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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