Find the particular solution for the initial value problem

\(\frac{{{d^2}y}}{{d{x^2}}}\, + \,12\frac{{dy}}{{dx}}\, + \,36y = 0\), with y(0) = 3 and \({\left. {\frac{{dy}}{{dx}}} \right|_{x = 0}}\) = -36.

1
\(\left( {3 - 18x} \right){e^{ - 6x}}\)
2
\(\left( {3 + 25x} \right){e^{ - 6x}}\)
3
\(\left( {3 - 12x} \right){e^{ - 6x}}\)
4
\(\left( {3 + 20x} \right){e^{ - 6x}}\)

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