$\mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{differential}\mathrm{equation}\log e\left(\frac{dy}{dx}\right)=3x+4y\mathrm{is}\mathrm{given}\mathrm{by}$

The correct option is:

$4e^{3x}+3e^{-4y}+C=0$

Step-by-step Explanation:

We are given the following first-order differential equation:

${\log }_e\left(\frac{dy}{dx}\right)=3x+4y$

To eliminate the logarithm, we can rewrite the equation in exponential form by raising both sides as powers of base e:

$\frac{dy}{dx}=e^{3x+4y}$

Using the laws of exponents, we can separate the variables on the right-hand side:

$\frac{dy}{dx}=e^{3x}·e^{4y}$

Now, we rearrange the terms to separate the variables x and y on opposite sides of the equation:

$\frac{dy}{e^{4y}}=e^{3x}dx$

which can be written as:

$e^{-4y}dy=e^{3x}dx$

Next, we integrate both sides of the equation:

$\int e^{-4y}dy=\int e^{3x}dx$

Carrying out the integration, we get:

$\frac{e^{-4y}}{-4}=\frac{e^{3x}}{3}+C_1$

where $C_1$ is the constant of integration.

We rearrange the equation by moving all terms to one side:

$\frac{e^{3x}}{3}+\frac{e^{-4y}}{4}+C_1=0$

To eliminate the fractions, we multiply the entire equation by the least common multiple of 3 and 4, which is 12:

$4e^{3x}+3e^{-4y}+12C_1=0$

Defining a new constant $C=12C_1$, we obtain the final general solution:

$4e^{3x}+3e^{-4y}+C=0$