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

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.