Problem 37 - Advanced Engineering Mathematics

Question

Separating variables, we have \\[\frac{d y}{y-y^{3}}=\frac{d y}{y(1-y)(1+y)}=\left(\frac{1}{y}+\frac{1 / 2}{1-y}-\frac{1 / 2}{1+y}\right) d y=d x\\]. Integrating, we get \\[\ln |y|-\frac{1}{2} \ln |1-y|-\frac{1}{2} \ln |1+y|=x+c\\]. When \(y>1,\) this becomes \\[\ln y-\frac{1}{2} \ln (y-1)-\frac{1}{2} \ln (y+1)=\ln \frac{y}{\sqrt{y^{2}-1}}=x+c.\\] Letting \(x=0\) and \(y=2\) we find \(c=\ln (2 / \sqrt{3}) .\) Solving for \(y\) we get \(y_{1}(x)=2 e^{x} / \sqrt{4 e^{2 x}-3},\) where \(x>\ln (\sqrt{3} / 2)\) When \(0\ln (\sqrt{3} / 2)\).

Step-by-Step Solution

Answer

Four cases were solved for different ranges of \(y\): \(y_1\), \(y_2\), \(y_3\), and \(y_4\).

Step 1: Separate Variables

We start by separating the given differential equation. We have \( \frac{d y}{y-y^{3}} = \frac{1}{y} + \frac{1}{2(1-y)} - \frac{1}{2(1+y)} \ dy = dx \). This equation is properly separated into a form where both sides of the equation depend on different variables.

Step 2: Integrate Both Sides

Integrating both sides, we get \( \int \left( \frac{1}{y} + \frac{1/2}{1-y} - \frac{1/2}{1+y} \right) dy = \int dx \). The left side becomes \( \ln |y| - \frac{1}{2} \ln |1-y| - \frac{1}{2} \ln |1+y| \) and the right side gives us \( x + c \), resulting in \( \ln |y| - \frac{1}{2} \ln |1-y| - \frac{1}{2} \ln |1+y| = x + c \).

Step 3: Solve for Different Cases

Based on different ranges of \( y \), solve the equation for four distinct cases.

Step 4: Sub-step 3.1: When \( y > 1 \)

Here, \( \ln y - \frac{1}{2} \ln (y-1) - \frac{1}{2} \ln (y+1) = \ln \frac{y}{\sqrt{y^2-1}} = x + c \). Given \( x = 0 \) and \( y = 2 \), we find \( c = \ln \frac{2}{\sqrt{3}} \). Solving for \( y \) gives \( y_{1}(x) = \frac{2 e^x}{\sqrt{4e^{2x} - 3}} \) and \( x > \ln \sqrt{\frac{3}{2}} \).

Step 5: Sub-step 3.2: When \( 0 < y < 1 \)

Here, \( \ln y - \frac{1}{2} \ln (1-y) - \frac{1}{2} \ln (1+y) = \ln \frac{y}{\sqrt{1-y^2}} = x + c \). Given \( x = 0 \) and \( y = \frac{1}{2} \), we find \( c = \ln \frac{1}{\sqrt{3}} \). Solving for \( y \), we get \( y_{2}(x) = \frac{e^x}{\sqrt{e^{2x} + 3}} \) with \( -\infty < x < \infty \).

Step 6: Sub-step 3.3: When \( -1 < y < 0 \)

In this case, \( \ln(-y) - \frac{1}{2} \ln(1-y) - \frac{1}{2} \ln(1+y) = \ln \frac{-y}{\sqrt{1-y^2}} = x + c \). For \( x = 0 \) and \( y = -\frac{1}{2} \), we have \( c = \ln \frac{1}{\sqrt{3}} \). Solving for \( y \), we find \( y_{3}(x) = \frac{-e^x}{\sqrt{e^{2x} + 3}} \) with \( -\infty < x < \infty \).

