Problem 17 - Calculus for Biology and Medicine

Question

In Problems , solve each autonomous differential equc tion. $$ \frac{d h}{d s}=2 h+1, \text { where } h(0)=4 $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Type of Differential Equation

The given differential equation is \( \frac{dh}{ds} = 2h + 1 \). This is a first-order linear autonomous differential equation, as it does not explicitly depend on \( s \).

Step 2: Separate Variables

Rearrange the equation to isolate terms involving \( h \) on one side:\[ \frac{dh}{2h + 1} = ds \].

Step 3: Integrate Both Sides

Integrate both sides of the equation:\[ \int \frac{1}{2h + 1} \, dh = \int ds \]The left side integrates to \( \frac{1}{2} \ln |2h + 1| \) and the right side to \( s + C \).

Step 4: Simplify The Solution

Solve for \( h \) to express the solution explicitly. Multiply the left side by 2:\[ \ln |2h + 1| = 2s + C \]Take the exponential of both sides to solve for \( h \):\[ |2h + 1| = e^{2s+C} \].For convenience, substitute \( C = \ln K \), then \( e^C = K \), giving us:\[ 2h + 1 = Ke^{2s} \].

Step 5: Solve for the Particular Solution Using Initial Condition

Substitute the initial condition \( h(0) = 4 \) to find \( K \):\[ 2(4) + 1 = K e^{2 \times 0} \rightarrow 9 = K \].This yields the particular solution:\[ 2h + 1 = 9e^{2s} \].

Step 6: Final Expression for \( h \)

Isolate \( h \) to find the explicit solution:\[ 2h = 9e^{2s} - 1 \]\[ h = \frac{9}{2}e^{2s} - \frac{1}{2} \].

Key Concepts

First-Order Linear Differential Equations

First-order linear differential equations are a type of differential equation where the highest derivative present is the first derivative. These equations are linear in terms of the unknown function and its derivative. They typically have the form:

• \( \frac{dy}{dx} + P(x)y = Q(x) \)

In the context of an autonomous differential equation, the function does not explicitly depend on the independent variable. Autonomous equations are a subcategory where the rate of change of a variable is dependent solely on the variable itself and possibly time, not on other external factors. This is evident in the given exercise, where the differential equation \(\frac{dh}{ds} = 2h + 1\) does not include \(s\). Thus, the function's behavior relies exclusively on the variable \(h\). Understanding first-order linear differential equations lays the groundwork for solving more complex equations using various techniques like separation of variables or the integrating factor method.

Integrating Factor Method

The integrating factor method is a powerful technique used to solve first-order linear differential equations. It allows one to transform a non-easily integrable equation into a form that can be solved with simple integration. This method is especially useful when the equation cannot be separated directly into functions of a single variable. Ordinarily, the differential equation should be written in standard form:

• \( \frac{dy}{dx} + P(x)y = Q(x) \)

The integrating factor \( \mu(x) \) is crucial and determined by:

• \( \mu(x) = e^{\int P(x) \, dx} \)

Multiplying the whole equation by \( \mu(x) \) can simplify it to make the left side resemble the derivative of a product, which then allows straightforward integration on both sides. While this method is not necessary for the exercise at hand due to its more amenable form for separation of variables, recognizing when to apply the integrating factor method is vital in acquiring a complete toolbox for differential equations.

Initial Value Problem

An initial value problem (IVP) in differential equations provides a specific value of the unknown function at a particular point. This known point, hence, is called an initial condition, and it enables finding a particular, unique solution out of many theoretical possibilities.
For example, the condition \(h(0) = 4\) specifies what the function \(h(s)\) equals when \(s = 0\). Solving the differential equation yields a general solution with a constant, and the initial value conditions help determine this constant. Solving an IVP helps to model real-world scenarios with defined starting points, such as population at the start of measurement or the initial temperature of a cooling object. In this problem, using the given information allows us to find \(K\), the constant that forms part of the particular solution \(2h + 1 = 9e^{2s}\). Successfully employing initial conditions is crucial for solving any problem that directly relates to a physical system or situation.

Separation of Variables

Separation of variables is a key technique for solving first-order differential equations, particularly adept for autonomous equations. When feasible, this method involves rearranging the equation to isolate one variable and its associated differential on one side of the equation, and the other variable on the other.
In the problem, the differential equation \(\frac{dh}{ds} = 2h + 1\) is rearranged as follows:

• \(\frac{dh}{2h + 1} = ds\)

Both sides are then integrated to solve the equation. The left-hand side involves integrating a function of \(h\), and the right-hand side with respect to \(s\). Through this process, the solution \(\frac{1}{2} \ln |2h + 1| = s + C \) is derived. Separation of variables is a straightforward and effective method when applicable, providing a clear path to solution by directly breaking apart the equation into integrable components related to both the variables involved.