Let's solve the differential equation: \[ \cos x (e^{2y} - y) y' = e^y \sin 2x, \quad y() = \] ### Step 1: Rearranging Variables Bring all \(y\) terms to one side and \(\) terms the other: \[ (e^{2y} - y) y' e^{-y} = \frac{\sin 2x}{\cos x} \] \[ (e^{2y} - y) e^{-y} y' = \frac{\sin 2x}{\cos x} \] Now, notice \(y' = \frac{dy}{dx}\): \[ (e^{2y} - y) e^{-y} \frac{dy}{dx} = \frac{\sin 2x}{\cos x} \] \[ (e^{2y} - y) e^{-y} dy = \frac{\sin 2x}{\cos x} dx \] ### Step 2: Simplifying Each Side #### Left Side: \[ (e^{2y} - y) e^{-y} = e^{2y} e^{-y} - y e^{-y} = e^{y} - y e^{-y} \] So, \[ (e^{y} - y e^{-y})\, dy = \frac{\sin 2x}{\cos x}\, dx \] #### Right Side: \[ \frac{\sin 2x}{\cos x} = \frac{2 \sin x \cos x}{\cos x} = 2 \sin x \] ### Step 3: Integrate Both Sides \[ \int (e^{y} - y e^{-y})\, dy = \int 2 \sin x\, dx \] #### Left Side: Let's split the integral: \[ \int e^{y} dy - \int y e^{-y} dy \] - \(\int e^{y} dy = e^y\) - For \(\int y e^{-y} dy\), use integration by parts: - Let \(u = y\), \(dv = e^{-y} dy\) - \(du = dy\), \(v = -e^{-y}\) \[ \int y e^{-y} dy = -y e^{-y} + \int e^{-y} dy = -y e^{-y} - e^{-y} \] So, \[ \int y e^{-y} dy = - (y + 1) e^{-y} \] Therefore, the left side is: \[ e^y - [-(y + 1) e^{-y}] = e^y + (y + 1) e^{-y} \] #### Right Side: \[ \int 2 \sin x dx = -2 \cos x \] ### Step 4: Combine Results \[ e^y + (y + 1) e^{-y} = -2 \cos x + C \] ### Step 5: Apply Initial Condition Given \(y() = \): \[ e^ + ( + 1)e^ = -2\cos + C \implies 1 + 1 = -2(1) + C \implies 2 = -2 + C \implies C = 4 \] ### Final Solution \[ \boxed{ e^y + (y + 1) e^{-y} = -2 \cos x + 4 } \] Or, if you want it in terms of \(y\): \[ e^y + (y + 1) e^{-y} + 2\cos x = 4 \]