Step 7: Sub-step 3.4: When \( y < -1 \)

For \( y < -1 \), \( \ln(-y) - \frac{1}{2} \ln(1-y) - \frac{1}{2} \ln(-1-y) = \ln \frac{-y}{\sqrt{y^2-1}} = x + c \). Setting \( x = 0 \) and \( y = -2 \), we find \( c = \ln \frac{2}{\sqrt{3}} \). Solving for \( y \), we obtain \( y_{4}(x) = \frac{-2 e^x}{\sqrt{4e^{2x} - 3}} \), with \( x > \ln \sqrt{\frac{3}{2}} \).

Key Concepts

Separation of Variables

Separation of Variables is a fundamental technique in solving differential equations, particularly when the variables can be distinctly separated. This means that each side of the equation becomes dependent on different variables, like expressing all terms involving \(y\) on one side and all terms involving \(x\) on the other. This approach simplifies the problem to a format where integration can be applied more straightforwardly.

To execute this, we perform the following steps:

• Rearrange the equation to isolate terms involving \(dy\) and \(dx\).
• Ensure that all instances of \(y\) and terms involving \(dy\) are on one side of the equation, with \(x\) and \(dx\) on the other.
• The equation is then ready for integration.

Using Separation of Variables allows for each variable to be considered individually, paving the way for manageable integration and eventual solution of the differential equation.

Integration Techniques

Integration Techniques are essential for solving equations obtained after separating variables. In our specific exercise, integrating both sides of the equation involves handling expressions with non-trivial denominators, often requiring partial fraction decomposition:

1. **Partial Fraction Decomposition**: This involves expressing a complex rational function as a sum of simpler fractions, each of which can be integrated directly. - For example, \( \frac{1}{y-y^3} \) is decomposed into \( \frac{1}{y} + \frac{1/2(1-y)} - \frac{1/2(1+y)} \).2. **Natural Logarithm Integration**: Once in simpler form, most terms become immediate candidates for integration using natural logarithms: - Like \( \int \frac{1}{y} dy = \ln|y| + C \), a fundamental integration result.3. **Antiderivatives**: These are the integral results used to express the solutions in terms of other functions and constants.Effective Integration Techniques transform the mathematics from complex manipulations into understandable expressions, leading to the mental leap from algebraic forms to logarithmic solutions, crucial in differential equation solving.

Initial Value Problems

Initial Value Problems (IVPs) in differential equations provide specific solutions under given initial conditions. This additional information allows for the determination of the constant of integration obtained from indefinite integrals. IVPs are typically presented as a differential equation accompanied by an initial condition.

Here's how they work:

• After integrating and obtaining the general solution, apply the given initial condition. For example, if \(x = 0\) and \(y = 2\), substitute these values to find the constant \(c\).
• The calculated constant makes the solution specific, aligning it with the context of the physical or geometrical situation.
• In the example, plugging the initial condition into the logarithmic equation allows solving for a precise expression of \(y\) depending on \(x\).

IVPs transform the general solution into a specific, unique solution, making them invaluable for practical applications in real-world situations.

Logarithmic Functions

Logarithmic Functions are pivotal in manipulating and expressing solutions to differential equations, especially those involving constant growth rates or ratios. In the context of the given problem, logarithms help simplify multiplication and exponentiation appearing in differential solutions.

Here's why they are useful:

• **Logarithm Properties**: They transform multiplicative relationships into additive ones, making complex calculations easier to handle. For instance, leveraging properties like \(\ln(a) + \ln(b) = \ln(ab)\).
• **Natural Logarithm**: The base for solving many continuous growth equations, such as those derived from separating variables in differential problems. It particularly shines where exponential growth or decay is involved.
• **Inverse Nature**: As logarithms are inverses of exponential functions, they are instrumental in solving equations where variables are in exponents or under a root, making them excellent for extracting function forms from implicit relations.

Understanding and applying logarithmic functions lead to cleaner, more interpretable solutions, which clarify the path from complex algebraic expressions to simple function relationships.