Question

The integrating factor of the differential equation $(e^x - \sin y) dx + \cos y \, dy = 0$ is

• A. $e^x$
• B. $e^{-x}$
• C. $e^x \sin y$
• D. $e^{-x} \cos y$

Hint:

Rules for Integrating Factors: 1. If $\frac{M_y - N_x}{N} = f(x)$, then $I.F. = e^{\int f(x) dx}$. 2. If $\frac{N_x - M_y}{M} = g(y)$, then $I.F. = e^{\int g(y) dy}$.

Answer 1

The Correct Option is B

Step 1: Concept
An equation $M(x, y) dx + N(x, y) dy = 0$ is exact if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. If it is non-exact, we find an integrating factor $\mu(x)$ such that multiplying the equation by $\mu(x)$ makes it exact.

Step 2: Key Formulas and Approach
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 the integrating factor is: \[ I.F. = e^{\int f(x) dx} \]

Step 3: Step-by-step Explanation

• Identify $M$ and $N$ from $(e^x - \sin y) dx + \cos y \, dy = 0$: \[ M = e^x - \sin y, \quad N = \cos y \]
• Compute partial derivatives: \[ \frac{\partial M}{\partial y} = -\cos y, \quad \frac{\partial N}{\partial x} = 0 \]
• Since $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$, the equation is not exact.
• Calculate the test ratio: \[ \frac{1}{N} \left( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} \right) = \frac{-\cos y - 0}{\cos y} = -1 \]
• Since $-1$ is a function depending solely on $x$ ($f(x) = -1$): \[ I.F. = e^{\int (-1) dx} = e^{-x} \]
• Verification: Multiply original equation by $e^{-x}$: \[ (1 - e^{-x} \sin y) dx + e^{-x} \cos y \, dy = 0 \] Now $\frac{\partial M'}{\partial y} = -e^{-x} \cos y = \frac{\partial N'}{\partial x}$, which is exact!

Step 4: Final Answer
The integrating factor of the differential equation is $e^{-x}$. Thus, Option (B) is correct.