the value of our function initially.
A common way to create new functions using Calculus is by defining the new function to be the solution(s) of a specified IVP.
Example 8.1.1.
We’ve seen that when we differentiate the formula \(y=x^2\) we get the differential equation \(\dx{y} = 2x\dx{x},\) or
\begin{equation}
\dfdx{y}{x}=2x.\tag{8.1}
\end{equation}
If we did not already know that the function
\(y(x)=x^2\) satisfies
equation (8.1) we could give a name to the solution (sqr
\((x)\text{,}\) perhaps?) and by fiat, define
\(\text{sqr}(x)\) to be whatever function solves this equation. But there is a problem. The solution of
equation (8.1) is a multifunction, remember? To choose a single branch we need to impose an initial condition.
Drill 8.1.2. Find the Pattern.
Solve the IVP that consists of the differential equation
\begin{equation*}
\dfdx{y}{x}=2
\end{equation*}
paired with each initial condition. Graph your solutions.
\(\displaystyle y(0)=0\)
\(\displaystyle y(0)=1\)
\(\displaystyle y(0)=-10\)
\(\displaystyle y(1)=2\)
\(\displaystyle y(-1)=1\)
\(\displaystyle y(0)=1\)
Find the solution of the the IVP
\begin{align*}
\dfdx{y}{x} =2x, \amp{}\amp{}y(x_0)=y_0
\end{align*}
based on your solutions in parts (i) through (vi). Assume that \(x_0\) and \(y_0\) are fixed, but unspecified constants.
Of course simply naming the function tells us nothing about it but the IVP itself can give us a some insight. For example since \(\dfdx{y}{x}=2x\lt 0\) when \(x\lt 0\text{,}\) the slope of the graph of \(y(x)\) is also negative when \(x\lt 0\text{,}\) Similarly, the slope of the graph of \(y(x)\) is positive when \(x\gt 0\text{.}\)
We already knew that since we have the formula \(y(x)=\text{sqr}(x)=x^2\) but this won’t be true in the next section.
