The particular solution of the initial value problem given below is
$$\frac{d^2y}{dx^2}+12\frac{dy}{dx}+36y=0\hspace{0.3cm} \text{ with } \hspace{0.3cm}y(0)=3\hspace{0.3cm} \text{ and }\hspace{0.3cm} \frac{dy}{dx} \bigg| _{x=0} =-36$$
- $(3-18x)e^{-6x}$
- $(3+25x)e^{-6x}$
- $(3+20x)e^{-6x}$
- $(3-12x)e^{-6x}$