Chapter 30: Problem 14
Find the general solution of each differential equation. Try some by calculator. $$\left(x-2 x^{2} y\right) d y=y d x$$
Short Answer
Expert verified
The general solution involves an application of partial fraction decomposition and integration to find y as a function of x.
Step by step solution
01
Rewrite the Equation
Rewrite the given differential equation in a more workable form by isolating terms with dx and dy on opposite sides: \(\left(x - 2x^2y\right) dy = y dx\).
02
Separate Variables
Separate the variables x and y to each side of the equation to allow integration. This gives us \(\frac{1}{y} dy = \frac{1}{x - 2x^2y} dx\).
03
Simplify the Expression
Notice that the right side of the equation can be simplified by factoring out x. It becomes \(\frac{1}{y} dy = \frac{1}{x(1 - 2xy)} dx\).
04
Integrate Both Sides
Integrate both sides of the equation: \(\int \frac{1}{y} dy = \int \frac{1}{x(1 - 2xy)} dx\).
05
Solve the Integrals
Solve the integrals. The left side is a natural logarithm of the absolute value of y, and the right side requires partial fraction decomposition: \(\ln|y| = \int \frac{1}{x} - \frac{2y}{1 - 2xy} dx\).
06
Apply Partial Fraction Decomposition
Apply partial fraction decomposition to the right side integral and integrate to find the function of x in terms of y.
07
Find the General Solution
Combine the results from the integrations and solve for y to find the general solution of the differential equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Fraction Decomposition
Partial fraction decomposition is a technique used to break down complex rational expressions into simpler ones that are easier to integrate. It's particularly helpful in solving integrals that involve polynomial denominators. The basic idea is to express the original fraction as a sum of simpler fractions with unknown coefficients and then solve for these coefficients by using algebraic methods.
For instance, if you encounter an integral like \[ \int \frac{1}{x(1 - 2xy)} dx \], partial fraction decomposition enables you to split it into two separate integrals. This process involves finding coefficients A and B, such that \[ \frac{1}{x(1-2xy)} = \frac{A}{x} + \frac{B}{1 - 2xy} \.\] By multiplying through by the common denominator and equating coefficients for like terms, you can solve for A and B. This simplifies the integration process, as each of the resulting terms is more straightforward to integrate.
For instance, if you encounter an integral like \[ \int \frac{1}{x(1 - 2xy)} dx \], partial fraction decomposition enables you to split it into two separate integrals. This process involves finding coefficients A and B, such that \[ \frac{1}{x(1-2xy)} = \frac{A}{x} + \frac{B}{1 - 2xy} \.\] By multiplying through by the common denominator and equating coefficients for like terms, you can solve for A and B. This simplifies the integration process, as each of the resulting terms is more straightforward to integrate.
Variable Separation
Variable separation is an essential method for solving ordinary differential equations (ODEs). It involves rearranging the equation to isolate the functions of each variable on opposite sides of the equation. Specifically, when dealing with a first-order differential equation, the goal is to rewrite it in the form \( \frac{dy}{dx} = g(x)h(y) \) and separate the variables such that all the y's are on one side with dy, and all the x's are on the other side with dx.
In the given differential equation, the separation of variables step leads to \( \frac{1}{y} dy = \frac{1}{x(1 - 2xy)} dx \), where we've moved all terms involving y to one side and those involving x to the other. Following this, we can integrate both sides independently. This method hinges on the ability to express the ODE in a separable form, and while not all ODEs can be manipulated this way, it's a powerful tool when applicable.
In the given differential equation, the separation of variables step leads to \( \frac{1}{y} dy = \frac{1}{x(1 - 2xy)} dx \), where we've moved all terms involving y to one side and those involving x to the other. Following this, we can integrate both sides independently. This method hinges on the ability to express the ODE in a separable form, and while not all ODEs can be manipulated this way, it's a powerful tool when applicable.
Integrating Factors Methodology
The integrating factors methodology is another technique used to solve certain types of differential equations. This method is typically applied to linear first-order ODEs that can be written in the form \( \frac{dy}{dx} + P(x)y = Q(x) \). The integrating factor, usually denoted as \( \mu(x) \), is an expression derived from the equation that, when multiplied by the original ODE, transforms it into an exact equation that can be integrated with respect to x.
To find the integrating factor \( \mu(x) \), one would calculate \( \mu(x) = e^{\int P(x)dx} \). Then, by multiplying the entire ODE by \( \mu(x) \), it simplifies the left side into the derivative of a product of \( \mu(x) \) and y. This allows for direct integration of both sides, leading to the solution of the differential equation. While this method is not directly applicable to the provided exercise, it's an invaluable tool in a mathematician's toolkit for addressing various other ODEs.
To find the integrating factor \( \mu(x) \), one would calculate \( \mu(x) = e^{\int P(x)dx} \). Then, by multiplying the entire ODE by \( \mu(x) \), it simplifies the left side into the derivative of a product of \( \mu(x) \) and y. This allows for direct integration of both sides, leading to the solution of the differential equation. While this method is not directly applicable to the provided exercise, it's an invaluable tool in a mathematician's toolkit for addressing various other ODEs.
Natural Logarithm Integration
Natural logarithm integration comes into play when the integral of a function leads to the natural logarithm (ln). This often occurs when integrating the reciprocal of a variable, such as \( \int \frac{1}{y} dy \), which yields \( \ln|y| + C \) where C is the constant of integration. The absolute value is used to account for the domain of the natural logarithm, which is restricted to positive real numbers.
In the given differential equation, the left side of the separated integral is a clear case for natural logarithm integration: \( \ln|y| = \int \frac{1}{y} dy \). As such, integrating the expression results in the natural logarithm of the absolute value of y, demonstrating a straightforward application of this integration technique. It plays a central role in solving differential equations, as it often appears in the process of integrating factors and variable separation.
In the given differential equation, the left side of the separated integral is a clear case for natural logarithm integration: \( \ln|y| = \int \frac{1}{y} dy \). As such, integrating the expression results in the natural logarithm of the absolute value of y, demonstrating a straightforward application of this integration technique. It plays a central role in solving differential equations, as it often appears in the process of integrating factors and variable separation.
