Question

The solution of the differential equation: \( x\cos y\,dy = (xe^x\log x + e^x)dx \) is

• A. \( \sin y - e^x\log x = c \)
• B. \( \sin y = e^x\log x + c \)
• C. \( \sin y = e^x + \log x + c \)
• D. \( \sin y = \frac{e^x}{x} + c \)

Hint:

In differential equations, look for expressions that are exact derivativesHere \( e^x\log x + \frac{e^x}{x} \) is directly the derivative of \( e^x\log x \).

Answer 1

The Correct Option is B

Step 1: Write the given differential equation.
\[ x\cos y\,dy = (xe^x\log x + e^x)dx \]

Step 2: Divide both sides by \( x \).
\[ \cos y\,dy = \left(e^x\log x + \frac{e^x}{x}\right)dx \]

Step 3: Observe the right-hand side.
The expression on the right side is:
\[ e^x\log x + \frac{e^x}{x} \]
This is the derivative of \( e^x\log x \).

Step 4: Verify using product rule.
\[ \frac{d}{dx}(e^x\log x) = e^x\log x + e^x\cdot \frac{1}{x} \]
\[ = e^x\log x + \frac{e^x}{x} \]

Step 5: Integrate both sides.
\[ \int \cos y\,dy = \int \left(e^x\log x + \frac{e^x}{x}\right)dx \]

Step 6: Perform integration.
\[ \int \cos y\,dy = \sin y \]
and
\[ \int \left(e^x\log x + \frac{e^x}{x}\right)dx = e^x\log x \]
So:
\[ \sin y = e^x\log x + c \]

Step 7: Final conclusion.
\[ \boxed{\sin y = e^x\log x + c} \]