https://file.aiquickdraw.com/mathbot/file/1727268455497qe27q3fj.jpg

The given differential equation in the image is:

\[
(x^3 + 2) y \, dx = x (y^4 + 3) \, dy
\]

To solve it using the method of separation of variables, let's proceed step by step:

### Step 1: Rearrange the equation
Separate the variables \(x\) and \(y\) to opposite sides.

\[
\frac{(x^3 + 2)}{x} \, dx = \frac{(y^4 + 3)}{y} \, dy
\]

### Step 2: Simplify both sides
Simplify the expressions on both sides:

\[
\frac{x^3 + 2}{x} \, dx = \frac{y^4 + 3}{y} \, dy
\]

This becomes:

\[
(x^2 + \frac{2}{x}) \, dx = (y^3 + \frac{3}{y}) \, dy
\]

### Step 3: Integrate both sides
Now, integrate both sides with respect to their respective variables.

- For the left side, we need to integrate \(x^2 + \frac{2}{x}\):

\[
\int (x^2 + \frac{2}{x}) \, dx = \frac{x^3}{3} + 2 \ln |x| + C_1
\]

- For the right side, we integrate \(y^3 + \frac{3}{y}\):

\[
\int (y^3 + \frac{3}{y}) \, dy = \frac{y^4}{4} + 3 \ln |y| + C_2
\]

### Step 4: Combine results
The general solution becomes:

\[
\frac{x^3}{3} + 2 \ln |x| = \frac{y^4}{4} + 3 \ln |y| + C
\]

Where \(C = C_2 - C_1\) is a constant.

### Step 5: Use initial condition
We are given the initial condition \(x = 1\), \(y = 1\). Substitute these into the equation to solve for \(C\):

\[
\frac{1^3}{3} + 2 \ln |1| = \frac{1^4}{4} + 3 \ln |1| + C
\]

This simplifies to:

\[
\frac{1}{3} + 0 = \frac{1}{4} + 0 + C
\]

\[
C = \frac{1}{3} - \frac{1}{4} = \frac{4}{12} - \frac{3}{12} = \frac{1}{12}
\]

### Final solution:

\[
\frac{x^3}{3} + 2 \ln |x| = \frac{y^4}{4} + 3 \ln |y| + \frac{1}{12}
\]

This is the solution to the given differential equation.

Would you like more details or clarification on any of the steps?

### 5 Related Questions:
1. How do you determine when to use variable separation in differential equations?
2. What other methods exist to solve differential equations besides separation of variables?
3. How can we use exact equations to solve this differential equation?
4. What role do logarithms play in solving differential equations?
5. Can initial conditions always be used to solve for the constant of integration?

### Tip:
Always check if an equation can be rearranged to separate variables before trying more complex methods like integrating factors or exact equations.

(x^3 + 2) y dx = x (y^4 + 3) dy at x=1, y=1. Solve using variable separation.

Learn how to solve the differential equation (x^3 + 2) y dx = x (y^4 + 3) dy using separation of variables. Includes step-by-step integration and the application of initial conditions at x=1, y=1.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation