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

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

Upstudy AI Solution

Answer

Solution

1. Check for Exactness

2. Find an Integrating Factor

3. Multiply Through by the Integrating Factor

4. Find the Potential Function

5. General Solution

Final Answer

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