https://file.aiquickdraw.com/mathbot/file/1727317793608kpnae57e.jpg

The given differential equation is:

\[
2x^2 y' + 8xy = 18y^3
\]

Let's solve this equation step by step.

### Step 1: Rewrite the equation
We start by simplifying and rewriting the equation:

\[
2x^2 y' + 8xy = 18y^3
\]

Divide through by \(2x^2\) to make the equation easier to work with:

\[
y' + \frac{8}{2x}y = \frac{18}{2x^2}y^3
\]

Simplifying the coefficients:

\[
y' + \frac{4}{x}y = \frac{9}{x^2}y^3
\]

### Step 2: Convert to standard form
This is a first-order differential equation of the form:

\[
y' + P(x)y = Q(x)y^n
\]

Here, \(P(x) = \frac{4}{x}\), \(Q(x) = \frac{9}{x^2}\), and \(n = 3\). This is a Bernoulli differential equation. We can solve it using the substitution \(v = y^{1-n} = y^{-2}\).

### Step 3: Perform substitution
Substitute \(v = y^{-2}\). Then, differentiate with respect to \(x\):

\[
\frac{dv}{dx} = -2y^{-3}y'
\]

Substituting into the original equation, we get:

\[
-2y^{-3}y' + \frac{4}{x}y = \frac{9}{x^2}y^3
\]

Multiply the equation by \(y^{-3}\):

\[
-2y^{-3}y' + \frac{4}{x}y^{-2} = \frac{9}{x^2}
\]

Using the substitution \(v = y^{-2}\), we simplify:

\[
-2 \frac{dv}{dx} + \frac{4}{x}v = \frac{9}{x^2}
\]

### Step 4: Simplify and solve the linear equation
Rearrange the equation:

\[
\frac{dv}{dx} - \frac{2}{x}v = -\frac{9}{2x^2}
\]

This is a linear first-order differential equation in \(v(x)\). The standard method of solving linear equations is to find the integrating factor. The integrating factor is:

\[
\mu(x) = e^{\int -\frac{2}{x} dx} = e^{-2 \ln x} = x^{-2}
\]

Multiply through by the integrating factor:

\[
x^{-2} \frac{dv}{dx} - \frac{2}{x^3} v = -\frac{9}{2x^4}
\]

This simplifies to:

\[
\frac{d}{dx}(x^{-2}v) = -\frac{9}{2x^4}
\]

### Step 5: Integrate both sides
Integrate both sides with respect to \(x\):

\[
x^{-2}v = \int -\frac{9}{2x^4} dx
\]

The right-hand side integrates to:

\[
x^{-2}v = \frac{9}{6x^3} + C = \frac{3}{2x^3} + C
\]

Multiply by \(x^2\) to solve for \(v\):

\[
v = \frac{3}{2x} + Cx^2
\]

### Step 6: Back-substitute for \(v\)
Recall that \(v = y^{-2}\), so:

\[
y^{-2} = \frac{3}{2x} + Cx^2
\]

Take the reciprocal to solve for \(y\):

\[
y = \left( \frac{3}{2x} + Cx^2 \right)^{-\frac{1}{2}}
\]

This is the general solution to the differential equation.

---

Would you like further details or clarifications on any of the steps?

Here are five related questions:
1. How do you identify a Bernoulli differential equation?
2. What is the purpose of using an integrating factor in solving linear differential equations?
3. Can you show how to verify a solution to a differential equation?
4. What other methods can be used to solve first-order nonlinear differential equations?
5. How does substitution help simplify differential equations?

**Tip:** When solving differential equations, carefully identify the type of equation (e.g., linear, separable, Bernoulli) as this dictates the solution method.

Find the general solution of the differential equation 2x^2 y' + 8xy = 18y^3

Learn how to solve the differential equation 2x^2 y' + 8xy = 18y^3 using Bernoulli's method. This step-by-step guide covers key concepts in differential equations, substitution techniques, and the integrating factor method. Suitable for undergraduate students studying calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation