Question

If \( \frac{dy}{dx} - y \log_e 2 = 2^{\sin x} (\cos x - 1) \log_e 2 \), then \( y \) is:

• A. \( 2^{\sin x} + c2^x \)
• B. \( 2^{\cos x} + c2^x \)
• C. \( 2^{\sin x} + c2^{-x} \)
• D. \( 2^{\cos x} + c2^{-x} \)

Hint:

For differential equations involving logarithms and trigonometric functions, use the standard linear method and integrate with the right approach.

Answer 1

The Correct Option is A

Step 1: The equation is: \[ \frac{dy}{dx} - y \log_e 2 = 2^{\sin x} (\cos x - 1) \log_e 2 \] \[ {This is a linear differential equation.} \] \[ {I.F. } = e^{-\int \log_e 2 \, dx} = e^{-x \log_e 2} = 2^{-x} \] \[ {Then the general solution is} \] \[ y 2^{-x} = \int 2^{-x} 2^{\sin x} (\cos x - 1) \log_e 2 \, dx + c \] \[ {Now let } \sin x - x = t \Rightarrow (\cos x - 1) dx = dt \] \[ \therefore y 2^{-x} = \log_e 2 \int 2^t \, dt + c \] \[ \therefore y 2^{-x} = 2^t + c \] \[ \therefore y = 2^{x+t} + c 2^x \] \[ y = 2^{\sin x} + c 2^x \] Thus, the solution is \( y = 2\sin x + c2^x \).