Let's solve the given differential equation step by step: Given: \[ 3e^x \tan y + (1-e^x) \sec^2 y \frac{dy}{dx} = 0 \] Let's rewrite it in a more standard form: \[ (1-e^x) \sec^2 y \frac{dy}{dx} = -3e^x \tan y \] \[ \sec^2 y \frac{dy}{dx} = \frac{-3e^x \tan y}{1-e^x} \] \[ \frac{dy}{dx} = \frac{-3e^x \tan y}{(1-e^x)\sec^2 y} \] Recall that \(\sec^2 y = 1/\cos^2 y\), and \(\tan y = \sin y/\cos y\), so: \[ \frac{dy}{dx} = \frac{-3e^x (\sin y/\cos y)}{(1-e^x)(1/\cos^2 y)} \] \[ = \frac{-3e^x \sin y \cos^2 y}{(1-e^x)\cos y} \] \[ = \frac{-3e^x \sin y \cos y}{1-e^x} \] This is a separable equation: \[ \frac{dy}{\sin y \cos y} = \frac{-3e^x}{1-e^x} dx \] Recall that \(\sin y \cos y = \frac{1}{2}\sin 2y\): \[ \frac{dy}{\frac{1}{2}\sin 2y} = \frac{-3e^x}{1-e^x} dx \] \[ \frac{2\,dy}{\sin 2y} = \frac{-3e^x}{1-e^x} dx \] Now integrate both sides: \[ \int \frac{2\,dy}{\sin 2y} = \int \frac{-3e^x}{1-e^x} dx \] ### Left Side: \[ \int \frac{2\,dy}{\sin 2y} = 2 \int \csc 2y \, dy \] Recall: \[ \int \csc(au) du = -\frac{1}{a} \ln\left| \csc(au) + \cot(au) \right| + C \] So, \[ 2 \int \csc 2y \, dy = -\ln\left| \csc 2y + \cot 2y \right| \] ### Right Side: Let \(u = 1 - e^x\), so \(du = -e^x dx\), thus \(e^x dx = -du\): \[ \int \frac{-3e^x}{1-e^x} dx = -3 \int \frac{e^x}{1-e^x} dx \] \[ = -3 \int \frac{-du}{u} \] \[ = 3 \int \frac{du}{u} \] \[ = 3 \ln |u| = 3 \ln |1-e^x| \] ### Combine Results: \[ -\ln\left| \csc 2y + \cot 2y \right| = 3 \ln |1-e^x| + C \] Or, \[ \ln\left| \csc 2y + \cot 2y \right| = -3 \ln |1-e^x| + C \] \[ \ln\left| \csc 2y + \cot 2y \right| = \ln |(1-e^x)^{-3}| + C \] Exponentiating both sides: \[ \left| \csc 2y + \cot 2y \right| = A (1-e^x)^{-3} \] Where \(A = e^C\) is an arbitrary constant. --- ## **Final Answer:** \[ \boxed{ \left| \csc 2y + \cot 2y \right| = A (1-e^x)^{-3} } \] Where \(A\) is an arbitrary constant.