Problem 126 Solve the equation $$ y\left... [FREE SOLUTION]

Chapter 6: Problem 126

Solve the equation $$ y\left(x^{3}-y\right) d x-x\left(x^{3}+y\right) d y=0 $$

Short Answer

Expert verified

The solution to the exact differential equation $y(x^3 - y)dx - x(x^3 + y) dy = 0$ is given by the general solution $\frac{1}{2}yx^4 - \frac{1}{2}y^2x - \frac{1}{4}x^4y = C$, where C is a constant.

Step by step solution

Identify the given equation as exact differential equation

The given equation is: $y(x^3 - y)dx - x(x^3 + y) dy = 0$. We need to check if it is an exact differential equation. An exact differential equation has the form $M(x, y) dx + N(x, y) dy = 0$ and is exact if the partial derivative of $M(x, y)$ with respect to y is equal to the partial derivative of $N(x, y)$ with respect to x. In other words, we need to check whether $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. Here, $M(x, y) = y(x^3 - y)$ and $N(x, y) = -x(x^3 + y)$.

Compute the partial derivatives and verify exactness

We compute the partial derivatives as follows: $\frac{\partial M}{\partial y} = \frac{\partial}{\partial y} [y(x^3 - y)] = x^3 - 2y$, $\frac{\partial N}{\partial x} = \frac{\partial}{\partial x} [-x(x^3 + y)] = x^3 - 2y$. Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, this is an exact differential equation.

Integrate M with respect to x and N with respect to y

Now we can integrate M with respect to x and N with respect to y, and compare the results to find the general solution. From M: $\int M dx = \int y(x^3 - y) dx = \frac{1}{2}yx^4 - \frac{1}{2}y^2x + f_1(y)$, From N: $\int N dy = \int -x(x^3 + y) dy = -\frac{1}{2}x^2y^2 - \frac{1}{4}x^4y + f_2(x)$,

Combine the integrated functions to form the general solution

By comparing the two results, we can see that $f_1(y) =-\frac{1}{4}x^4y$, and $f_2(x) = \frac{1}{2}yx^4 - \frac{1}{2}y^2x$. The general solution for this exact differential equation is: $\frac{1}{2}yx^4 - \frac{1}{2}y^2x - \frac{1}{4}x^4y = C$, where C is a constant.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Differential Equations

A partial differential equation (PDE) involves functions of multiple variables and their partial derivatives. The exercise at hand distinguishes itself as an exact differential equation, which is a specific type of PDE.

With exact differential equations, both the terms can be expressed using partial derivatives. Here, **M(x, y)** and **N(x, y)** are two functions. For our equation, these functions are **M(x, y) = y(x^3 - y)** and **N(x, y) = -x(x^3 + y)**.

To confirm if an equation is exact, we check:

The partial derivative of **M(x, y)** with respect to **y**.
The partial derivative of **N(x, y)** with respect to **x**.

For exactness, these derivatives should be equal. Once verified, it classifies the equation as an exact PDE, simplifying the path to finding a solution.

Integration Techniques

To solve exact differential equations, a systematic use of integration is required. Initially, we integrate **M** with respect to **x** and **N** with respect to **y**.

**Why two integrations?**
Integrating both parts separately helps to uncover the potential functions that form a general solution.

The integrations undertaken are:

The integration of **M(x, y)dx**: \$\int y(x^3 - y) \, dx = \frac{1}{2}yx^4 - \frac{1}{2}y^2x + f_1(y)$
The integration of **N(x, y)dy**: \$\int -x(x^3 + y) \, dy = -\frac{1}{2}x^2y^2 - \frac{1}{4}x^4y + f_2(x)$

By approaching integration this way, we break down complex equations into manageable forms, later combining results to construct the equation's general solution.

General Solution

After performing integrations, the last step is to deduce the general solution. Combining the integrated expressions ensures that they align under one cohesive equation.

**How is this done?**
By equating expressions derived from both integrations and adjusting for any arbitrary functions **f_1(y)** and **f_2(x)**.

The results:

Identifying that the potential functions must satisfy both integral expressions.
Matching terms from two expressions to achieve one complete equation.

The final outcome is the general solution: \[\frac{1}{2}yx^4 - \frac{1}{2}y^2x - \frac{1}{4}x^4y = C\]
Here, **C** is a constant, representing an infinite set of solutions that satisfy the initial conditions. This showcases how exact differential solutions chant the harmony of integration results into one succinct expression.

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91影视

Solve the equation $$ y\left(x^{3}-y\right) d x-x\left(x^{3}+y\right) d y=0 $$

Short Answer

Step by step solution

Identify the given equation as exact differential equation

Compute the partial derivatives and verify exactness

Integrate M with respect to x and N with respect to y

Combine the integrated functions to form the general solution

Key Concepts

Partial Differential Equations

Integration Techniques

General Solution

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