Question

Solving a Homogeneous Differential Equation In Exercises \(75-80\) , solve the homogeneous differential equation in terms of \(x\) and \(y .\) A homogeneous differential equation is an equation of the form \(M(x, y) d x+N(x, y) d y=0,\) where \(M\) and \(N\) are homogeneous functions of the same degree. To solve an equation of this form by the method of separation of variables, use the substitutions \(y=v x\) and \(d y=x d v+v d x\) $$ \left(x^{3}+y^{3}\right) d x-x y^{2} d y=0 $$

Short answer

Please refer to the detailed steps to solve the integral.

Step-by-step solution

Identify the homogeneous differential equation

The given equation \( \left(x^{3}+y^{3}\right) d x-x y^{2} d y=0 \) is a homogeneous differential equation.

Substitute y and dy

We substitute \(y= vx\) and \(dy = x dv+ v dx\) into our equation to convert it into a form easy to integrate. Our equation transforms as follows: \(\left(x^{3}+(vx)^{3}\right)dx-x(vx)^{2}\left(x dv + v dx \right)=0\). Now it can be simplified to start separating variables.

Simplification and Separation of Variables

Simplify the equation into \(\left(x^{3} + v^{3}x^{3}\right)dx - v^{2}x^{4}(dv + v dx)=0\). Then, distribute the \(dx\) and separate the terms to obtain \(x^{3}(1 + v^{3})dx - v^{2}x^{4}dv - v^{3}x^{4}dx = 0\). Now, rearrange to collect all \(dx\) and \(dv\) terms separately: \(x^{3}(1 + v^{3})dx + v^{3}x^{4}dx = v^{2}x^{4}dv\). Combine like terms and simplify: \[x^{3}(1 + v^{3} + v^{3}v)dx = v^{2}x^{4}dv\] After canceling \(x^3\) from both sides, we obtain \[(1 + 2v^{3})dx = v^{2}x dv\] Now the variables \(x\) and \(v\) are separated.

Integration

The separation of variables allows us to integrate the equation. We obtain the following: \[\int \frac{1}{1 + 2v^{3}} dv = \int \frac{v^{2}}{x} dx\] By integrating the left side with respect to \(v\) and the right side with respect to \(x\), we can find a general solution.

Solving the integral

Unfortunately, the left hand side integral doesn't have standard integral form that can be solved easily. This will become a differential equation itself. Solve these equations further to get a possible function.

Final Solution

The final solution could be expressed as a function of v(x) by the equation that you get after solving step 5. Remember to substitute back \(v = \frac{y}{x}\) to get the solution in terms of \(x\) and \(y\) only.

Key concepts

Separation of Variables

In solving homogeneous differential equations, a key technique used is separation of variables. This method allows us to transform equations so that each variable can be manipulated independently on either side of the equation.
Imagine splitting the equation into two separate parts: one involving only the variable \( x \) and the other involving only \( v \) (or sometimes \( y \) and a function of \( x \)).
In our example, after substituting \( y = vx \) and making necessary simplifications, we found the equation:

• \((1 + 2v^{3})dx = v^{2}x \, dv\)

Here, variables \( x \) and \( v \) are apart, enabling us to integrate each side with respect to its own variable.

Integration

Integration is the process of finding the function that originally generated a given derivative.
Once we've separated the variables in our equation, we proceed to integrate each side.
For the equation

• \(\int \frac{1}{1 + 2v^{3}} \, dv = \int \frac{v^{2}}{x} \, dx \)

We aim to find an antiderivative for each side. This involves recognizing and perhaps transforming the functions to apply integration techniques.
Sometimes the integrals can be solved directly, but other times they require more specialized methods or even form a topic of further study.

Substitution Method

Substitution is a powerful technique to simplify complex equations, especially homogeneous ones.
In our exercise, we substitute \( y = vx \) and \( dy = x \, dv + v \, dx \) to transform the original equation into one that can be more easily managed and separated.
This substitution works because homogeneous functions are of the same degree, allowing us to express \( y \) as a product of another variable \( v \) and \( x \).

• After substitution, the equation \( (x^3 + (vx)^3)dx - x(vx)^2(x \, dv + v \, dx) = 0 \) becomes more tractable.

Through substitution, patterns emerge in the equation, often leading us to simpler forms or direct integrations.

Degree of Function

The degree of a function in the context of homogeneous differential equations refers to the power to which variables are raised.
In this exercise, both functions \( M(x, y) \) and \( N(x, y) \) are homogeneous and of the same degree. This means that each term in the function has variables which contribute equally to the total degree.

• For instance, \( x^3 + y^3 \) is a homogeneous function of degree 3.

This characteristic is what allows substitution techniques like \( y = vx \) to work smoothly because the transformed equation reflects this consistency in degree, leading to cleaner separation of variables.
Recognizing the degree is crucial as it guides both substitution choices and the path to simplification.