Question

The solution of the differential equation \[ \frac{dy}{dx} + y \cos x = \frac{1}{2} \sin 2x \] is

• A. \( y e^{\sin x} = e^{\sin x + 1} + c \)
• B. \( y e^{\sin x} = e^{\sin x (sin x - 1)} + c \)
• C. \( y e^{\sin 2x} = e^{\sin 2x (sin x - 1)} + c \)
• D. \( y e^{\cos x} = e^{\sin x (cos x - 1)} + c \)

Hint:

When solving first-order linear differential equations, use the method of integrating factors to simplify the equation and find the general solution.

Answer 1

The Correct Option is B

We are given the differential equation: \[ \frac{dy}{dx} + y \cos x = \frac{1}{2} \sin 2x \] This is a first-order linear differential equation. To solve this equation, we use the method of integrating factors. The standard form of a linear first-order differential equation is: \[ \frac{dy}{dx} + P(x)y = Q(x) \] Here, \( P(x) = \cos x \) and \( Q(x) = \frac{1}{2} \sin 2x \). Step 1: Find the integrating factor The integrating factor is given by: \[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \cos x \, dx} = e^{\sin x} \] Step 2: Multiply through by the integrating factor Multiplying the entire differential equation by \( e^{\sin x} \), we get: \[ e^{\sin x} \frac{dy}{dx} + y e^{\sin x} \cos x = \frac{1}{2} e^{\sin x} \sin 2x \] Now, the left side is the derivative of \( y e^{\sin x} \), so the equation becomes: \[ \frac{d}{dx} \left( y e^{\sin x} \right) = \frac{1}{2} e^{\sin x} \sin 2x \] Step 3: Integrate both sides Now, we integrate both sides: \[ y e^{\sin x} = \int \frac{1}{2} e^{\sin x} \sin 2x \, dx + c \] After performing the integration, the solution is: \[ y e^{\sin x} = e^{\sin x (sin x - 1)} + c \] Thus, the correct answer is option (B).