Problem 26 - Calculus for Biology and Medicine

Question

Use the partial-fraction method to solve $$\frac{d y}{d x}=y(1-y)$$ where \(y_{0}=2\) for \(x_{0}=0\).

Step-by-Step Solution

Answer

The solution is \(y(x) = \frac{2e^x}{1+2e^x}\).

Step 1: Separate Variables

We start by separating the variables so all terms involving \(y\) are on one side and all terms involving \(x\) are on the other side. The given differential equation is \(\frac{dy}{dx} = y(1-y)\). We can rewrite it as \(\frac{dy}{y(1-y)} = dx\).

Step 2: Perform Partial Fraction Decomposition

The left-hand side \(\frac{1}{y(1-y)}\) needs to be split into partial fractions. We set \(\frac{1}{y(1-y)} = \frac{A}{y} + \frac{B}{1-y}\). Solving for \(A\) and \(B\), we find that \(A=1\) and \(B=1\). Thus, \(\frac{1}{y(1-y)} = \frac{1}{y} + \frac{1}{1-y}\).

Step 3: Integrate Both Sides

Integrate both sides. The equation becomes \(\int \left(\frac{1}{y} + \frac{1}{1-y}\right) dy = \int dx\). This results in \(\ln|y| - \ln|1-y| = x + C\).

Step 4: Simplify Using Logarithmic Properties

Simplify the expression using logarithmic properties: \(\ln\left|\frac{y}{1-y}\right| = x + C\).

Step 5: Solve for y

Exponentiate both sides to solve for \(y\): \(\frac{y}{1-y} = Ce^{x}\), which leads to \(y = \frac{Ce^x}{1+Ce^x}\).

Step 6: Apply Initial Condition

Use the initial condition \(y(0) = 2\) to find \(C\). Plugging into \(y = \frac{Ce^0}{1+Ce^0} = 2\), we get \(C = 2\). Therefore, \(y = \frac{2e^x}{1+2e^x}\).

Step 7: Final Verification

Check if the solution satisfies the initial differential equation and initial condition. Substitute \(y = \frac{2e^x}{1+2e^x}\) back into the original equation to verify.

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a valuable technique used in integration, specifically when dealing with rational functions. In simple terms, it's the process of breaking down a complicated fraction into a sum of simpler fractions. These simpler fractions are easier to integrate individually.

In the exercise, we started with the expression \( \frac{1}{y(1-y)} \). By using partial fraction decomposition, we rewrite it into two separate fractions: \( \frac{A}{y} + \frac{B}{1-y} \).

To figure out the values of \( A \) and \( B \) in this particular case, we equate the original expression to the decomposed form and solve. For \( \frac{1}{y(1-y)} = \frac{A}{y} + \frac{B}{1-y} \), multiplying through by \( y(1-y) \) gives us \( 1 = A(1-y) + By \). By setting corresponding coefficients equal, we find that \( A = 1 \) and \( B = 1 \). Therefore, the decomposition is \( \frac{1}{y} + \frac{1}{1-y} \).

This simplification makes it much easier to integrate each term separately later on.

Separation of Variables

The separation of variables is a fundamental method used to solve differential equations, especially those which are initially presented as derivatives. It involves rearranging the equation so that all terms involving one variable (like \( y \)) are isolated on one side, and all terms involving the other variable (like \( x \)) are isolated on the opposite side. This sets the stage for straightforward integration by providing two sides that are ready to integrate.

In our exercise, the original equation was \( \frac{dy}{dx} = y(1-y) \). The goal was to separate the \( y \) terms from the \( x \) terms. Thus, we manipulated the equation to get \( \frac{dy}{y(1-y)} = dx \). Now, each side of the equation is ready to be integrated independently, which simplifies solving the equation further.

Separation of variables is often the first key step in making a differential equation "solvable," laying the groundwork for integration techniques to be used effectively.

Integration Techniques

Integration techniques form the backbone of solving differential equations after separating variables. Once variables have been separated, the next step is to tackle each side of the equation by integrating.

In our context, once we had \( \int \left( \frac{1}{y} + \frac{1}{1-y} \right) dy = \int dx \), we proceeded to integrate both sides. For the left-hand side, splitting the integrals gives us \( \int \frac{1}{y} \, dy \) and \( \int \frac{1}{1-y} \, dy \). These are standard integrals which lead to natural logarithmic functions:

• The integral of \( \frac{1}{y} \, dy \) is \( \ln|y| \).
• The integral of \( \frac{1}{1-y} \, dy \) is \( -\ln|1-y| \) (noticing the negative sign due to differentiation chain rule).

Thus the integration results in \( \ln|y| - \ln|1-y| \), which simplifies further using the logarithmic property: \( \ln \left| \frac{y}{1-y} \right| \).

Finally, the integration constant \( C \) appears when integrating \( dx \), resulting in \( x + C \). By applying logarithmic properties and initial conditions, we solve for \( y \), arriving at our final solution.