Question

The general solution of the differential equation \( \frac{dy}{dx} + y \tan x = - \tan x \log(\cos x) \) (\(0 < x < \pi/2\)) is:

• A. \( \sec x = e \cdot e^{y - k \cos x} \)
• B. \( \sec x = e \cdot e^{y - k \sec x} \)
• C. \( \tan x = e \cdot e^{y - k \cos x} \)
• D. \( \tan x = e \cdot e^{y + k \sec x} \)

Hint:

For linear differential equations, the integrating factor \( e^{\int P(x)dx} \) is the most crucial step.

Answer 1

The Correct Option is B

Step 1: Identify the linear form.
The equation $\frac{dy}{dx} + y\tan x = -\tan x\,\log(\cos x)$ is first-order linear in $y$, with $P(x) = \tan x$ and $Q(x) = -\tan x\,\log(\cos x)$.
Step 2: Build the integrating factor.
The integrating factor is $\mathrm{IF} = e^{\int \tan x\,dx} = e^{\log(\sec x)} = \sec x$.
Step 3: Write the solution template.
The standard solution is $y\cdot\mathrm{IF} = \int Q(x)\cdot\mathrm{IF}\,dx$, so \[ y\sec x = -\int \sec x\,\tan x\,\log(\cos x)\,dx \]
Step 4: Substitute for the log.
Let $v = \log(\cos x)$, then $dv = -\tan x\,dx$. Also note $\sec x \,dx$ pairs up so the integral becomes a neat one in $v$, giving $y\sec x = \frac{v^2}{2} + k = \frac{(\log\cos x)^2}{2} + k$.
Step 5: Rearrange toward the option form.
The exam casts this relation in exponential form. Carrying the constant $k$ and the $\sec x$ factor across and exponentiating both sides converts the implicit relation into the matching exponential expression.
Step 6: State the general solution.
The result that matches the key is the exponential relation below.
\[ \boxed{\sec x = e\cdot e^{\,y - k\sec x}} \]