Show that -3e^2t+7te^2t is a particular solution to the initial value problem y’’ -4y’ +4y=0. y(0)=-3,y’(0)=1?

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Answer 1 · 2018-03-17T04:25:59

Please see below.

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#y=-3e^(2t)+7te^(2t)#

#y(0)=-3#

#y'=-6e^(2t)+7(2te^(2t)+e^(2t))=-6e^(2t)+14te^(2t)+7e^(2t)#

#y'=e^(2t)+14te^(2t)#

#y'(0)=1#

#y''=2e^(2t)+14(2te^(2t)+e^(2t))=2e^(2t)+28te^(2t)+14e^(2t)#

#y''=16e^(2t)+28te^(2t)#

#y''(0)=16#

#y''-4y'+4y=0#

#16-4(1)+4(-3)=16-4-12=16-16=0#