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.The integrating factor (I.F.) is calculated as: \( e^{-\int \log_e 2 \, dx} = e^{-x \log_e 2} = 2^{-x} \).The general solution is given by:\[ y 2^{-x} = \int 2^{-x} 2^{\sin x} (\cos x - 1) \log_e 2 \, dx + c \]Let \( t = \sin x - x \). Differentiating with respect to x yields \( dt = (\cos x - 1) dx \).Substituting these into the integral, we get:\[ \therefore y 2^{-x} = \log_e 2 \int 2^t \, dt + c \]Evaluating the integral:\[ \therefore y 2^{-x} = 2^t + c \]Rearranging to solve for y:\[ \therefore y = 2^{x+t} + c 2^x \]Substituting \( t = \sin x - x \) back:\[ y = 2^{\sin x} + c 2^x \]Thus, the solution is \( y = 2\sin x + c2^x \).