Use Bernoulli Equation Method to solve (8y^2-xy)dx+x^2dy=0

To solve the differential equation \((8y^2 - xy)dx + x^2dy = 0\) using the Bernoulli Equation method, we first rewrite the equation in a form that suggests a possible transformation. The Bernoulli equation is typically of the form:

\[
\frac{dy}{dx} + P(x) y = Q(x) y^n
\]

However, this differential equation does not immediately appear in the standard form of Bernoulli's equation. Let's manipulate it into a solvable form.

### Step 1: Rewrite the equation

First, express the differential equation in terms of \(\frac{dy}{dx}\). Start by dividing the whole equation by \(dx\):

\[
(8y^2 - xy) + x^2 \frac{dy}{dx} = 0
\]

Rearrange this into a form for \(\frac{dy}{dx}\):

\[
x^2 \frac{dy}{dx} = -(8y^2 - xy)
\]

This simplifies to:

\[
\frac{dy}{dx} = -\frac{8y^2 - xy}{x^2}
\]

### Step 2: Identify Bernoulli form

We aim to bring this equation into a Bernoulli form. Factor out \(y\) from the right-hand side:

\[
\frac{dy}{dx} = -\frac{y(8y - x)}{x^2}
\]

This is now a Bernoulli equation of the form:

\[
\frac{dy}{dx} + \frac{x}{x^2}y = \frac{8}{x^2} y^2
\]

Which simplifies to:

\[
\frac{dy}{dx} + \frac{1}{x} y = \frac{8}{x^2} y^2
\]

This is in the form of a Bernoulli equation where \(n = 2\).

### Step 3: Solve using substitution

Let \(v = y^{1-n} = y^{-1}\), which implies:

\[
\frac{dv}{dx} = -y^{-2} \frac{dy}{dx}
\]

Now substitute into the original equation. Since \(n = 2\), the transformation becomes:

\[
\frac{dv}{dx} - \frac{1}{x} v = -\frac{8}{x^2}
\]

### Step 4: Solve the linear equation

This is a linear differential equation in \(v\):

\[
\frac{dv}{dx} - \frac{1}{x} v = -\frac{8}{x^2}
\]

To solve this, we find the integrating factor \( \mu(x) = e^{\int -\frac{1}{x} dx} = \frac{1}{x} \). Multiply through by the integrating factor:

\[
\frac{1}{x} \frac{dv}{dx} - \frac{1}{x^2} v = -\frac{8}{x^3}
\]

This simplifies to:

\[
\frac{d}{dx} \left( \frac{v}{x} \right) = -\frac{8}{x^3}
\]

### Step 5: Integrate both sides

Integrate both sides with respect to \(x\):

\[
\frac{v}{x} = \int -\frac{8}{x^3} dx
\]

This gives:

\[
\frac{v}{x} = \frac{4}{x^2} + C
\]

Thus, multiplying through by \(x\):

\[
v = \frac{4}{x} + Cx
\]

### Step 6: Back-substitute for \(v\)

Recall that \(v = y^{-1}\), so:

\[
y^{-1} = \frac{4}{x} + Cx
\]

Finally, solving for \(y\):

\[
y = \frac{1}{\frac{4}{x} + Cx}
\]

This is the general solution to the differential equation.

### Would you like further details or have any questions?

Here are 5 related questions that expand on this topic:

1. How do you solve Bernoulli equations when the exponent \(n\) is different?
2. What are the common techniques to solve non-linear first-order differential equations?
3. How do you determine integrating factors for linear differential equations?
4. How does the Bernoulli equation relate to separable differential equations?
5. What are some applications of Bernoulli’s equation in fluid dynamics?

**Tip**: When solving differential equations, always check if a substitution or transformation simplifies the equation into a familiar form, like a Bernoulli or linear form.

Use Bernoulli Equation Method to solve (8y^2 - xy)dx + x^2dy = 0

Learn how to solve the differential equation (8y^2 - xy)dx + x^2dy = 0 using the Bernoulli equation method. This step-by-step guide covers advanced calculus concepts, including nonlinear first-order differential equations, substitution methods, and integrating factors. Ideal for advanced high school or undergraduate calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation