y′=−2c1sin2x+2c2cos2xy'=-2c_1sin2x+2c_2cos2xy′=−2c1sin2x+2c2cos2x
y′′=−4c1cos2x−4c2sin2xy''=-4c_1cos2x-4c_2sin2xy′′=−4c1cos2x−4c2sin2x
Then:
y′′+4y=−4c1cos2x−4c2sin2x+4c1cos2x+4c2sin2x=0y'' + 4y =-4c_1cos2x-4c_2sin2x+ 4c_1 cos 2x +4 c_2 sin 2x=0y′′+4y=−4c1cos2x−4c2sin2x+4c1cos2x+4c2sin2x=0
Answer: C. y′′+4y=0y'' + 4y = 0y′′+4y=0
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments