Problem 34 - Advanced Engineering Mathematics

Question

Separating variables we obtain \(\frac{d y}{(y-1)^{2}}=d x .\) Then \(-\frac{1}{y-1}=x+c \quad\) and \(\quad y=\frac{x+c-1}{x+c}\). Setting \(x=0\) and \(y=1.01\) we obtain \(c=-100 .\) The solution is \\[y=\frac{x-101}{x-100}.\\]

Step-by-Step Solution

Answer

The solution is \( y = \frac{x - 101}{x - 100} \).

Step 1: Identify the Differential Equation

We start with the given differential equation \( \frac{d y}{(y-1)^{2}}=d x \). The goal is to solve this equation and find the expression for \( y \) in terms of \( x \) and any constants involved.

Step 2: Integrate Both Sides

Integrate both sides of the equation separately. On the left, we integrate with respect to \( y \): \[-\int \frac{1}{(y-1)^2} \, dy = \int \, dx.\] The left-hand side evaluates to \(-\frac{1}{y-1}\) and the right-hand side to \(x + c\), resulting in the equation:\[-\frac{1}{y-1} = x + c.\]

Step 3: Solve for \( y \)

Rearrange the equation \(-\frac{1}{y-1} = x + c \) to express \( y \) in terms of \( x \) and \( c \).Multiply both sides by \(-(y - 1)\):\[1 = -(y - 1)(x + c).\]Re-arrange to get:\[y - 1 = -\frac{1}{x + c}.\]Add 1 to both sides:\[y = 1 - \frac{1}{x + c} = \frac{(x+c)-1}{x+c}.\]

Step 4: Use Initial Conditions to Find Constant \( c \)

We have \( y = \frac{x+c-1}{x+c} \). Given the initial condition \( x = 0 \) and \( y = 1.01 \), substitute these values into the equation:\[1.01 = \frac{0 + c - 1}{0 + c}.\]Simplify to solve for \( c \):\[1.01c = c - 1.\]This results in:\[0.01c = -1,\]\[c = -100.\]

Step 5: Substitute \( c \) Back into the Solution

Substitute \( c = -100 \) back into the expression \( y = \frac{x+c-1}{x+c} \) to find the specific solution:\[y = \frac{x - 100 - 1}{x - 100}.\]Therefore, simplify the expression:\[y = \frac{x - 101}{x - 100}.\]

Key Concepts

Separation of Variables

Separation of variables is a vital technique for solving first-order differential equations. It involves rearranging an equation such that each variable and its differentials are on opposite sides of the equation. This allows integration of each side separately. Let's consider the equation from the exercise: \( \frac{d y}{(y-1)^{2}}=d x \).

• First, we identify fractional components that contain the variable \( y \). The term \( \frac{1}{(y-1)^{2}} \) in the equation is isolated.
• Next, we multiply both sides of the equation to separate \( dy \) and \( dx \). This gives us \( \frac{dy}{(y-1)^2} = dx \).

Notice how each variable now appears with its own differential, making it easier to integrate each side of the equation independently. This step is crucial as it sets the groundwork for integrating both sides and finding the general solution for \( y \) in terms of \( x \).
Separation of variables streamlines the solving process, especially when non-linear elements appear in the equation.

Integration Techniques

Integration is the mathematical process used to find the antiderivative, or integral, of a function. In the context of differential equations, such as our example \( \frac{dy}{(y-1)^2} = dx \), integration is necessary for solving once the variables have been separated.

• Begin by integrating the left-hand side with respect to \( y \). The integral of \( -\frac{1}{(y-1)^2} \) with respect to \( y \) yields \( -\frac{1}{y-1} \).
• Next, integrate the right-hand side with respect to \( x \), which simply results in \( x + c \), where \( c \) is the constant of integration.

These integration techniques allow us to find a relationship between \( y \) and \( x \). They are essential tools in calculus that enable us to solve equations involving rates of change.
Mastering integration not only aids in solving differential equations but is also a fundamental skill in analyzing various mathematical and real-world problems.

Initial Conditions

Initial conditions are specific values given for the variables in a differential equation scenario, often at a particular point in time. These values enable us to find the particular solution to the differential equation by determining the constant of integration.Let’s look at our exercise:

• We have the general solution for \( y \) as \( y = \frac{x+c-1}{x+c} \).
• The initial conditions provided are \( x = 0 \) and \( y = 1.01 \).

Substituting \( x = 0 \) and \( y = 1.01 \) into the solution equation allows us to solve for the constant \( c \). This gives: \( 1.01 = \frac{c - 1}{c} \). Solving this, we find \( c = -100 \).
Using initial conditions is crucial as it allows for finding unique solutions tailored to specific problem settings. They ensure that the solution fits within the defined parameters or conditions of a physical scenario.

Solution to Differential Equations

After integrating and using initial conditions, we arrive at the final solution to the differential equation. This solution represents a relationship between dependent and independent variables formed under given conditions and constraints.For the equation we've been discussing, the specific solution is found to be:\[ y = \frac{x - 101}{x - 100} \]This expression gives the value of \( y \) for any \( x \), considering the initial conditions already used to solve for \( c \). The solution provides a complete picture of how the dependent variable changes with respect to the independent variable:

• It includes the effects of all constants derived from integration and initial conditions.
• It allows us to predict the behavior of the system described by the differential equation.

Understanding this final solution helps in analyzing how systems evolve over time, which is essential in many fields like physics, engineering, and economics. Interpretations drawn from solutions to differential equations provide insights into the dynamics of such systems under study.