Problem 18 - Advanced Engineering Mathematics

Question

Each \(D E\) in Problems \(15-22\) is a Bernoulli equation. In Problems 15-20, solve the given differential equation by using an appropriate substitution. $$ x \frac{d y}{d x}-(1+x) y=x y^{2} $$

Step-by-Step Solution

Answer

The solution is \( y = \frac{x}{-1-x+Ce^{-x}} \).

Step 1: Identify the Bernoulli Equation

The given differential equation is \(x \frac{dy}{dx} - (1+x)y = xy^2\). A Bernoulli equation is of the form \( \frac{dy}{dx} + P(x)y = Q(x)y^n \). By rearranging our equation, it matches the form with \( P(x) = \frac{-(1+x)}{x} \), \( Q(x) = 1 \), and \( n = 2 \).

Step 2: Apply the Appropriate Substitution

For Bernoulli equations, let \( u = y^{1-n} = y^{-1} \). Then \( y = \frac{1}{u} \) and \( \frac{dy}{dx} = -\frac{1}{u^2}\frac{du}{dx} \). Substitute into the original equation.

Step 3: Transform to a Linear Equation

Substituting \( u = y^{-1} \) and \( \frac{dy}{dx} = -\frac{1}{u^2}\frac{du}{dx} \) in the original equation gives: \(-x \frac{1}{u^2} \frac{du}{dx} - (1+x)\frac{1}{u} = \frac{x}{u^2} \). Simplify to obtain the linear form: \( \frac{du}{dx} + \frac{1+x}{x} u = -1 \).

Step 4: Find the Integrating Factor

To solve the linear differential equation, find the integrating factor. Here, \( P(x) = \frac{1+x}{x} = \frac{1}{x} + 1 \). The integrating factor \( \mu(x) = e^{\int \left(\frac{1}{x} + 1\right) dx} = e^{\ln x + x} = xe^x \).

Step 5: Solve the Linear Differential Equation

Multiply through by the integrating factor: \( xe^x \frac{du}{dx} + (1+x)ue^x = -xe^x \). The equation becomes: \( \frac{d}{dx}(ue^{x}x) = -xe^{x} \). Integrate both sides with respect to \( x \).

Step 6: Integrate Both Sides

The left side integrates to \( ue^xx = \int -xe^{x} dx = -(x+1)e^x + C \). Solve for \( u \): \( u = -\frac{1}{x} - \frac{1}{x^2} + \frac{Ce^{-x}}{x} \).

Step 7: Substitute Back to Original Variable y

Remember \( u = y^{-1} \), so \( y = \frac{1}{u} = \left( -\frac{1}{x} - \frac{1}{x^2} + \frac{Ce^{-x}}{x} \right)^{-1} \). Simplify to get \( y = \frac{x}{-1-x+Ce^{-x}} \).

Step 8: Provide the General Solution

The general solution of the differential equation is \( y = \frac{x}{-1-x+Ce^{-x}} \), where \( C \) is an arbitrary constant.

Key Concepts

Differential Equations

Differential equations are mathematical expressions that relate a function with its derivatives. These equations are vital in modeling the behavior of physical systems in fields such as physics, engineering, and biology.
For example, when describing how the temperature changes over time in a room, you may use a differential equation to show the relationship between time and temperature. Doing so helps in predicting future states given initial conditions.

• Ordinary Differential Equations (ODEs) involve functions of a single variable and their derivatives.
• Partial Differential Equations (PDEs) involve several variables and derivatives with respect to those variables.

The solution to a differential equation gives insight into how the system behaves under certain conditions. In practice, solving them can be complex, requiring various techniques and approaches.

Linear Differential Equations

Linear differential equations are a special group with unique properties making them often simpler to handle than general ones. They can be written in the form \( \frac{dy}{dx} + P(x)y = Q(x) \), where both \( P(x) \) and \( Q(x) \) are functions of the independent variable only.

• A key feature of linear equations is their proportionality property. The principle of superposition states that a linear combination of solutions is also a solution.
• These equations can be either homogeneous, with \( Q(x) = 0 \), or non-homogeneous, where \( Q(x) eq 0 \).

Solving these typically involves finding an integrating factor, a function that simplifies the equation to make it solvable through integration. Mastery of this type requires recognizing the structure and appropriately applying known solution techniques.

Integrating Factor

The integrating factor is an approach to solve linear first-order differential equations. It helps to convert the equation into an easily integrable form. The general idea is to multiply the entire equation by a specially chosen function, tailored to make one side a perfect differential.

• The function, often denoted as \( \mu(x) \), is derived from the coefficient of \( y \) in the linear form, \( \frac{dy}{dx} + P(x)y = Q(x) \).
• Calculate \( \mu(x) = e^{\int P(x) \, dx} \). Once found, multiply through by this to allow integration of both sides.

This method is powerful, offering a systematic pathway to solve equations that otherwise might be tough. It’s like catching the right "rhythm" of the equation to dance it into simplicity.

Mathematical Substitution

Mathematical substitution is a technique that simplifies complex equations by replacing variables with defined expressions. This turns the problem into one that’s more manageable.

• In solving Bernoulli equations, a common substitution involves transforming the original equation into a linear form.
• For a Bernoulli equation \( \frac{dy}{dx} + P(x)y = Q(x)y^n \), letting \( u = y^{1-n} \) simplifies it. Consequently, \( y = u^{\frac{1}{1-n}} \) and derivatives adjust accordingly.

This method often serves as a precursor to more familiar techniques like separation of variables or the use of integrating factors. It requires a keen eye for recognizing patterns in the equation that hint toward possible simplifications.