solve using First Order DEs solutions by substitutions: xy dx + (2x^2 + y^2)dy = 0. Answer must be: 3x^2 y^4 + y^6 = C

To solve the differential equation \(xy \, dx + (2x^2 + y^2) \, dy = 0\) using the method of substitution, we follow these steps:

### Step 1: Rearrange the equation
The given equation is:
\[
xy \, dx + (2x^2 + y^2) \, dy = 0
\]
We can rewrite it as:
\[
xy \, dx = -(2x^2 + y^2) \, dy
\]
Or:
\[
\frac{dx}{dy} = -\frac{2x^2 + y^2}{xy}
\]

### Step 2: Choose an appropriate substitution
To simplify this equation, let's use the substitution \(v = \frac{x}{y}\), which implies \(x = vy\).

Differentiating \(x = vy\) with respect to \(y\):
\[
\frac{dx}{dy} = v + y \frac{dv}{dy}
\]

### Step 3: Substitute into the differential equation
Substituting \(x = vy\) and \(\frac{dx}{dy} = v + y \frac{dv}{dy}\) into the differential equation:
\[
v + y \frac{dv}{dy} = -\frac{2(vy)^2 + y^2}{vy}
\]

Simplify the expression:
\[
v + y \frac{dv}{dy} = -\frac{2v^2y^2 + y^2}{vy} = -\frac{y^2(2v^2 + 1)}{vy}
\]
\[
v + y \frac{dv}{dy} = -\frac{y(2v^2 + 1)}{v}
\]

This reduces to:
\[
v + y \frac{dv}{dy} = -\frac{2v^2 + 1}{v}
\]

Now, simplify it further:
\[
y \frac{dv}{dy} = -\frac{2v^2 + 1}{v} - v
\]
\[
y \frac{dv}{dy} = -\frac{2v^2 + 1 + v^2}{v}
\]
\[
y \frac{dv}{dy} = -\frac{3v^2 + 1}{v}
\]

This simplifies to:
\[
y \frac{dv}{dy} = -3v - \frac{1}{v}
\]
\[
\frac{v \, dv}{3v^2 + 1} = -\frac{dy}{y}
\]

### Step 4: Integrate both sides
Now, integrate both sides to solve for \(v\) and \(y\):

The left side requires partial fractions to integrate:
\[
\int \frac{v \, dv}{3v^2 + 1} = \int -\frac{dy}{y}
\]

Let's integrate the left side:
\[
\frac{1}{6} \ln(3v^2 + 1) = -\ln|y| + C
\]

Exponentiate both sides to remove the logarithms:
\[
3v^2 + 1 = \frac{C}{y^6}
\]

Substitute back \(v = \frac{x}{y}\):
\[
3\left(\frac{x}{y}\right)^2 + 1 = \frac{C}{y^6}
\]
\[
\frac{3x^2}{y^2} + 1 = \frac{C}{y^6}
\]

Multiply by \(y^6\) to clear the fraction:
\[
3x^2y^4 + y^6 = C
\]

### Final Answer
The solution to the differential equation is:
\[
3x^2y^4 + y^6 = C
\]

Would you like more details or have any questions?

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**Related Questions:**

1. How do you determine the appropriate substitution in first-order differential equations?
2. Why is partial fraction decomposition useful in integration?
3. What is the role of the constant of integration in solving DEs?
4. How do you verify a solution obtained from a first-order differential equation?
5. Can this method be applied to other types of first-order non-linear differential equations?

**Tip:** When dealing with complex differential equations, it helps to simplify the problem using substitutions that align variables or help eliminate terms.

Learn how to solve the differential equation xy dx + (2x^2 + y^2) dy = 0 using the method of substitution. Detailed step-by-step solution provided with integration and substitution techniques. Suitable for advanced college-level students and anyone studying differential equations.

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