Problem 19 - Advanced Engineering Mathematics

Question

(a) \(\operatorname{From} 2 W^{2}-W^{3}=W^{2}(2-W)=0\) we see that \(W=0\) and \(W=2\) are constant solutions. (b) Separating variables and using a CAS to integrate we get \\[ \frac{d W}{W \sqrt{4-2 W}}=d x \quad \text { and } \quad-\tanh ^{-1}\left(\frac{1}{2} \sqrt{4-2 W}\right)=x+c \\] Using the facts that the hyperbolic tangent is an odd function and \(1-\tanh ^{2} x=\operatorname{sech}^{2} x\) we have \\[ \begin{aligned} \frac{1}{2} \sqrt{4-2 W} &=\tanh (-x-c)=-\tanh (x+c) \\ \frac{1}{4}(4-2 W) &=\tanh ^{2}(x+c) \\ 1-\frac{1}{2} W &=\tanh ^{2}(x+c) \\ \frac{1}{2} W &=1-\tanh ^{2}(x+c)=\operatorname{sech}^{2}(x+c) \end{aligned} \\] Thus, \(W(x)=2 \operatorname{sech}^{2}(x+c)\) (c) Letting \(x=0\) and \(W=2\) we find that \(\operatorname{sech}^{2}(c)=1\) and \(c=0.\)

Step-by-Step Solution

Answer

The solutions are \(W = 0\), \(W = 2\), and \(W(x) = 2\operatorname{sech}^2(x)\).

Step 1: Identify Constant Solutions

From the equation \(2W^2 - W^3 = W^2(2 - W) = 0\), identify the constant solutions. Set each factor of the product equal to zero: \(W^2 = 0\) gives \(W = 0\), and \(2 - W = 0\) gives \(W = 2\). Thus, the constant solutions are \(W = 0\) and \(W = 2\).

Step 2: Separate Variables

To solve the differential equation, separate the variables. Rewrite it as \( \frac{dW}{W\sqrt{4 - 2W}} = dx \). This allows us to integrate both sides with respect to their respective variables.

Step 3: Integrate the Equation

Using a Computer Algebra System (CAS), integrate both sides. The result is \(-\tanh^{-1}\left(\frac{1}{2}\sqrt{4 - 2W}\right) = x + c\), where \(c\) is the integration constant.

Step 4: Simplify Using Hyperbolic Identities

Use the identity for the hyperbolic tangent function, taking advantage of the fact that it's odd: \(-\tanh(-x-c) = \tanh(x+c)\), which reduces to \(-\tanh(x+c)\). Transforming this gives \(\frac{1}{2}\sqrt{4 - 2W} = -\tanh(x+c)\).

Step 5: Solve for \(W(x)\)

Square both sides to simplify: \(\frac{1}{4}(4 - 2W) = \tanh^2(x+c)\). Rearrange to find \(1 - \frac{1}{2}W = \tanh^2(x+c)\). Thus, \(\frac{1}{2}W = 1 - \tanh^2(x+c) = \operatorname{sech}^2(x+c)\). Therefore, we have \(W(x) = 2\operatorname{sech}^2(x+c)\).

Step 6: Determine \(c\) Using Initial Condition

Apply the initial condition \(x = 0\) and \(W = 2\). Substitute these values into the solution \(W(x) = 2\operatorname{sech}^2(x+c)\), which gives \(2 = 2\operatorname{sech}^2(c)\). From \(\operatorname{sech}^2(c) = 1\), determine \(c = 0\) as the solution.

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but are based on hyperbolas rather than circles. They include functions like the hyperbolic sine, cosh (cosine), and one we use often: the hyperbolic tangent, \( \tanh(x) \). These functions are important in solving differential equations because they help express and simplify complex relationships.

When solving the differential equation in the exercise, we use the identity \( 1 - \tanh^2(x) = \operatorname{sech}^2(x) \). This property is crucial as it allows us to express solutions in terms of hyperbolic functions, which are often more easily interpretable. Importantly, the hyperbolic tangent function, \( \tanh(x) \), like its trigonometrical counterpart, \( \tan(x) \), is an odd function. This means \( \tanh(-x) = -\tanh(x) \).

• An "odd function" symmetry is essential here as it aids in making substitution in our equation to arrive at a cleaner solution.
• Understanding hyperbolic identities makes it easier to manipulate equations and reach final solutions.

Variable Separation

Variable separation is a method used to solve differential equations by separating variables on each side of the equation. This technique lets us rewrite the equation so that all the terms involving one variable and its differential are on one side, while the terms involving the other variable and its differential are on the other side of the equation.

In the given exercise, we separate the variables by rewriting the equation as \( \frac{dW}{W \sqrt{4 - 2W}} = dx \). The aim here is to integrate both sides independently, enabling us to solve the equation. This step is crucial because it simplifies the complexity of the equation. Once the variables are separated, the integration process becomes straightforward.

• Ensure each side of the equation only contains one variable and its differential.
• This technique often prepares the differential equation for nearer to a solution or applying a specific mathematical method, such as hyperbolic functions.

Constant Solutions

In the context of differential equations, constant solutions are solutions that do not depend on the variable, often remaining the same over time. Detecting such solutions at the outset can simplify the problem greatly. They often serve as boundary conditions or initial states in more complex scenarios.

From our example, the initial expression \( 2W^2 - W^3 = W^2(2 - W) = 0 \) leads us to discover that \( W = 0 \) and \( W = 2 \) are constant solutions. We find these by setting the factors of the product equal to zero. Finding constant solutions first helps to determine if the equation is consistent or if any specific initial value or boundary condition can be immediately deduced.

• Constant solutions provide benchmarks or reference points in a sea of possible solutions.
• Identifying them can simplify further computations or provide crucial insights into a system's behavior at equilibrium.

Computer Algebra System (CAS)

A Computer Algebra System (CAS) is a software or program designed to perform symbolic mathematics. These tools are invaluable when solving complex integrals or differential equations as they automate difficult algebraic computations, saving time and reducing errors.

In our solution process, the CAS played a key role by integrating both sides of the separated equation: \(-\tanh^{-1}\left(\frac{1}{2}\sqrt{4 - 2W}\right) = x + c\). This integration is manual but guided by the CAS for accuracy. Given the complexity of such integrals, using a CAS ensures that transformations using identities are precise.

• CAS tools can handle large calculations and algebraic manipulations swiftly.
• They help visualize mathematical relationships that can be hard to interpret manually.
• It's important to understand the equations conceptually even when using CAS to guide solutions.