Explanation

(c)
Given, $\log{\left( \frac{dy}{dx} \right) =}3x + 4y$
$\Rightarrow$ $\frac{dy}{dx} = e^{3x}\ e^{4y}$
$\Rightarrow$ $e^{- 4y}\ dy = e^{3x}\ dx$
On integrating both sides, we get
$\frac{e^{- 4y}}{- 4} = \frac{e^{3x}}{3} + c$
At $x = 0,\ y = 0$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ - \frac{1}{4} = \frac{1}{3} + c$
$\Rightarrow$ $c = - \frac{7}{12}$
$\therefore$ Solution is
$\frac{e^{- 4y}}{- 4} = \frac{e^{3x}}{3} - \frac{7}{12}$
$\Rightarrow$ $4e^{3x} + 3e^{- 4y} = 7$