Initial Value Problems

Definition

Definition

An initial value problem is a differential equation (i.e., an equation involving \(f'\)) combined with an initial condition (i.e., \(f(a) = b\)).

The goal of an initial value problem is to find the unique function that satisfies the differential equation and the initial condition.

How to Solve an Initial Value Problem

Use the following 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

An initial value problem

Solve the following initial value problem:

\[\begin{align*} f'(x) &= 4x^3 -3x^2 +x+9\\ f(1) &= 6 \end{align*}\]

Step 1:   Find the general solution to the differential equation.

\[\begin{align*} f(x) &= \int f'(x) ~dx \\ &= \int 4x^3 -3x^2 +x+9 ~dx\\ &= x^4-x^3+\frac{x^2}{2}+9 x + C \end{align*}\]

Step 2:   Use the initial condition to solve for \(C\).

\[\begin{align*} 6 &= f(1) \\ &= 1 - 1 + \frac{1}{2} + 9 +C && \text{plug $x=1$ into our result from Step 1}\\ &= \frac{1}{2}+9 +C && \text{simplify} \end{align*}\]

Therefore

\[C = 6 - \frac{1}{2} - 9 = - \frac{7}{2}.\]

Step 3:   Combine the results from Steps 1 and 2 to get the solution to the initial value problem.

\[f(x) = x^4 - x^3 + \frac{x^2}{2} + 9x - \frac{7}{2}\]

Example 2

Find maximum profit given the marginal profit function

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

\[-\frac{8}{100}x + 24 = 0\]

which is when

\[x = \frac{24\cdot 100}{8} = 300\]

Notice that \(P''(x) = -8/100\) is always negative. Therefore, \(P(x)\) is always concave down which means \(x = 300\) 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.

\[\begin{align*} P'(x) &= -\frac{8}{100}x +24 && \text{stated in the problem}\\ P(0) &= -1,500 && \text{since fixed costs are \$1,500} \end{align*}\]

Note that the initial profit is negative since fixed costs count against profit.

Step 3:   Solve the initial value problem.

\[\begin{align*} P(x) &= \int P'(x) ~dx \\ &= \int -\frac{8}{100}x+24 ~ dx \\ &= -\frac{8}{100}\cdot \frac{x^2}{2} + 24x +C \\ &= -\frac{4}{100}x^2 +24x + C \\ \end{align*}\]

\[\begin{align*} P(0) &= -1500 \\ &= -\frac{4}{100}(0)^2 + 24(0) +C \\ &= C \end{align*}\]

Therefore, \(P(x) = -\dfrac{4}{100}x^2 + 24x -1500\).

Step 4:   Compute the maximum profit.

From Step 1, we know that profit is maximized when \(x=300\).

\[\begin{align*} P(300) &= -\frac{4}{100}(300)^2 + 24(300) - 1500 \\ %&= -\frac{4}{100}(300)(300) + 24(300) - 1500 \\ %&= -4(3)(300) + 24(300) - 1500 \\ &= -3600 + 7200 - 1500 \\ &= 2100 \end{align*}\]