Let's solve the given **exact differential equation** step by step, as requested: ### Given Equation \[ (2y^2 + ye^{xy}) + (4xy + xe^{xy} + 2y)y' = 0 \] Or, written in differential form: \[ (2y^2 + ye^{xy})\,dx + (4xy + xe^{xy} + 2y)\,dy = 0 \] Let \( M(x, y) = 2y^2 + ye^{xy} \) and \( N(x, y) = 4xy + xe^{xy} + 2y \). --- ## **Step 1: Check for Exactness** An equation \( M(x, y)dx + N(x, y)dy = 0 \) is **exact** if: \[ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \] Calculate each: - \(\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(2y^2 + ye^{xy}) = 4y + e^{xy} + xye^{xy}\) - \(\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(4xy + xe^{xy} + 2y) = 4y + e^{xy} + xye^{xy}\) **Explanation:** Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the equation is **exact**. --- ## **Step 2: Find the Solution** Since it's exact, there exists a function \( F(x, y) \) such that: \[ \frac{\partial F}{\partial x} = M(x, y), \quad \frac{\partial F}{\partial y} = N(x, y) \] **Integrate \( M(x, y) \) with respect to \( x \):** \[ F(x, y) = \int (2y^2 + ye^{xy})\,dx = 2y^2x + y\int e^{xy}\,dx \] Let’s solve \( \int e^{xy}\,dx \): Let \( u = xy \implies du = y\,dx \implies dx = \frac{du}{y} \) \[ \int e^{xy}\,dx = \int e^u \cdot \frac{du}{y} = \frac{1}{y} \int e^u du = \frac{1}{y} e^u = \frac{1}{y} e^{xy} \] So, \[ F(x, y) = 2y^2x + y \left(\frac{1}{y} e^{xy}\right) + h(y) = 2y^2x + e^{xy} + h(y) \] Now, differentiate \( F(x, y) \) with respect to \( y \): \[ \frac{\partial F}{\partial y} = 4xy + xe^{xy} + h'(y) \] Set this equal to \( N(x, y) \): \[ 4xy + xe^{xy} + h'(y) = 4xy + xe^{xy} + 2y \implies h'(y) = 2y \implies h(y) = y^2 \] **Final Solution:** \[ F(x, y) = 2y^2x + e^{xy} + y^2 = C \] --- ## **Final Answer:** \[ \boxed{2y^2x + e^{xy} + y^2 = C} \] --- ### **Summary of Steps:** 1. **Checked exactness** by verifying \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). 2. **Integrated** to find the potential function and determined the solution.