Problem 8 - Calculus: Early Transcendentals

Question

Solve the differential equation. \( \frac {dH}{dR} = \frac {RH^2 \sqrt{1 + R^2}}{\ln H} \)

Step-by-Step Solution

Answer

H in relation to R is described by: \( H(\ln H - 1) = \frac{1}{3}(1 + R^2)^{3/2} + C \).

Step 1: Identify Variables and Isolate Terms

First, identify the variables involved: the dependent variable is \(H\) and the independent variable is \(R\). Notice that the equation is in the form of a separable differential equation. To solve it, start by rearranging terms to isolate expressions involving \(H\) and \(R\). Divide both sides by \(H^2\) and multiply both sides by \(\ln H\) to get:\[ \ln H \, dH = \frac{RH^2}{H^2} \sqrt{1 + R^2} \, dR \] which simplifies to \[ \ln H \, dH = R\sqrt{1 + R^2} \, dR \]

Step 2: Integrate Both Sides

Integrate both sides separately. Start with the left side: the integral of \(\ln H \, dH\):\[ \int \ln H \, dH = H(\ln H - 1) + C_1\] (integration by parts used here, with \(u = \ln H\) and \(dv = dH\))Now, integrate the right side: \[ \int R \sqrt{1 + R^2} \, dR \] Use substitution: let \(u = 1 + R^2\), so \(du = 2R \, dR\). The integral becomes:\\[ \frac{1}{2} \int \sqrt{u} \, du = \frac{1}{3}u^{3/2} + C_2 = \frac{1}{3}(1 + R^2)^{3/2} + C_2\]

Step 3: Equate and Simplify

Equating the integrated results from both sides,\[ H(\ln H - 1) + C_1 = \frac{1}{3}(1 + R^2)^{3/2} + C_2 \]Simplify the equation by consolidating the constants:\[ H(\ln H - 1) = \frac{1}{3}(1 + R^2)^{3/2} + C \]where \(C = C_2 - C_1\).

Step 4: Solve for H

While it's not possible to explicitly solve for \(H\) in terms of \(R\) straightforwardly because the equation involves a transcendental function (\(\ln H\)), this implicit solution describes \(H\) in relation to \(R\). To find a particular solution, additional conditions (initial conditions) for \(H\) would be needed. However, this is the general implicit form of the solution.

Key Concepts

Separable Differential Equations

A separable differential equation is one that can be rewritten so that all terms involving the dependent variable and its differential are on one side of the equation and all terms involving the independent variable are on the other side. In the context of our exercise, this involves transforming the given differential equation by isolating all terms involving the function and its derivative from the terms involving the independent variable, which is crucial for applying integration techniques.

To do this:

• Identify the dependent and independent variables; in this exercise, they are respectively \(H\) and \(R\).
• Reorganize the equation to separate these variables. This often requires algebraic manipulation, such as multiplying or dividing by the differential elements.

Once you have separation, each side of the equation can be integrated with respect to its corresponding variable, paving the way for solving the equation.

Integration by Parts

Integration by Parts is a crucial technique for integrating products of functions. It comes from the product rule for differentiation and provides a formula for integrating complex expressions that can be split into two functions.

The formula for integration by parts is:\[ \int u \, dv = uv - \int v \, du \]
In our problem, this method was used to integrate \( \ln H \, dH \) on the left side, facilitating the solution of the differential equation. Here are the steps:

• Identify \( u \) and \( dv \). For \( \ln H \, dH \), we choose \( u = \ln H \) and \( dv = dH \).
• Differentiate \( u \) to find \( du \) and integrate \( dv \) to find \( v \).
• Apply the formula to find the integral and simplify the expression if possible.

This method effectively transforms an otherwise difficult integration into a simpler problem that is easier to solve.

Substitution Method

The substitution method simplifies complex integrals by replacing variables with a new variable, often transforming a challenging integral into a more manageable form. It's one of the most versatile and widely-used techniques in integration.

For the exercise, substitution was applied to the right side of the equation to integrate \( R \sqrt{1 + R^2} \, dR \).

• Choose a substitution: Let \( u = 1 + R^2 \). This simplifies the integral's structure.
• Find \( du \): Differentiate the substitution equation, giving \( du = 2R \, dR \), or \( R \, dR = \frac{1}{2}du \).
• Re-write the integral in terms of \( u \) and integrate.

By substituting \( u \) and changing the integral's format, the integral becomes much more straightforward, leading to a quicker and less error-prone solution.

Transcendental Functions

Transcendental functions, such as exponential, logarithmic, and trigonometric functions, are non-algebraic by nature and require specific techniques for their manipulation and integration.

In this particular exercise, the transcendental function involved is the natural logarithm, \( \ln H \), which complicates solving the differential equation because it doesn't easily allow for expressing \(H\) explicitly in terms of \(R\).
This implicitly leaves an equation of the form:

• Incorporate the transcendental expression into the general solution and account for it when integrating or when initial conditions are provided.
• Recognize that when an expression involving a transcendental function emerges, it often signifies the difficulty or impossibility of finding simple algebraic solutions.

The transcendental nature of the logarithm function in the final solution means that additional conditions or numerical methods may be needed for an explicit solution.