Question

If the solution $y(x)$ of the given differential equation $(e^y+1) \cos x dx + e^y \sin x dy = 0$ passes through the point $(\pi/2, 0)$ then the value of $y(\pi/6)$ is

• A. $\log 2$
• B. $\log 3$
• C. $e^2$
• D. $e^{-3}$

Hint:

Always find the constant of integration $c$ using the given point before solving for a specific value of $y$.

Answer 1

The Correct Option is B

Step 1: Concept
This is an exact differential equation or can be solved by separating variables.

Step 2: Meaning
Rearrange: $e^y \sin x dy = -(e^y+1) \cos x dx \implies \frac{e^y}{e^y+1} dy = -\frac{\cos x}{\sin x} dx$.

Step 3: Analysis
Integrate both sides: $\log(e^y+1) = -\log(\sin x) + \log c \implies \log[(e^y+1)\sin x] = \log c \implies (e^y+1)\sin x = c$. Using point $(\pi/2, 0)$: $(e^0+1)\sin(\pi/2) = c \implies (1+1)(1) = c \implies c = 2$.

Step 4: Conclusion
Equation: $(e^y+1)\sin x = 2$. For $x = \pi/6$: $(e^y+1)(1/2) = 2 \implies e^y+1 = 4 \implies e^y = 3 \implies y = \log 3$. Final Answer: (B)