Problem 10
Question
For \(y^{\prime}+\frac{2}{x} y=\frac{3}{x}\) an integrating factor is \(e^{\int(2 / x) d x}=x^{2}\) so that \(\frac{d}{d x}\left[x^{2} y\right]=3 x\) and \(y=\frac{3}{2}+c x^{-2}\) for \(0< x<\infty\).
Step-by-Step Solution
Verified Answer
The solution is \( y = \frac{3}{2} + Cx^{-2} \).
1Step 1: Identify the form of the differential equation
The given differential equation is first-order linear and can be written as: \( y' + P(x)y = Q(x) \), where \( P(x) = \frac{2}{x} \) and \( Q(x) = \frac{3}{x} \).
2Step 2: Determine the integrating factor
The integrating factor \( \mu(x) \) is determined by the formula \( e^{\int P(x) \, dx} \). Here \( P(x) = \frac{2}{x} \), so we calculate the integral: \[ \mu(x) = e^{\int \frac{2}{x} \, dx} = e^{2 \ln|x|} = x^2. \]
3Step 3: Apply the integrating factor
Multiply the entire differential equation by the integrating factor \( x^2 \): \[ x^2 y' + x^2 \cdot \frac{2}{x}y = x^2 \cdot \frac{3}{x}. \] This simplifies to \[ \frac{d}{dx}[x^2 y] = 3x. \]
4Step 4: Integrate both sides
Integrate the equation from the previous step: \[ \int \frac{d}{dx}[x^2 y] \, dx = \int 3x \, dx. \] The left side simplifies to \( x^2 y \), and the right side integrates to \( \frac{3}{2}x^2 + C \). Hence, \( x^2 y = \frac{3}{2}x^2 + C \).
5Step 5: Solve for \( y \)
Divide both sides by \( x^2 \) to solve for \( y \): \[ y = \frac{3}{2} + \frac{C}{x^2}. \] Here, \( C \) is an arbitrary constant determined by initial conditions, though none are given in the problem.
Key Concepts
Integrating FactorDifferential EquationsSolution Techniques
Integrating Factor
The concept of an integrating factor is a crucial tool used to solve first-order linear differential equations. In such differential equations, the aim is to transform the equation in a way that allows easy integration. This is where the integrating factor comes into play.
For a differential equation of the form \( y' + P(x)y = Q(x) \), the integrating factor \( \mu(x) \) is given by the formula:
In our given problem, the integrating factor was \( x^2 \), because the function \( P(x) = \frac{2}{x} \) integrates to \( 2 \ln |x| \), making \( \mu(x) = e^{2 \ln |x|} = x^2 \). This factor, when multiplied through the entire equation, aligns the equation nicely for further steps.
For a differential equation of the form \( y' + P(x)y = Q(x) \), the integrating factor \( \mu(x) \) is given by the formula:
- \( \mu(x) = e^{\int P(x) \, dx} \)
In our given problem, the integrating factor was \( x^2 \), because the function \( P(x) = \frac{2}{x} \) integrates to \( 2 \ln |x| \), making \( \mu(x) = e^{2 \ln |x|} = x^2 \). This factor, when multiplied through the entire equation, aligns the equation nicely for further steps.
Differential Equations
Differential equations involve derivatives and are used to describe various phenomena that change over time or space. First-order linear differential equations, like the one in our example, involve only the first derivative of the unknown function. These equations typically have the structure:
For example, when faced with the task of solving the differential equation \( y' + \frac{2}{x} y = \frac{3}{x} \), we follow a systematic process that ultimately allows us to express \( y \) in terms of \( x \) by using specific solution techniques, such as integrating factors.
- \( y' + P(x)y = Q(x) \)
- \( y' \) is the first derivative of \( y \) with respect to \( x \).
- \( P(x) \) and \( Q(x) \) are functions of \( x \).
For example, when faced with the task of solving the differential equation \( y' + \frac{2}{x} y = \frac{3}{x} \), we follow a systematic process that ultimately allows us to express \( y \) in terms of \( x \) by using specific solution techniques, such as integrating factors.
Solution Techniques
Solving first-order linear differential equations often involves a systematic procedure that starts with identifying the form of the equation, determining an integrating factor, and then applying it. Let’s break this process down:
- Identify the Equation Form: Recognize that the equation is in the form \( y' + P(x)y = Q(x) \). This is crucial for selecting the correct method.
- Find the Integrating Factor: Calculate \( \mu(x) = e^{\int P(x) \, dx} \). Here, it's \( x^2 \).
- Multiply Through: Apply the integrating factor across the entire equation to transform it into a form where the left side is a product's derivative: \( \frac{d}{dx}[x^2y] = 3x \).
- Integrate: Both sides are integrated with respect to \( x \), yielding a new equation where the left side integrates directly to a simple expression.
- Solve for \( y \): The obtained expression is then solved for \( y \), often resulting in an equation that includes an arbitrary constant \( C \), accounting for initial conditions.
Other exercises in this chapter
Problem 9
For \(y^{\prime}-\frac{1}{x} y=x \sin x\) an integrating factor is \(e^{-\int(1 / x) d x}=\frac{1}{x}\) so that \(\frac{d}{d x}\left[\frac{1}{x} y\right]=\sin x
View solution Problem 9
Let \(M=y^{3}-y^{2} \sin x-x\) and \(N=3 x y^{2}+2 y \cos x\) so that \(M_{y}=3 y^{2}-2 y \sin x=N_{x} .\) From \(f_{x}=y^{3}-y^{2} \sin x-x\) we obtain \(f=x y
View solution Problem 11
(a) The initial-value problem is \(d h / d t=-8 A_{h} \sqrt{h} / A_{w}, h(0)=H\) Separating variables and integrating we have \\[ \frac{d h}{\sqrt{h}}=-\frac{8
View solution Problem 11
Assume that \(A=A_{0} e^{k t}\) and \(k=-0.00012378\). If \(A(t)=0.145 A_{0}\) then \(t \approx 15,600\) years.
View solution