Question

A particular solution of \(3 \mathrm{e}^x \tan y \mathrm{~d} x+\left(1-\mathrm{e}^x\right) \sec ^2 y \mathrm{~d} y=0\) with \(y(1)=\frac{\pi}{4}\) is

• A. \(\mathrm{tan}y=(\frac{1-\mathrm{e}^{3}}{1-\mathrm{e}^{x}})^{3}\)
• B. \(\mathrm{tan}y=(\frac{1-\mathrm{e}^{2}}{1-\mathrm{e}^{x}})^{3}\)
• C. \(\mathrm{tan}y=(\frac{1-\mathrm{e}}{1-\mathrm{e}^{x}})^{3}\)
• D. \(\mathrm{tan}y=(\frac{1-\mathrm{e}^{x}}{1-\mathrm{e}})^{3}\)

Hint:

Separate the variables, integrate both sides and use the given condition to fix the constant.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The equation \(3e^x\tan y\,dx+(1-e^x)\sec^2y\,dy=0\) is variable separable.

Step 2: Separate and integrate
\(\frac{\sec^2y}{\tan y}dy=\frac{-3e^x}{1-e^x}dx\)
Integrating: \(\log|\tan y|=3\log|1-e^x|+\log C\), so \(\tan y=C(1-e^x)^3\).

Step 3: Use the condition
At \(x=1,\ y=\frac{\pi}{4}\): \(1=C(1-e)^3\), so \(C=\frac{1}{(1-e)^3}\).

Step 4: Result
\(\tan y=\left(\frac{1-e^x}{1-e}\right)^3\), which is option (D).

Final Answer:
The particular solution is \(\tan y=\left(\frac{1-e^x}{1-e}\right)^3\). \[ \boxed{\text{(D) }\tan y=\left(\dfrac{1-\mathrm{e}^{x}}{1-\mathrm{e}}\right)^{3}} \]