Problem 22 - Advanced Engineering Mathematics

Question

\(\left(e^{x}+e^{-x}\right) \frac{d y}{d x}=y^{2}\)

Step-by-Step Solution

Answer

The solution is \( y = -\frac{1}{2 \tanh^{-1}(e^x) + C} \).

Step 1: Identify the Type of Equation

The given equation \( \left(e^{x}+e^{-x}\right) \frac{d y}{d x}=y^{2} \) is a first-order differential equation. On inspection, it appears to be separable because both sides of the equation are products of functions of \( x \) and \( y \) separately.

Step 2: Separate Variables

Rearrange the equation to separate the variables. Divide both sides by \( y^2 \) and multiply both sides by \( dx \) to get: \[ \frac{1}{y^2} dy = \frac{1}{e^x + e^{-x}} dx. \] Now, the variables are separated with all \( y \) terms on one side and all \( x \) terms on the other.

Step 3: Integrate Both Sides

Integrate both sides of the equation to find \( y \). The left side becomes: \( \int \frac{1}{y^2} dy = -\frac{1}{y} \). The right side involves the integral of \( \frac{1}{e^x + e^{-x}} dx \), which simplifies to: \( \int \frac{2}{e^{2x} + 1} dx \) through substitution, actually related to \( \tanh^{-1}(e^x) \).

Step 4: Compute each Integral

Compute the antiderivatives for each side: \[ -\frac{1}{y} = 2 \tanh^{-1}(e^x) + C, \] where \( C \) is the constant of integration. Thus, both sides of the equation have been integrated.

Step 5: Solve for y

To solve for \( y \), rearrange the expression: \[ y = -\frac{1}{2 \tanh^{-1}(e^x) + C}. \] This expresses \( y \) in terms of \( x \) and the constant \( C \).

Step 6: Final Answer

The solution to the differential equation is \[ y(x) = -\frac{1}{2 \tanh^{-1}(e^x) + C}. \] This expression gives the function \( y \) in terms of \( x \) using \( C \), which would be determined by initial conditions, if given.

Key Concepts

Separable Equations

Separable equations are a particular subset of differential equations where the variables can be separated on different sides of the equation. The benefit of this form is that it allows each side to depend on only one variable. In the context of our given problem, we are dealing with a first-order differential equation that can be rearranged in such a way.

The original equation is \( \left(e^{x}+e^{-x}\right) \frac{d y}{d x}=y^{2} \). This equation consists of two parts involving functions of \( y\) and \( x \). By rearranging terms to get all \( y \) terms on one side and all \( x \) terms on the other, one can deal with simpler forms of integration for each variable.

• The "separable" method often involves multiplying or dividing each side by certain expressions to achieve this separation.
• For this equation, one would divide by \( y^2 \) and multiply by \( dx \) to get \( \frac{1}{y^2} dy = \frac{1}{e^x + e^{-x}} dx \).

Breaking down equations in this way allows us to individually handle the integration of both \( y\) and \( x\) functions, streamlining the problem-solving process.

Integration

Integration is a fundamental operation in calculus, often used to solve equations that involve rates of change, such as differential equations. In separable differential equations, once the variables are separated, the next essential step is to integrate each side independently.

For our problem, we need to integrate \( \int \frac{1}{y^2} dy \) and \( \int \frac{1}{e^x + e^{-x}} dx \).

• The integral of \( \frac{1}{y^2} \) with respect to \( y \) gives \( -\frac{1}{y} \). This is a straightforward integration which assumes a basic understanding of power rules.
• The integral \( \frac{1}{e^x + e^{-x}} \) requires a substitution which can transform it into a more recognizable form. Substitution can simplify complex integrals by changing variables or functions to make them easier to integrate.

These integrations ultimately lead to a function describing \( y \) in terms of \( x \), plus a constant \( C \), an important part of indefinite integration.

Initial Conditions

Initial conditions in the context of differential equations are specific values provided that allow us to solve for constants involved in the integration. When solving an ordinary differential equation, the integration process introduces constants like \( C \), which need to be determined from initial or boundary conditions.

In our exercise, the constant \( C \) represents this unknown value post-integration. Without specific initial conditions, such as a point \((x_0, y_0)\) through which the solution must pass, \( C \) remains undetermined.

• Applying initial conditions is crucial when finding particular solutions to differential equations.
• These conditions convert the general solution, with the arbitrary constant, into a specific one that fits the given circumstances.

If we had an initial condition like \( y(a) = b \), plugging \( x = a \) and \( y = b \) into the general solution would solve for \( C \), yielding a particular solution.

Hyperbolic Functions

Hyperbolic functions, like \( \tanh \), are analogs to trigonometric functions but based on hyperbolas instead of circles. They are often utilized in calculus and differential equations due to their unique properties and relationships with exponential functions.

In our problem, the integration of \( \frac{1}{e^x + e^{-x}} \) led to an expression involving \( \tanh^{-1}(e^x) \).

• The hyperbolic tangent inverse function, \( \tanh^{-1}(x) \), is the inverse function of \( \tanh(x) \), which resembles the regular tangent but for hyperbolic functions.
• These functions can simplify solving equations that would otherwise require complex algebraic manipulation.

This is a useful way to relate certain exponential equations into simpler terms that might be more familiar from other areas of mathematics, aiding in deductive reasoning for solving intricate integrals.