Question #094f0

2 Answers
Jun 25, 2017

y= c_1 cos 2t +c_2 sin 2t

Explanation:

The solution to a second degree DE is determined by the roots of its characteristic equation. In the instant case the characteristic equation of DE is m^2 +4=0 . Its roots are +2i and -2i, which are distincts and imaginary.

There would be two solutions to DE, y_1= c_1 cos 2t and y_2=c_2 sin 2t

The general solution would thus be y= c_1 cos 2t +c_2 sin 2t. The value of constants c_1 and c_2 are determined by the initial values conditions attached to the DE.
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