The general solution of eˣ cos y dx – eˣ sin y dy = 0 is

The correct option is eˣ cos y = k.

Let's find the general solution of the given differential equation step-by-step.

The given differential equation is:

$$e^x\cos ydx-e^x\sin ydy=0$$

We can solve this differential equation by rewriting it to separate the variables or by recognizing it as an exact differential equation. Let's use the method of separation of variables.

First, we rearrange the equation by moving the term containing dy to the right-hand side:

$$e^x\cos ydx=e^x\sin ydy$$

Since $e^x\ne 0$ for any real x, we can divide both sides of the equation by $e^x$:

$$\cos ydx=\sin ydy$$

Now, we separate the variables by dividing both sides by $\cos y$ (assuming $\cos y\ne 0$):

$$dx=\frac{\sin y}{\cos y}dy$$

Next, we integrate both sides of the equation:

$$\int 1dx=\int \frac{\sin y}{\cos y}dy$$

The integral of the left side is simply x. For the right side, we can use the substitution $u=\cos y$, which gives $du=-\sin ydy$, or $-du=\sin ydy$. Substituting these into the integral:

$$x=\int \frac{-du}{u}$$

$$x=-\ln \left|u\right|+C$$

where C is the constant of integration. Substituting back $u=\cos y$:

$$x=-\ln |\cos y|+C$$

Now we rewrite this equation to solve for the constant. Add $\ln |\cos y|$ to both sides:

$$x+\ln |\cos y|=C$$

To eliminate the logarithm, we exponentiate both sides with base e:

$$e^{x+\ln |\cos y|}=e^C$$

Using the laws of exponents, the left side simplifies to:

$$e^x\cdot e^{\ln |\cos y|}=e^C$$

$$e^x|\cos y|=e^C$$

Removing the absolute value signs introduces a ± sign on the right-hand side:

$$e^x\cos y=\pm e^C$$

Let $k=\pm e^C$ be a new arbitrary constant. This simplifies our equation to the final general solution:

$$e^x\cos y=k$$