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:

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

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

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: From Step 1, we know that profit is maximized when \(x=300\).
\[\begin{align*} P(300) &= -\frac{4}{100}(300)^2 + 24(300) - 1500 \\ %&= -\frac{4}{100}(300)(300) + 24(300) - 1500 \\ %&= -4(3)(300) + 24(300) - 1500 \\ &= -3600 + 7200 - 1500 \\ &= 2100 \end{align*}\]