$$ ( x ^ { 3 } + y ^ { 3 } ) d x = ( x ^ { 2 } y + x y ^ { 2 } ) d y $$

Question

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To solve the differential equation:

$$
(x^3 + y^3) \, dx = (x^2 y + x y^2) \, dy
$$

we can follow these steps:

### **1. Rewrite the Equation**

First, bring all terms to one side to set the equation to zero:

$$
(x^3 + y^3) \, dx - (x^2 y + x y^2) \, dy = 0
$$

### **2. Check for Exactness**

A differential equation of the form $$ M \, dx + N \, dy = 0 $$ is **exact** if:

$$
\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}
$$

Here,
- $$ M = x^3 + y^3 $$
- $$ N = -(x^2 y + x y^2) $$

Compute the partial derivatives:

$$
\frac{\partial M}{\partial y} = 3y^2
$$
$$
\frac{\partial N}{\partial x} = -(2xy + y^2)
$$

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

### **3. Make the Equation Homogeneous**

To handle non-exact equations, we can look for an integrating factor or use substitution techniques. In this case, the equation is **homogeneous**, so we'll use the substitution:

$$
y = v x \quad \Rightarrow \quad dy = v \, dx + x \, dv
$$

### **4. Substitute and Simplify**

Replace $$ y $$ and $$ dy $$ in the original equation:

$$
(x^3 + (v x)^3) \, dx = (x^2 \cdot v x + x \cdot v^2 x^2) (v \, dx + x \, dv)
$$

Simplify each term:

$$
x^3 (1 + v^3) \, dx = x^3 (v + v^2) (v \, dx + x \, dv)
$$

Divide both sides by $$ x^3 $$:

$$
(1 + v^3) \, dx = (v + v^2)(v \, dx + x \, dv)
$$

Expand the right-hand side:

$$
(1 + v^3) \, dx = (v^2 + v^3) \, dx + (v + v^2) x \, dv
$$

Subtract $$ (v^2 + v^3) \, dx $$ from both sides:

$$
(1 - v^2) \, dx = (v + v^2) x \, dv
$$

### **5. Separate Variables**

Divide both sides by $$ x(1 - v^2) $$:

$$
\frac{dx}{x} = \frac{v + v^2}{1 - v^2} \, dv
$$

Simplify the right-hand side:

$$
\frac{v + v^2}{1 - v^2} = \frac{v(1 + v)}{(1 - v)(1 + v)} = \frac{v}{1 - v}
$$

Thus, the equation becomes:

$$
\frac{dx}{x} = \frac{v}{1 - v} \, dv
$$

### **6. Integrate Both Sides**

Integrate both sides:

$$
\int \frac{1}{x} \, dx = \int \frac{v}{1 - v} \, dv
$$

**Left Side:**

$$
\int \frac{1}{x} \, dx = \ln |x| + C_1
$$

**Right Side:**

$$
\int \frac{v}{1 - v} \, dv
$$

Let’s set $$ u = 1 - v $$, then $$ du = -dv $$. Rewrite the integral:

$$
\int \frac{v}{1 - v} \, dv = -\int \frac{(1 - u)}{u} \, du = -\int \left(1 - \frac{1}{u}\right) \, du = -u + \ln |u| + C_2
$$

Substitute back $$ u = 1 - v $$:

$$
- (1 - v) + \ln |1 - v| + C_2 = v - 1 + \ln |1 - v| + C_2
$$

### **7. Combine the Results**

Combine the integrated forms:

$$
\ln |x| = v - 1 + \ln |1 - v| + C
$$

where $$ C = C_2 - C_1 $$ is the constant of integration.

### **8. Substitute Back $$ v = \frac{y}{x} $$**

Recall that $$ v = \frac{y}{x} $$. Substitute back:

$$
\ln x = \frac{y}{x} - 1 + \ln \left(1 - \frac{y}{x}\right) + C
$$

Rearrange terms:

$$
\ln x + 1 - \frac{y}{x} = \ln \left(1 - \frac{y}{x}\right) + C
$$

For simplicity, we can write the solution in implicit form:

$$
(x - y) e^{\frac{y}{x}} = K
$$

where $$ K = e^{C} $$ is a new constant.

### **9. Final General Solution**

$$
(x - y) e^{\frac{y}{x}} = C
$$

where $$ C $$ is an arbitrary constant.

---

**Summary:**

The general solution to the differential equation:

$$
(x^3 + y^3) \, dx = (x^2 y + x y^2) \, dy
$$

is:

$$
(x - y) e^{\frac{y}{x}} = C
$$

where $$ C $$ is an arbitrary constant.
The general solution to the differential equation $$ (x^3 + y^3) \, dx = (x^2 y + x y^2) \, dy $$ is: $$ (x - y) e^{\frac{y}{x}} = C $$ where $$ C $$ is an arbitrary constant.

\( ( x ^ { 3 } + y ^ { 3 } ) d x = ( x ^ { 2 } y + x y ^ { 2 } ) d y \)

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