Problem 22 - Advanced Engineering Mathematics

Question

From \(y^{\prime}+y=y^{-1 / 2}\) and \(w=y^{3 / 2}\) we obtain \(\frac{d w}{d x}+\frac{3}{2} w=\frac{3}{2} .\) An integrating factor is \(e^{3 x / 2}\) so that \(e^{3 x / 2} w=\) \(e^{3 x / 2}+c\) or \(y^{3 / 2}=1+c e^{-3 x / 2} .\) If \(y(0)=4\) then \(c=7\) and \(y^{3 / 2}=1+7 e^{-3 x / 2}\).

Step-by-Step Solution

Answer

The solution is \( y^{3/2} = 1 + 7e^{-3x/2} \).

Step 1: Rewrite the given transformation

We start with the transformation where a substitution is provided. Given the original differential equation: \[ y' + y = y^{-1/2} \]And the substitution \( w = y^{3/2} \), this implies that \( y = (w)^{2/3} \) and thus its derivative with respect to \( x \) becomes \( \, dy/dx = \frac{2}{3} w^{1/3} \cdot dw/dx \). This allows us to express the given substitution in terms of \( w \) and its derivative.

Step 2: Substitute in the Differential Equation

With \( y = (w)^{2/3} \), differentiate both sides to convert the equation:\[ y' = \frac{2}{3} (w)^{-1/3} \cdot \frac{dw}{dx} \] By substituting \( y \) and \( y' \) into the original equation:\[\frac{2}{3} (w)^{-1/3} \frac{dw}{dx} + (w)^{2/3} = ((w)^{2/3})^{-1/2} \] Further simplifying yields:\[\frac{2}{3} \frac{dw}{dx} = (w)^{-1/2} \] Now multiply through by \( (w)^{1/3} \) to give:\[ \frac{d w}{dx} + \frac{3}{2} w = \frac{3}{2} \]

Step 3: Solve using Integrating Factor

The resulting linear first order equation is:\[ \frac{dw}{dx} + \frac{3}{2}w = \frac{3}{2} \]We find the integrating factor, \( \mu(x) = e^{\int \frac{3}{2} \, dx} = e^{3x/2} \).Multiply through by the integrating factor:\[ e^{3x/2}\frac{dw}{dx} + \frac{3}{2}e^{3x/2}w = e^{3x/2} \cdot \frac{3}{2} \]This results in:\[ \frac{d}{dx} (e^{3x/2}w) = e^{3x/2}\cdot\frac{3}{2} \] Integrating both sides:\[ e^{3x/2}w = e^{3x/2} + C \]

Step 4: Solve for Original Function

Simplify the equation obtained by integration:\[ w = 1 + Ce^{-3x/2} \] Using the transformation \( w = y^{3/2} \), we have:\[ y^{3/2} = 1 + Ce^{-3x/2} \] Substitute the initial condition given, \( y(0) = 4 \). Evaluate the expression:\[ 4^{3/2} = 1 + C \] so \( C = 7 \). This leads to the equation:\[ y^{3/2} = 1 + 7e^{-3x/2} \].

Key Concepts

Integrating Factor

The integrating factor is a powerful tool for solving linear first-order differential equations. It is especially useful when dealing with non-homogeneous terms. In this particular problem, the equation is in the form: \[ \frac{dw}{dx} + \frac{3}{2}w = \frac{3}{2} \] This is a typical linear differential equation with constant coefficients.

• An integrating factor, \( \mu(x) \), is used to simplify such equations.
• It is usually found with the expression: \( e^{\int P(x) \, dx} \), where \( P(x) \) comes from the equation \( \frac{dw}{dx} + P(x)w = g(x) \).
• For this exercise, \( P(x) = \frac{3}{2} \), leading to the integrating factor \( e^{3x/2} \).
• This integrating factor transforms the equation into an exact derivative, enabling straightforward integration.

Multiply through by the integrating factor to form a simple expression: \[ \frac{d}{dx} (e^{3x/2}w) = e^{3x/2} \cdot \frac{3}{2} \] This step is crucial because it turns the left-hand side into an easily integrable form.

Initial Value Problems

Initial value problems (IVP) represent scenarios where the value of the solution is specified at a particular point. For this problem, the initial condition given is \( y(0) = 4 \).

• It provides a specific solution that satisfies both the differential equation and this initial state.
• After finding a general solution taking the form \( y^{3/2} = 1 + Ce^{-3x/2} \), substituting \( y(0) = 4 \) helps determine \( C \).
• Calculate \( 4^{3/2} \): The number is 8, which needs to match \( 1 + C \).
• This gives \( C = 7 \). So, the equation now represents the particular solution that fits the initial condition.

This type of problem is common in scenarios modeling real-world processes where the system's state is known at a starting condition. This allows us to predict the system's future behavior.

Linear Differential Equations

Linear differential equations are a cornerstone of mathematical modeling. They appear in various fields including physics, engineering, and economics. A linear differential equation of the first order typically has the form: \[ \frac{dy}{dx} + P(x)y = g(x) \] Key characteristics include:

• The dependent variable and its derivatives appear linearly (no powers, products, or functions of \( y \)).
• These equations can be homogeneous if \( g(x) = 0 \) or non-homogeneous if \( g(x) eq 0 \).
• The exercise provided becomes linear through substitution: transforming \( y^{\prime} + y = y^{-1/2} \) to \( \frac{dw}{dx} + \frac{3}{2}w = \frac{3}{2} \).
• Solving involves finding a particular solution that satisfies the initial conditions.

Understanding how to solve linear differential equations is essential for navigating more complex models. This knowledge is foundational to predicting the behavior of natural systems and engineered processes.