Subtopic: Linear and non-linear first order Ordinary Differential Equations (ODE)
Question Statement:
For the first-order differential equation $M(x, y)\,dx + N(x, y)\,dy = 0$, which of the following statements is/are TRUE?
Options:
(A) The equation is exact if and only if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$
(B) If $\frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}\right) = f(x)$ is a function of $x$ alone, then $e^{\int f(x)\,dx}$ is an integrating factor
(C) If $\frac{1}{M}\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) = g(y)$ is a function of $y$ alone, then $e^{\int g(y)\,dy}$ is an integrating factor
(D) An exact differential equation can always be solved by finding a potential function $\phi(x, y)$ such that $d\phi = M\,dx + N\,dy = 0$
Answer Key:
A, B, C, D
Step-by-Step Solution & Explanation
1. **Exactness Criterion (Option A)**: In a simply connected domain, $M\,dx + N\,dy = 0$ is exact if and only if the cross partials match: $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ (True).
2. **Integrating Factor in $x$ (Option B)**: When $\frac{1}{N}(M_y - N_x) = f(x)$, multiplying by $\mu(x) = e^{\int f(x)\,dx}$ yields an exact differential equation (True).
3. **Integrating Factor in $y$ (Option C)**: When $\frac{1}{M}(N_x - M_y) = g(y)$, multiplying by $\mu(y) = e^{\int g(y)\,dy}$ yields an exact equation (True).
4. **Potential Function (Option D)**: Exactness implies the existence of $\phi(x, y)$ such that $\nabla \phi = \langle M, N \rangle$, with general solution $\phi(x, y) = C$ (True).