Question

The general solution of the differential equation $$ \frac{dy}{dx} = \frac{2y^2 + 1}{2y^3 - 4xy + y}. $$ 

• A. \( 4xy^2 + 2x = y^4 + y^2 + c \)
• B. \( 2xy^2 + x = y^4 - y^2 + c \)
• C. \( 4xy^2 - 2x = y^4 + y^2 + c \)
• D. \( 4xy^2 + 2x = y^4 - y^2 + c \)

Hint:

Check for exactness in differential equations before solving. - Use direct integration for exact equations.

Answer 1

The Correct Option is A

Step 1: Express the equation in differential form
Rewriting: \[ (2y^3 - 4xy + y) dx - (2y^2 + 1) dy = 0. \] Step 2: Check for exactness
Computing partial derivatives, \[ M(x, y) = 2y^3 - 4xy + y, \quad N(x, y) = -(2y^2 + 1). \] Since \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the equation is exact. Step 3: Solve using integration
\[ \int (2y^3 - 4xy + y) dx = F(x, y). \] \[ \int (2y^2 + 1) dy = G(x, y). \] Solving, we get: \[ 4xy^2 + 2x = y^4 + y^2 + c. \] Thus, the correct answer is \( \boxed{4xy^2 + 2x = y^4 + y^2 + c} \).