Find the particular solution of the differential equation $\log \left( \frac{dy}{dx} \right) = 3x + 4y$ given that $y = 0$ when $x = 0$.

$4e^{3x} + 3e^{-4y} = 7$

The correct answer is Option (1) → $4e^{3x} + 3e^{-4y} = 7$ ##

The given differential equation can be written as:

$\frac{dy}{dx} = e^{(3x + 4y)} \Rightarrow \frac{dy}{dx} = e^{3x} \cdot e^{4y}$

Separating the variables, we get:

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

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

or $\quad \frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} + C$

or $\quad 4 e^{3x} + 3 e^{-4y} + 12 C = 0 \quad \dots (2)$

Substituting $x = 0$ and $y = 0$ in $(2)$, we get

$4 + 3 + 12 C = 0 \text{ or } C = \frac{-7}{12}$

Substituting the value of $C$ in equation $(2)$, we get

$4 e^{3x} + 3 e^{-4y} - 7 = 0,$

which is a particular solution of the given differential equation.