The differential equation $(2xy^3 + y \cos x) \, dx + (3x^2 y^2 + \sin x + k) \, dy = 0$ is exact for any constant real value of $k$. The potential function $u(x, y) = C$ representing the general solution to this exact equation is:

(A) $x^2 y^3 + y \sin x + k y = C$

(B) $2x^2 y^3 + y \cos x + k y = C$

(C) $x^2 y^2 + \sin x + k x = C$

(D) $x y^3 + y \sin x = C$

Answer Key:

Verified in official key

### 1. Strategy & Theoretical Principles For an ODE $M(x, y) \, dx + N(x, y) \, dy = 0$ to be exact, it must satisfy: $$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$$ Then there exists a potential function $u(x, y)$ such that $\frac{\partial u}{\partial x} = M$ and $\frac{\partial u}{\partial y} = N$, with solution $u(x, y) = C$. ### 2. Analytical Verification of Exactness - $M(x, y) = 2xy^3 + y \cos x \implies \frac{\partial M}{\partial y} = 6xy^2 + \cos x$ - $N(x, y) = 3x^2 y^2 + \sin x + k \implies \frac{\partial N}{\partial x} = 6xy^2 + \cos x$ Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the equation is exact. ### 3. Step-by-Step Calculation of Potential Function Integrate $M$ with respect to $x$ (treating $y$ as constant): $$u(x, y) = \int (2xy^3 + y \cos x) \, dx = x^2 y^3 + y \sin x + g(y)$$ Differentiating with respect to $y$ and equating to $N$: $$\frac{\partial u}{\partial y} = 3x^2 y^2 + \sin x + g'(y) = 3x^2 y^2 + \sin x + k$$ $$g'(y) = k \implies g(y) = k y$$ Therefore, the potential function is: $$u(x, y) = x^2 y^3 + y \sin x + k y = C$$ ### 4. Distractor Forensics - **Option A:** Correct potential function. - **Option B:** Failed to integrate $2x$ to $x^2$. - **Option C:** Incorrect powers. ### 5. Exam Trap & Standard Reference - **Trap Alert:** Remember to differentiate the integrated function with respect to $y$ to determine $g'(y)$. - **Standard Reference:** Kreyszig, E., *Advanced Engineering Mathematics*, Chapter 1.

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