Problem 29 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The solution is \(y = \frac{2e^x}{3-2e^x}\).

Step 1: Separate Variables

Start by rewriting the differential equation to separate the variables. Divide both sides of the equation by \(y(1+y)\) and multiply both sides by \(dx\): \[ \frac{1}{y(1+y)} \, dy = dx \] This separates the \(y\) terms from the \(x\) terms.

Step 2: Use Partial Fraction Decomposition

Decompose \( \frac{1}{y(1+y)} \) into partial fractions: \[ \frac{1}{y(1+y)} = \frac{A}{y} + \frac{B}{1+y} \]Multiply through by \(y(1+y)\) and equate the numerators: \[ 1 = A(1+y) + By \] Solve for \(A\) and \(B\) by substituting convenient values. Setting \(y = 0\), gives \(A=1\), and setting \(y = -1\), gives \(B = -1\). Therefore, we have: \[ \frac{1}{y(1+y)} = \frac{1}{y} - \frac{1}{1+y} \]

Step 3: Integrate Both Sides

Integrate both sides of the equation: \[ \int \left( \frac{1}{y} - \frac{1}{1+y} \right) \, dy = \int dx \]This gives: \[ \ln|y| - \ln|1+y| = x + C \] which simplifies to: \[ \ln \left| \frac{y}{1+y} \right| = x + C \]

Step 4: Solve For Constant Using Initial Condition

Use the initial condition \(y(0)=2\) to find the constant \(C\). Substitute \(x = 0\) and \(y = 2\) into the equation: \[ \ln \left| \frac{2}{1+2} \right| = 0 + C \] \[ \ln \left( \frac{2}{3} \right) = C \] So, \(C = \ln \left( \frac{2}{3} \right)\).

Step 5: Solve for \(y\) in Terms of \(x\)

Rewrite the equation to solve for \(y\): \[ \ln \left| \frac{y}{1+y} \right| = x + \ln \left( \frac{2}{3} \right) \] Exponentiate both sides to eliminate the logarithm: \[ \left| \frac{y}{1+y} \right| = e^x \cdot \frac{2}{3} \] Since \(y = 2\) initially, consider the possibility \( \frac{y}{1+y} = \frac{2}{3}e^x \). Solve for \(y\): \[ y = \frac{2e^x}{3-2e^x} \]

Key Concepts

Separable Differential Equations

Separable differential equations are a class of equations that can be rewritten in a form that allows for the separation of the variables. This typically involves expressing the equation such that all instances of the dependent variable and its derivatives are on one side, and all instances of the independent variable are on the other.
For example, with a differential equation like \( \frac{dy}{dx} = y(1+y) \), you can rearrange terms to separate them as follows:

• Move all terms involving \(y\) to one side: \( \frac{1}{y(1+y)} \, dy \).
• Move the differential of \(x\) to the other side: \( = dx \).

This separation process is crucial as it allows us to tackle each side independently when solving. It effectively reduces the problem to integrating both sides separately, making the solution much more approachable.

Initial Value Problem

An initial value problem (IVP) in differential equations is a type of problem where you're given an ordinary differential equation along with initial conditions. These conditions specify the value of the unknown function at a particular point, thus allowing us to find a particular solution.
In our original problem, the initial condition is \(y(0)=2\). This condition provides the essential information to determine the constant of integration that appears after the integration process.

• After finding the general solution, substitute the initial condition into the equation: \( \ln \left| \frac{2}{1+2} \right| = 0 + C \).
• This yields the value of \(C\), ensuring that the solution satisfies the initial condition.

This step is vital because without it, we would have an infinite number of possible solutions, and the initial condition helps us pin down exactly which one matches the given requirements.

Integration Techniques

Integration is a fundamental technique in solving differential equations, particularly when dealing with separable equations. Once the variables have been separated, the next step is to integrate each side of the resultant equation.
For the function \( \frac{1}{y(1+y)} \), partial fraction decomposition becomes necessary to facilitate integration.

• First, decompose the fraction: \( \frac{1}{y(1+y)} = \frac{1}{y} - \frac{1}{1+y} \).
• This breakdown allows integration to be performed more easily: \( \int \left( \frac{1}{y} - \frac{1}{1+y} \right) \, dy = \int dx \).

These integrals yield logarithmic forms: \( \ln|y| - \ln|1+y| \) which combine using properties of logarithms. Understanding how to manipulate and integrate using partial fractions is critical for resolving the expressions that arise from separating variables.