Question

Solve each differential equation by first finding an integrating factor. $$ \left(2 x y^{2}+y\right) d x+\left(2 y^{3}-x\right) d y=0 $$

Short answer

Rewrite the given equation as a first-order linear differential equation: $$ \frac{dy}{dx} + \left(\frac{-x}{2y^3-x} \right)y = \frac{2xy^2+y}{2y^3-x} $$ Find the integrating factor, \(\mu (x) = e^{2y^3 - x}\). Multiply the equation by the integrating factor and integrate: $$ \int \left( e^{2y^3-x} \frac{dy}{dx} -(e^{2y^3-x})y \right) dx = \int (e^{2y^3-x})(2xy^2+y)dx $$ Finally, solve for y(x) approximately using numerical methods or software tools.

Step-by-step solution

Rewrite as a first-order linear differential equation

First, we rewrite the given differential equation as a first-order linear differential equation. We have: $$ (2xy^2+y)dx + (2y^3-x)dy=0 $$ This can be written as: $$ \frac{dy}{dx} + \left(\frac{-x}{2y^3-x} \right)y = \frac{2xy^2+y}{2y^3-x} $$

Determine the integrating factor

The integrating factor is given by: $$ \mu(x) = e^{\int P(x)dx} $$ where \(P(x) = \frac{-x}{2y^3-x}\). To find the integrating factor, we need to integrate only with respect to \(x\), treating \(y\) as constant. So, $$ \mu(x) = e^{\int \frac{-x}{2y^3-x} dx} $$

Multiply the equation by the integrating factor

Before integrating, we must simplify the integrand. This can be achieved by the following substitution: $$ u(x,y) = 2y^3 - x $$ $$ \frac{du}{dx} = -1 $$ Substitute \(u\) into the integrating factor: $$ \mu(x) = e^{\int \frac{-x}{u} dx} $$ Now, substitute \(x=-u+2y^3\), and \(dx = -du\), and multiply the equation by the integrating factor: $$ \mu(x) \frac{dy}{dx} + \mu(x) \left(\frac{-x}{2y^3-x} \right)y = \mu(x) \frac{2xy^2+y}{2y^3-x} $$

Integrate the new equation and simplify

Calculate \(\mu(x)\) by integrating \(-x/u\): $$ \mu(x) = e^{\int \frac{-x}{u} dx} = e^{\int \frac{-x}{-u+2y^3} du} = e^{\int 1 du} = e^u = e^{2y^3-x} $$ Then multiply our differential equation by the found integrating factor: $$ e^{2y^3-x} \frac{dy}{dx} -(e^{2y^3-x})y = (e^{2y^3-x})(2xy^2+y) $$ Now integrate both sides with respect to x: $$ \int \left( e^{2y^3-x} \frac{dy}{dx} -(e^{2y^3-x})y \right) dx = \int (e^{2y^3-x})(2xy^2+y)dx $$

Solve for the function y(x)

Since the integration is quite complex, we can solve for the function y(x) approximately by numerical methods (such as Euler's method, Runge-Kutta methods, etc.) or use a software tool (like Mathematica, MATLAB, etc.) to find an explicit or implicit solution for y(x).

Key concepts

Integrating Factor

When solving first-order linear differential equations, an integrating factor is a crucial concept that simplifies the process. It is essentially a function, usually denoted by \(\mu(x)\), used to make the given equation exact. The integrating factor is calculated by an exponential function raised to the integral of the coefficient of the dependent variable. Here's how we find it:

• Identify the coefficient \( P(x) \) of the dependent variable in the differential equation.
• Compute the integrating factor \(\mu(x) = e^{\int P(x)dx}\).
• Multiply the entire differential equation by \(\mu(x)\) to obtain an exact equation.

In our exercise solution, after isolating \(\frac{dy}{dx}\), we determined \( P(x) \) and integrated it with respect to \( x \) to find the integrating factor. Multiplication by \(\mu(x)\) transformed the differential equation into one that is easier to integrate and brings us a step closer to solving for \( y(x) \).

First-order Linear Differential Equation

A first-order linear differential equation has a standard form \(\frac{dy}{dx} + P(x)y = Q(x)\), where \(P(x)\) and \(Q(x)\) are functions of \(x\) only. The 'first-order' denotes that the equation involves the first derivative \(\frac{dy}{dx}\) without any higher derivatives, while 'linear' refers to the variable \(y\) and its derivative appearing linearly (that is, raised only to the first power and not multiplied together).

To solve these equations, we rearrange the terms to isolate the derivative, calculate the integrating factor, use it to exact the equation, and then integrate both sides. In the provided step-by-step solution, we have rewritten a seemingly complex equation into the aforementioned standard form, which allowed for the subsequent steps in finding the integrating factor and moving towards a solution.

Numerical Methods for Differential Equations

Not all differential equations can be solved analytically, and that is where numerical methods for differential equations come into play. These methods provide approximations to the solutions and are particularly useful when dealing with complex or non-linear equations.

Some common numerical methods include:

• Euler's Method: A basic approach using tangent lines to approximate solutions.
• Runge-Kutta Methods: More advanced techniques that provide better accuracy over Euler's Method.
• Finite Difference Methods: They discretize the equation over a grid to find solutions.

Our challenge in the exercise's final step indicates the difficulty in finding an analytical solution, nudging us towards numerical methods or computational software to find \( y(x) \).

Ordinary Differential Equations

Ordinary differential equations (ODEs) are equations involving derivatives of a function of a single variable. They are termed 'ordinary' to distinguish them from partial differential equations (PDEs), which involve partial derivatives of a function of multiple variables.

Essentially, ODEs express the relationship between a function and its derivatives. They are widely used to model various physical phenomena, such as motion, growth, decay, and are classified based on their order, linearity, and other characteristics.

In the given exercise, we worked with an ODE as it involved derivatives with respect to a single independent variable \( x \). Understanding ODEs forms the backbone for tackling more complex problems in engineering, physics, and other scientific fields.