$$ ( 6 x ^ { 2 } + 2 y ^ { 2 } ) d x - ( x ^ { 2 } + 4 x y ) d y = 0 $$

Question

UpStudy AI · Accepted Answer

To solve the differential equation:

$$
(6x^{2} + 2y^{2})\,dx - (x^{2} + 4xy)\,dy = 0
$$

we can follow these steps:

### 1. **Check for Exactness**

First, write the equation in the standard form $$ M(x, y)\,dx + N(x, y)\,dy = 0 $$:

$$
M = 6x^{2} + 2y^{2}, \quad N = -(x^{2} + 4xy)
$$

Compute the partial derivatives:

$$
\frac{\partial M}{\partial y} = 4y, \quad \frac{\partial N}{\partial x} = - (2x + 4y)
$$

Since $$ \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x} $$, the equation is **not exact**.

### 2. **Find an Integrating Factor**

To make the equation exact, we look for an integrating factor $$ \mu(x) $$ that depends only on $$ x $$.

Calculate:

$$
\frac{M_y - N_x}{N} = \frac{4y - (-2x - 4y)}{-(x^{2} + 4xy)} = \frac{2x + 8y}{-x^{2} - 4xy} = \frac{-2(x + 4y)}{x(x + 4y)} = \frac{-2}{x}
$$

Since this expression depends only on $$ x $$, the integrating factor is:

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

### 3. **Multiply Through by the Integrating Factor**

Multiply the entire differential equation by $$ \mu(x) = x^{-2} $$:

$$
\left(6x^{2} + 2y^{2}\right)x^{-2} dx - \left(x^{2} + 4xy\right)x^{-2} dy = 0
$$

Simplify:

$$
(6 + 2\frac{y^{2}}{x^{2}})\,dx - \left(1 + 4\frac{y}{x}\right)\,dy = 0
$$

Now, the equation is **exact**.

### 4. **Find the Potential Function**

Find a function $$ \psi(x, y) $$ such that:

$$
\frac{\partial \psi}{\partial x} = 6 + 2\frac{y^{2}}{x^{2}}, \quad \frac{\partial \psi}{\partial y} = -1 - 4\frac{y}{x}
$$

Integrate $$ \frac{\partial \psi}{\partial x} $$ with respect to $$ x $$:

$$
\psi(x, y) = \int \left(6 + 2\frac{y^{2}}{x^{2}}\right) dx = 6x - 2\frac{y^{2}}{x} + h(y)
$$

Differentiate $$ \psi $$ with respect to $$ y $$:

$$
\frac{\partial \psi}{\partial y} = -4\frac{y}{x} + h'(y)
$$

Set this equal to $$ -1 - 4\frac{y}{x} $$:

$$
-4\frac{y}{x} + h'(y) = -1 - 4\frac{y}{x} \implies h'(y) = -1 \implies h(y) = -y + C
$$

Thus, the potential function is:

$$
\psi(x, y) = 6x - \frac{2y^{2}}{x} - y + C
$$

### 5. **General Solution**

Set $$ \psi(x, y) = \text{constant} $$:

$$
6x - \frac{2y^{2}}{x} - y = C
$$

where $$ C $$ is an arbitrary constant.

### **Final Answer**

After simplifying, a general solution is 6 x – y – (2 y²)⁄x equals a constant. Thus,

6x − y − 2y²/x = C
The general solution is $$ 6x - y - \frac{2y^{2}}{x} = C $$.

\( ( 6 x ^ { 2 } + 2 y ^ { 2 } ) d x - ( x ^ { 2 } + 4 x y ) d y = 0 \)