Problem 35 - Calculus for Biology and Medicine

Question

(a) Use partial fractions to show that $$\int \frac{d u}{u^{2}-a^{2}}=\frac{1}{2 a} \ln \left|\frac{u-a}{u+a}\right|+C$$ (b) Use your result in (a) to find a solution of $$\frac{d y}{d x}=y^{2}-4$$ that passes through (i) \((0,0)\), (ii) \((0,2)\), and (iii) \((0,4)\).

Step-by-Step Solution

Answer

(a) The integral evaluates to \( \frac{1}{2a} \ln \left| \frac{u-a}{u+a} \right| + C \). (b) Solutions: (i) \(y=0\), (ii) No real solution, and (iii) \(y=4\).

Step 1: Setup for Partial Fraction Decomposition

The integral to evaluate is \( \int \frac{du}{u^2 - a^2} \). Recognize \( u^2 - a^2 \) is a difference of squares and factor it as \((u-a)(u+a)\). Decompose the fraction \( \frac{1}{u^2 - a^2} \) into partial fractions: \( \frac{1}{u^2 - a^2} = \frac{A}{u-a} + \frac{B}{u+a} \).

Step 2: Determine Coefficients for Partial Fractions

Clear the denominators by multiplying through by \((u-a)(u+a)\) giving \(1 = A(u+a) + B(u-a) \). Solve for \(A\) and \(B\) by substituting specific values for \(u\). Set \(u = a\) to find \(B\), and set \(u = -a\) to find \(A\).

Step 3: Solve for A and B

Substitute \(u = a\) to get \(1 = B(2a)\) which gives \(B = \frac{1}{2a}\). Substitute \(u = -a\) to get \(1 = A(-2a)\) and solve for \(A\), giving \(A = -\frac{1}{2a}\).

Step 4: Write the Decomposed Form

Substitute \(A\) and \(B\) back into the partial fraction decomposition: \( \frac{1}{u^2 - a^2} = \frac{-1/2a}{u-a} + \frac{1/2a}{u+a} \).

Step 5: Integrate the Decomposed Expression

Integrate term by term: \( \int \frac{-1/2a}{u-a} \, du + \int \frac{1/2a}{u+a} \, du \). These are standard logarithmic integrals: \( \frac{-1}{2a} \ln|u-a| + \frac{1}{2a} \ln|u+a| + C \).

Step 6: Simplify the Result

Combine the logarithms using properties of logarithms: \[ \frac{1}{2a} \ln \left| \frac{u+a}{u-a} \right| + C = \frac{1}{2a} \ln \left| \frac{u-a}{u+a} \right| + C \]. This result matches the provided form.

Step 7: Solve the Differential Equation Using the Result

The differential equation is \( \frac{dy}{dx} = y^2 - 4 \). Recognize this is of the form \( \frac{1}{y^2 - 4} dy = dx \). Integrate both sides: \( \int \frac{1}{y^2 - 4} \, dy = \int \, dx \). Use part (a) where \( a = 2 \), giving \( \frac{1}{4} \ln \left| \frac{y-2}{y+2} \right| = x + C \).

Step 8: Determine Constant C for Initial Conditions

Use initial condition (i) \((0,0)\): \( \frac{1}{4} \ln \left| \frac{0-2}{0+2} \right| = 0 + C \) gives \(C = 0\) since \( \ln 1 = 0\). Initial condition (ii) \((0,2)\) leads to an undefined log equation (no real solution), and initial condition (iii) \((0,4)\): \( \frac{1}{4} \ln \left| \frac{4-2}{4+2} \right| = 0 + C \) also gives \(C = 0\).

Key Concepts

Partial Fractions

Partial fractions are a crucial method in calculus for simplifying complex rational expressions into simpler forms that are easier to integrate or differentiate. This decomposition involves expressing a rational function, such as \( \frac{1}{u^2 - a^2} \), into a sum of fractions with simpler denominators.
Let's take a closer look at the process:

• Identify the form of the denominator: In this case, \( u^2 - a^2 \) is a difference of squares, which can be factored into \((u-a)(u+a)\).
• Set up the partial fraction decomposition: Our goal is to express \( \frac{1}{u^2 - a^2} \) as \( \frac{A}{u-a} + \frac{B}{u+a} \).
• Determine the coefficients: To find \( A \) and \( B \), clear the denominators by multiplying through by \( (u-a)(u+a) \), resulting in an equation that you can solve by choosing strategic values for \( u \). Setting \( u = a \) and \( u = -a \) simplifies this process, allowing us to solve for \( B \) and \( A \) respectively.

Understanding partial fractions is crucial for simplifying complicated integrals, and is foundational for solving differential equations efficiently.

Integral Calculus

Integral calculus focuses on the concept of integration, which is essentially the opposite operation to differentiation. It involves finding the total accumulation of quantities, such as area under a curve. In this exercise, we build on the partial fraction decomposition result to integrate the function.
After decomposition into partial fractions (\( \frac{-1/2a}{u-a} + \frac{1/2a}{u+a} \)), each term can be integrated separately. This is where the powerful technique of recognizing the 'standard forms' of integrals comes into play.

• Integrating basic logarithmic forms: The terms like \( \frac{1}{u} \) integrate directly to natural logarithms. Thus, \( \int \frac{-1/2a}{u-a} \, du \) becomes \( \frac{-1}{2a} \ln|u-a| \) and similarly for the second term.
• Applying properties of logarithms: Once you have the logarithmic sums, use logarithmic identities to combine them. For example, \( \ln|u-a| - \ln|u+a| = \ln \left| \frac{u-a}{u+a} \right| \). This simplifies the final expression significantly.

Mastering the basic forms of integrals and properties of logarithms greatly aids in solving more complicated calculus problems.

Differential Equations

Differential equations involve equations with unknown functions and their derivatives. They are prevalent in many fields such as physics, engineering, and economics. In the exercise, we tackled the differential equation \( \frac{dy}{dx} = y^2 - 4 \).
This is a separable differential equation, which means it can be expressed as \( \frac{1}{y^2 - 4} dy = dx \), allowing us to integrate both sides separately:

• Integration using results from partial fractions: We use the previously derived integral of \( \int \frac{1}{y^2 - 4} \, dy = \frac{1}{4} \ln \left| \frac{y-2}{y+2} \right| + C \). This highlights how interconnected integral calculus and differential equations can be.
• Finding particular solutions: Substituting initial conditions such as \( (0,0) \) and \( (0,4) \) into the integrated expression helps find specific solutions or values for \( C \). However, not all conditions provide valid real solutions, as seen with the point \( (0,2) \).

Differential equations, especially separable ones, often rely heavily on integration techniques to resolve them into meaningful solutions that correspond to real-life scenarios.