Explanation

(a)
Given, $\frac{dy}{dx} = \tan\theta = 2x + 3y$
Put $2x + 3y = z\ \ \ \Rightarrow 2 + 3\ \ \ \ \frac{dy}{dx} = \frac{dz}{dx}$
$\Rightarrow$ $\frac{dy}{dx} = \left( \frac{dz}{dx} - 2 \right)\ \frac{1}{3}$
$\therefore$ $\frac{dz}{dx} - 2 = 3z\ \Rightarrow \frac{dz}{3z + 2} = dx$
On integrating, we get
$\frac{\log{(3z + 2)}}{3} = x + C$
$\Rightarrow$ $\frac{\log{(6x + 9y + 2)}}{3} = x + C$
Since, it passes through (1,2).
$\therefore$ $\frac{\log{(6 + 18 + 2)}}{3} = 1 + C$
$\Rightarrow$ $C = \frac{\log 26}{3} - 1$
$\therefore$ $\frac{\log{(6x + 9y + 2)}}{3} = x + \frac{\log 26}{3} - 1$
$\Rightarrow$ $\log{\left( \frac{6x + 9y + 2}{26} \right) = 3(x - 1)}$
$\Rightarrow$ $6x + 9y + 2 = 26e^{3(x - 1)}$