The particular solution of the initial value problem given below is:

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

1
(3 – 18x) e−6x
2
(3 + 25x) e−6x
3
(3 + 20x) e−6x
4
(3 − 12x) e−6x

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