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Magazine Chapter 9: Problem 15
Solve each first-order linear differential equation. $$ y^{\prime}-\frac{2}{x} y=6 x^{3}-9 x^{2} $$
Short Answer
Expert verified
The solution is \( y = 3x^4 - 9x^3 + Cx^2 \).
Step by step solution
01
Identify the Type of Differential Equation
The given equation is a first-order linear differential equation, of the form \( y' + P(x)y = Q(x) \). Here, \( P(x) = -\frac{2}{x} \) and \( Q(x) = 6x^3 - 9x^2 \).
02
Find the Integrating Factor
To solve the equation, find the integrating factor \( \mu(x) \) which is given by \( \mu(x) = e^{\int P(x) \, dx} = e^{\int -\frac{2}{x} \, dx} = e^{-2\ln|x|} = x^{-2} \).
03
Multiply Through by the Integrating Factor
Multiply the entire differential equation by \( x^{-2} \) to get \( x^{-2}y' + x^{-2}(-\frac{2}{x})y = (6x^3 - 9x^2)x^{-2} \), which simplifies to \( x^{-2}y' - 2x^{-3}y = 6x - 9 \).
04
Write the Left Side as a Derivative
The left side of the equation can be rewritten as the derivative of a product: \( \frac{d}{dx}(x^{-2}y) \). The equation now reads \( \frac{d}{dx}(x^{-2}y) = 6x - 9 \).
05
Integrate Both Sides
Integrate both sides with respect to \( x \): \[\int \frac{d}{dx}(x^{-2}y) \, dx = \int (6x - 9) \, dx \]This gives \( x^{-2}y = 3x^2 - 9x + C \), where \( C \) is the constant of integration.
06
Solve for \( y \)
To find \( y \), multiply through by \( x^2 \): \[ y = x^2(3x^2 - 9x + C) \]This simplifies to \( y = 3x^4 - 9x^3 + Cx^2 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integrating Factor
The method of integrating factors is vital for solving first-order linear differential equations. It's a tool that allows us to simplify these equations into a solvable form. An integrating factor is a specially chosen function that, when multiplied with the original differential equation, transforms it into an equation that can easily be integrated.
Let's break it down into straightforward steps:
Let's break it down into straightforward steps:
- Identify the function \( P(x) \) from the standard form of the differential equation \( y' + P(x)y = Q(x) \). In our example, \( P(x) = -\frac{2}{x} \).
- The integrating factor \( \mu(x) \) is given by the exponential function: \( \mu(x) = e^{\int P(x) \, dx} \). This involves integrating \( P(x) \).
- For our problem, integrating \( -\frac{2}{x} \) gives us \( -2\ln|x| \), and the exponential on that is \( x^{-2} \).
Derivative of a Product
Once you have an integrating factor applied, you can simplify the equation by recognizing that the left side becomes the derivative of a product. This identification is crucial because it turns a complex expression into something familiar and more manageable.
Here's how this works:
Here's how this works:
- The product in question arises between the integrating factor and the dependent variable \( y \). For our equation, that product is \( x^{-2}y \).
- You will rewrite the left-hand side of the equation as the derivative \( \frac{d}{dx}(x^{-2}y) \). This derivative expression makes clearer the relationship between the terms.
Constant of Integration
When you integrate both sides of the equation, a constant of integration naturally appears. This constant, denoted as \( C \), is indispensable whenever you integrate an indefinite integral. It represents any family of solutions differing by this constant value.
In our solution:
In our solution:
- After integrating, you find \( x^{-2}y = 3x^2 - 9x + C \). Here, \( C \) captures all possible vertical shifts of your solution.
- Each differential equation solution might fit different initial conditions, and the constant allows that adaptability. It will modify until it meets specified conditions, if any are provided.
