Chapter 8: Problem 1
Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$x^{2} y^{\prime}-x y=1 / x$$
Short Answer
Expert verified
The differential equation is linear first-order, and the solution is \( y = -\frac{1}{3x^{2}} + Cx \).
Step by step solution
01
- Identify the Type of Differential Equation
Identify the type of the given differential equation: \[x^{2} y^{\text{'}} - x y = \frac{1}{x}\]Rewriting it as: \[y^{\text{'}} - \frac{y}{x} = \frac{1}{x^{3}}\]This is a linear first-order differential equation.
02
- Standard Form
Rewrite the differential equation in standard form:\[y^{\text{'}} + P(x)y = Q(x)\]Here, \( P(x) = -\frac{1}{x} \) and \( Q(x) = \frac{1}{x^{3}} \). The equation is then:\[y^{\text{'}} - \frac{y}{x} = \frac{1}{x^{3}}\].
03
- Find the Integrating Factor
Find the integrating factor \( \mu(x) \) using the formula:\[ \mu(x) = e^{\int P(x) dx} \]Given \( P(x) = -\frac{1}{x} \), we have:\[ \mu(x) = e^{\int -\frac{1}{x} dx} = e^{-\ln|x|} = \frac{1}{x} \].
04
- Multiply Through by the Integrating Factor
Multiply the entire differential equation by the integrating factor \( \mu(x) = \frac{1}{x} \):\[ \frac{1}{x} y^{\text{'}} - \frac{y}{x^{2}} = \frac{1}{x^{4}} \].This simplifies to:\[ (\frac{y}{x})^{\text{'}} = \frac{1}{x^{4}} \].
05
- Integrate Both Sides
Integrate both sides with respect to \(x\):\[ \int (\frac{y}{x})^{\text{'}} dx = \int \frac{1}{x^{4}} dx \].We get:\[ \frac{y}{x} = \int x^{-4} dx = \frac{x^{-3}}{-3} + C = -\frac{1}{3x^{3}} + C \].
06
- Solve for y
Solve for \(y\) by multiplying both sides by \( x \):\[ y = x \left(-\frac{1}{3x^{3}} + C\right) = -\frac{1}{3x^{2}} + Cx \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integrating Factor
An integrating factor is a function we use to simplify the solution of a linear first-order differential equation. By multiplying the entire differential equation by this factor, we can turn the left-hand side of the equation into an exact derivative of a product of functions, making it much easier to integrate.
To find the integrating factor, use the formula:
\[ \mu(x) = e^{\textstyle{\int P(x) \ dx}} \]
\( P(x) \) is derived from the standard form of the linear differential equation. In our example, we had:
\[ P(x) = -\frac{1}{x} \]
which leads to:
\[ \mu(x) = e^{\textstyle{\int -\frac{1}{x} \ dx}} = e^{\textstyle{-\ln|x|}} = \frac{1}{x} \]
This integrating factor transforms the equation into a form where it can be easily integrated and solved.
To find the integrating factor, use the formula:
\[ \mu(x) = e^{\textstyle{\int P(x) \ dx}} \]
\( P(x) \) is derived from the standard form of the linear differential equation. In our example, we had:
\[ P(x) = -\frac{1}{x} \]
which leads to:
\[ \mu(x) = e^{\textstyle{\int -\frac{1}{x} \ dx}} = e^{\textstyle{-\ln|x|}} = \frac{1}{x} \]
This integrating factor transforms the equation into a form where it can be easily integrated and solved.
Differential Equation Types
Differential equations come in different types, suitable for various methods of solution. Some common ones include:
In the given problem, we identified the equation:
\[ x^{2} y^{\text{'}} - x y = \frac{1}{x} \]
as a linear first-order differential equation. It's recognized by the form \( y' + P(x)y = Q(x) \), characterized by its first-order derivative and linearity in both \( y \) and \( y' \), making it amenable to methods like using integrating factors.
- Separable Differential Equations: Can be written as \( f(y)dy = g(x)dx \).
- Linear First-Order: Of the form \( y' + P(x)y = Q(x) \).
- Linear Second-Order: Typically looks like \( y'' + P(x)y' + Q(x)y = R(x) \).
In the given problem, we identified the equation:
\[ x^{2} y^{\text{'}} - x y = \frac{1}{x} \]
as a linear first-order differential equation. It's recognized by the form \( y' + P(x)y = Q(x) \), characterized by its first-order derivative and linearity in both \( y \) and \( y' \), making it amenable to methods like using integrating factors.
Solving Differential Equations
Solving differential equations involves several steps tailored to the type of equation. For a linear first-order differential equation like the one in the exercise, the steps are:
In the exercise:
\[ \int (\frac{y}{x})^{\text{'}} dx = \int \frac{1}{x^{4}} dx \]
This solves to:
\[ y = x (\frac{x^{-3}}{-3} + C) = -\frac{1}{3x^{2}} + Cx \]
Each step builds on the previous one, leading to the solution of the differential equation.
- Identify the Equation Type: Determine if the differential equation is linear first-order, separable, etc.
- Rewrite in Standard Form: Express it in the form \( y' + P(x)y = Q(x) \).
- Find the Integrating Factor: Compute\( \mu(x) = e^{\textstyle{\int P(x) \ dx}} \).
- Multiply Through by the Integrating Factor: This simplifies it to the form \( d/dx (\mu(x)y) = \mu(x)Q(x) \).
- Integrate Both Sides: Find the solution by integrating both sides with respect to \( x \).
- Solve for \( y \): Solve the integral to find \( y \) explicitly.
In the exercise:
\[ \int (\frac{y}{x})^{\text{'}} dx = \int \frac{1}{x^{4}} dx \]
This solves to:
\[ y = x (\frac{x^{-3}}{-3} + C) = -\frac{1}{3x^{2}} + Cx \]
Each step builds on the previous one, leading to the solution of the differential equation.