Problem 15 - Calculus for Biology and Medicine

Question

Solve the given autonomous differential equations. \(\frac{d h}{d s}=2 h+1\), where \(h(0)=4\)

Step-by-Step Solution

Answer

The solution is \( h(s) = \frac{9e^{2s} - 1}{2} \).

Step 1: Separate Variables

Begin by separating the variables in the differential equation. We have \( \frac{dh}{ds} = 2h + 1 \). Rearrange it to have all the \(h\) terms on one side and \(s\) terms on the other:\[ \frac{dh}{2h+1} = ds.\]

Step 2: Integrate Both Sides

Integrate both sides of the equation. The left-hand side integral is with respect to \(h\) and the right-hand side is with respect to \(s\):\[ \int \frac{dh}{2h+1} = \int ds.\]The left-hand side results in \( \frac{1}{2} \ln|2h+1| \) and the right-hand side is simply \(s + C\), where \(C\) is the integration constant. Thus,\[ \frac{1}{2} \ln|2h+1| = s + C.\]

Step 3: Solve for the Integration Constant

Use the initial condition \(h(0) = 4\) to find the integration constant \(C\). Substitute \(s = 0\) and \(h = 4\) into the equation:\[ \frac{1}{2} \ln|2(4) + 1| = 0 + C.\]This becomes \( \frac{1}{2} \ln 9 = C\), so \( C = \frac{1}{2} \ln 9\).

Step 4: Solve for h

Substitute the value of \(C\) back into the equation:\[ \frac{1}{2} \ln|2h+1| = s + \frac{1}{2} \ln 9.\]Multiply both sides by 2 to get rid of the fraction:\[ \ln|2h+1| = 2s + \ln 9.\]Exponentiate both sides to solve for \(2h+1\):\[ |2h+1| = 9e^{2s}.\]

Step 5: Solve for Final Solution h(s)

Since \(2h + 1\) is positive, we have:\[ 2h + 1 = 9e^{2s}.\]Solve for \(h\):\[ 2h = 9e^{2s} - 1.\]Thus,\[ h = \frac{9e^{2s} - 1}{2}.\]

Key Concepts

Separation of Variables

Separation of variables is a straightforward yet powerful method used to solve differential equations. The technique allows us to rearrange the equation so that each variable and its derivative is on opposite sides. In our exercise, we started with the autonomous differential equation \( \frac{dh}{ds} = 2h + 1 \).
The aim here is to manipulate the terms so that all expressions involving \( h \) are on one side, and those with \( s \) remain on the other. This gives us: \[ \frac{dh}{2h+1} = ds. \]This rearrangement is crucial because it lays the groundwork for integration. Once separated, we can proceed to integrate each side independently, continuing towards the solution.

• Ensure correct rearrangement: Swap terms carefully.
• Check units: The terms should logically represent corresponding quantities.

This method highlights a systematic approach to solving differential equations, emphasizing step-by-step manipulation.

Initial Conditions

Initial conditions play a pivotal role in determining the unique solution for a differential equation. They are set conditions provided in the problem that the solution must satisfy. For our example, the initial condition given is \( h(0) = 4 \).
When solving the equation \( \frac{1}{2} \ln|2h+1| = s + C \),the initial condition helps in evaluating the integration constant \( C \).
By substituting these initial conditions \( s = 0 \) and \( h = 4 \), into the equation:\[ \frac{1}{2} \ln|2(4) + 1| = C, \]we find that \( C = \frac{1}{2} \ln 9 \).

• Activate known values: Plug into the separated equation.
• Address initial conditions early: It secures an uncomplicated pathway to the solution.

Applying the appropriate initial conditions means every part of the solution adheres to the specific situation described by the problem.

Integration Constant

The integration constant \( C \) arises frequently when solving indefinite integrals within differential equations. In the context of our example, it emerges as part of the integral of the separated equation : \( \int \frac{dh}{2h+1} = \int ds. \)
Here, integration leads to a general solution, \( \frac{1}{2} \ln|2h+1| = s + C \),where \( C \) is essential for defining a particular solution. Until we apply an initial condition, \( C \) remains an unknown entity.
Using the given \( h(0) = 4 \),helps determine \( C \), leading to a precise specific solution suited for our scenario.

• Integration constant: It represents infinitesimal shifts in solution curves.
• Determine \( C \) through initial conditions: It personalizes the general solution.

The integration constant solidifies potentially infinite solutions into a singular one, rendering the equation's solution exact and relevant to its constraints.