Find the general solution of this differential equation with separable variables: ((3e^x)sin y dx + (1 - e^x)cos ydy = 0)

1. Break Down the Problem

We have the differential equation: $(3e^x \sin y) \, dx + (1 - e^x) \cos y \, dy = 0.$ We need to rearrange it into a form where we can separate the variables $x$ and $y$.

2. Relevant Concepts

Since the equation is separable, we want to rearrange it to the form: $\frac{dy}{dx} = -\frac{(3e^x \sin y)}{(1 - e^x) \cos y}.$

3. Analysis and Detail

Rearranging the equation gives us: $\frac{\cos y}{\sin y} \, dy = -\frac{3e^x}{1 - e^x} \, dx.$

Next, we identify that $\frac{\cos y}{\sin y} = \cot y$, hence we rewrite it as: $\cot y \, dy = -\frac{3e^x}{1 - e^x} \, dx.$

Now we separate the variables: $\int \cot y \, dy = -3 \int \frac{e^x}{1 - e^x} \, dx.$

Calculating the integrals:

1. Left-hand side: $\int \cot y \, dy = \ln |\sin y| + C_1.$

2. Right-hand side: For the right-hand side, we can simplify: $\frac{e^x}{1 - e^x} = -1 + \frac{1}{1 - e^x}.$ So, $-3 \int \frac{e^x}{1 - e^x} \, dx = -3 \left(-x - \ln |1 - e^x| \right) + C_2 = 3x + 3\ln |1 - e^x| + C_2.$

4. Verify and Summarize

So far, we have: $\ln |\sin y| = 3x + 3 \ln |1 - e^x| + C,$ where $C = C_2 - C_1$ is a constant.

Final Answer

Exponentiating both sides, we find: $|\sin y| = K |1 - e^x|^3 e^{3x},$ where $K = e^C$ is a constant. Thus, the general solution is: $\sin y = K (1 - e^x)^3 e^{3x} \quad \text{or} \quad \sin y = -K (1 - e^x)^3 e^{3x}.$