Problem 25 - Advanced Engineering Mathematics

Question

From \(\frac{1}{y} d y=\frac{1-x}{x^{2}} d x=\left(\frac{1}{x^{2}}-\frac{1}{x}\right) d x\) we obtain \(\ln |y|=-\frac{1}{x}-\ln |x|=c\) or \(x y=c_{1} e^{-1 / x} .\) Using \(y(-1)=-1\) we find \(c_{1}=e^{-1} .\) The solution of the initial-value problem is \(x y=e^{-1-1 / x}\) or \(y=e^{-(1+1 / x)} / x\).

Step-by-Step Solution

Answer

The solution to the initial-value problem is \( y = \frac{e^{-(1+1/x)}}{x} \).

Step 1: Understanding the Differential Equation

The equation given is \( \frac{1}{y} dy = \left( \frac{1-x}{x^2} \right) dx \). This is a separable differential equation, where you can integrate terms involving \( y \) separately from those involving \( x \).

Step 2: Separate Variables

Rewrite the equation as \( \frac{1}{y} dy = \left( \frac{1}{x^2} - \frac{1}{x} \right) dx \). The differential equation is now ready for integration.

Step 3: Integrate Both Sides

Integrate the left side and the right side separately. For the left side, \( \int \frac{1}{y} dy = \ln |y| + C_1 \). For the right side, \( \int \left( \frac{1}{x^2} - \frac{1}{x} \right) dx = -\frac{1}{x} - \ln |x| + C_2 \).

Step 4: Combine Constants and Rearrange

Equate the integral results: \( \ln |y| = -\frac{1}{x} - \ln |x| + C \). Rearrange to obtain \( x y = c_1 e^{-1/x} \) where \( c_1 = e^C \).

Step 5: Apply Initial Conditions

Given \( y(-1) = -1 \), substitute \( x = -1 \) and \( y = -1 \) into \( x y = c_1 e^{-1/x} \) to find \( c_1 \). Calculate \( (-1)(-1) = c_1 e^{-1/(-1)} \Rightarrow 1 = c_1 e^1 \), hence \( c_1 = e^{-1} \).

Step 6: Write the General Solution

Substitute \( c_1 = e^{-1} \) into the equation, so the solution is \( x y = e^{-1 - 1/x} \). The solution can also be written as \( y = \frac{e^{-(1+1/x)}}{x} \).

Key Concepts

Initial Value Problems

Initial Value Problems are a fundamental concept in the study of differential equations. They provide extra information that allows us to find specific solutions to differential equations, rather than just a general solution. In an initial value problem, we have a differential equation and additional condition(s) provided, typically the value of the function and/or its derivatives at a specific point. This allows us to determine the constants that may arise during the process of integration.

• For example, in our problem, the initial condition was given as \( y(-1) = -1 \), allowing us to find the constant \( c_1 = e^{-1} \).
• This initial condition plays a crucial role in ensuring that the solution not only satisfies the differential equation but also passes through the specified point.

This step is crucial because without initial conditions, solutions to differential equations can include arbitrary constants making them less specific. Initial conditions help us find unique solutions that fit a particular situation.

Integration Techniques

Integration Techniques are essential tools that allow us to solve separable differential equations efficiently. When handling these types of equations, we often aim to isolate the variables so that each side of the equation contains exclusively terms involving one variable.

• The separability of variables lets us arrange an equation such as \( \frac{1}{y} dy = \left( \frac{1}{x^2} - \frac{1}{x} \right) dx \), so we can integrate each side independently.
• In our exercise, the integral of the left side yields \( \ln |y| + C_1 \), and for the right side, the integration results in \( -\frac{1}{x} - \ln |x| + C_2 \).

A key point in integration techniques for differential equations is combining constants from both integrals. Once the integration is done, the equations are combined, where different constant terms from both integrations are usually united into a single constant. This forms part of the entire solution process.

Exponential Functions

Exponential Functions often appear in the solutions of differential equations and can easily handle expressions involving growth and decay, among other mathematical phenomena. In the context of separable differential equations, integrating terms can bring about exponential forms as part of the solution.

• In the exercise, the appearance of the exponential function \( e^{-1/x} \) was a result of integrating the fraction \( \frac{1-x}{x^2} \).
• These functions are incredibly handy, as they simplify the representation of complex solutions and help us understand the behavior of equations under varied conditions.

Exponential growth and decay are common themes that arise in initial value problems, and mastering the manipulation of such functions is crucial. Understanding the role these functions play is imperative as they can significantly affect the nature of the solution in terms of stability and long-term behavior.