Initial Value Problem#
An initial value problem is a differential equation (i.e., an equation involving \(f'\)) combined with an initial condition (i.e., \(f(a) = b\)).
Use the following strategy to solve an initial value problem.
Strategy to Solve an Initial Value Problem
Find the general solution to the differential equation by using the techniques of integration.
\[\begin{align*} f(x) &= \int f'(x) ~dx \\ &= F(x) + C \end{align*}\]where \(F\) is an antiderivative of \(f'\).
Use the initial condition, \(f(a) = b\), to solve for \(C\).
Example 1#
Solve the following initial value problem:
Step 1: Find the general solution to the differential equation.
Step 2: Use the initial condition to solve for \(C\).
Therefore
Step 3: Combine the results from Steps 1 and 2 to get the solution to the initial value problem.
Example 2#
The estimated marginal profit associated with producing/selling jasmine rice is \(P'(x) = -0.08x + 24\) dollars per pound per month where \(x\) is the production level in pounds per month. The fixed cost of producing/selling rice is $1,500 a month. What is the maximum monthly profit?
Step 1: Find the value of \(x\) that maximizes profit.
\(P'(x)\) equals zero when
which is when
Notice that \(P''(x) = -8/100\) is always negative. Therefore, \(P(x)\) is always concave down and \(x = 300\) corresponds to an absolute maximum profit.
Ultimately, we will evaluate \(P(300)\) to find the maximum monthly profit, but we don’t currently have a formula for \(P(x)\). Therefore, our next goal is to find \(P(x)\).
Step 2: Identify the initial value problem.
Note that the initial profit is negative since fixed costs count against profit.
Step 3: Solve the initial value problem:
Therefore, \(P(x) = -\dfrac{4}{100}x^2 + 24x -1500\).