https://r2.erweima.ai/mathbot/image/17222636958552jv0yz6e.png

Resuelva la ecuación y encuentre su solución general

To solve the differential equation given by:

\[ y' = \frac{x^4 + 3x^2 y^2 + y^4}{x^3 y} \]

we can rewrite the equation as follows:

\[ y' = \frac{x^4}{x^3 y} + \frac{3x^2 y^2}{x^3 y} + \frac{y^4}{x^3 y} \]

This simplifies to:

\[ y' = \frac{x}{y} + \frac{3y}{x} + \frac{y^3}{x^3} \]

To solve this differential equation, we need to identify its type and find a suitable method for solving it. This equation is not straightforwardly separable, linear, or exact. We might need to use a substitution or find an integrating factor. However, let's first rewrite the equation in a more manageable form.

We rewrite \( y' = \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{x}{y} + \frac{3y}{x} + \frac{y^3}{x^3} \]

Let's consider a substitution to simplify the equation. One potential substitution is to let \( y = vx \), where \( v \) is a function of \( x \). Then \( y' = v + x\frac{dv}{dx} \).

Substituting \( y = vx \) and \( y' = v + x\frac{dv}{dx} \) into the original equation:

\[ v + x\frac{dv}{dx} = \frac{x}{vx} + \frac{3(vx)}{x} + \frac{(vx)^3}{x^3} \]

Simplifying the right-hand side:

\[ v + x\frac{dv}{dx} = \frac{1}{v} + 3v + v^3 \]

Rewriting:

\[ v + x\frac{dv}{dx} = \frac{1}{v} + 3v + v^3 \]

Subtract \( v \) from both sides:

\[ x\frac{dv}{dx} = \frac{1}{v} + 2v + v^3 \]

This is a first-order separable differential equation. We can separate the variables and integrate:

\[ x \frac{dv}{dx} = \frac{1}{v} + 2v + v^3 \]

\[ \int \frac{v}{1 + 2v^2 + v^4} \, dv = \int \frac{dx}{x} \]

The left integral might need a partial fraction decomposition or a trigonometric substitution to solve, and the right integral is straightforward.

To solve the left integral:

\[ \int \frac{dv}{\frac{1}{v} + 2v + v^3} = \int \frac{dv}{v + v^3} \]

This can be rewritten as:

\[ \int \frac{dv}{v + v^3} = \int \frac{dv}{v(1 + v^2)} \]

Using partial fraction decomposition:

\[ \frac{1}{v(1 + v^2)} = \frac{A}{v} + \frac{Bv + C}{1 + v^2} \]

Solving for \( A \), \( B \), and \( C \), we get:

\[ 1 = A(1 + v^2) + (Bv + C)v \]
\[ 1 = A + Av^2 + Bv^2 + Cv \]

Matching coefficients, we find:

\[ A = 1 \]
\[ B = 0 \]
\[ C = -1 \]

So the partial fractions are:

\[ \frac{1}{v(1 + v^2)} = \frac{1}{v} - \frac{1}{1 + v^2} \]

Integrating each term separately:

\[ \int \frac{dv}{v} - \int \frac{dv}{1 + v^2} = \ln|v| - \arctan(v) \]

The right integral:

\[ \int \frac{dx}{x} = \ln|x| \]

Putting it all together:

\[ \ln|v| - \arctan(v) = \ln|x| + C \]

Rewriting in terms of \( y \) and \( x \):

\[ \ln \left| \frac{y}{x} \right| - \arctan \left( \frac{y}{x} \right) = \ln|x| + C \]

This is the implicit solution of the given differential equation.

Do you want details or have any questions?

Here are 8 related questions that expand on this topic:
1. How would you approach solving this differential equation if it included an additional term in the numerator?
2. What techniques can be used to solve nonlinear differential equations?
3. How can partial fraction decomposition help in solving integrals involving rational functions?
4. What are the typical methods for solving separable differential equations?
5. How does the substitution \( y = vx \) help in solving this type of differential equation?
6. What is the significance of finding an integrating factor in solving differential equations?
7. How can you verify that the solution obtained for a differential equation is correct?
8. How would the solution process change if the differential equation was higher-order?

**Tip:** When solving differential equations, always look for substitutions that can simplify the equation into a more familiar form, such as a separable or linear differential equation.

Learn how to solve the differential equation y' = (x/y) + (3y/x) + (y^3/x^3) using substitution and integrating techniques. This problem involves advanced concepts in differential equations, including the substitution method and partial fraction decomposition. Suitable for advanced college-level students and learners.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation