Solve the initial value problem \( \frac{d y}{d x}+y \cos x=4 \cos x, \quad y(0)=6 \) \( y(x)= \)

To solve the initial value problem \( \frac{dy}{dx} + y \cos x = 4 \cos x \), we will use an integrating factor. The integrating factor \( \mu(x) \) is given by \( e^{\int \cos x \, dx} = e^{\sin x} \). Multiplying the entire equation by the integrating factor: \[ e^{\sin x} \frac{dy}{dx} + y e^{\sin x} \cos x = 4 e^{\sin x} \cos x \] The left side can be simplified to: \[ \frac{d}{dx}(y e^{\sin x}) = 4 e^{\sin x} \cos x \] Now, we integrate both sides: \[ y e^{\sin x} = \int 4 e^{\sin x} \cos x \, dx \] For the right side: Using substitution \( u = \sin x \) gives \( du = \cos x \, dx \), transforming the integral to: \[ \int 4 e^u \, du = 4 e^u + C = 4 e^{\sin x} + C \] Thus, we have: \[ y e^{\sin x} = 4 e^{\sin x} + C \] Solving for \( y \): \[ y = 4 + Ce^{-\sin x} \] To find \( C \), use the initial condition \( y(0) = 6 \): \[ 6 = 4 + Ce^{-\sin(0)} \implies 6 = 4 + C \implies C = 2 \] Thus, the solution is: \[ y = 4 + 2 e^{-\sin x} \]