Easy calculus question. I will rate and like. Thank you.

Community Answer

Step 1Given differential equation is y"=-4yStep 2It con be written asy^('')+4y=0a.) y=sin 2t{:[" LHS "{:=y^('')+4y=(d)/(dt)((d)/(dt)(sin 2t)))+4(sin 2t)],[=(d)/(dt)(2cos 2t)+4sin 2t],[=-4sin 2t+4sin 2t=0=" RHS "]:}Hencey= sinet is a solution of given equation.{:[y=-(2)/(3)t^(3)],[:.y^('')=(d)/(dt)(-(2)/(3)xx3t^(2))=(d)/(dt)(-2t^(2))=-4t],[" LHs "=-4t+4(-(2)/(3)t^(3))=-4t-(8)/(3)t^(3)!=0],[:.y=-(2)/(3)t^(3)" is not a sulution of "]:}b ) give ... See the full answer