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if y=3e^(2x)cos(2x-3) verify that d^2y/dx^2-4dy/dx+8y=0

plz help me i tried all i could but it become too complicated for me

here
set u=3e^(2x) v=cos(2x-3)

du/dx=6e^(2x)
i used chain rule
dv/dx=-2sin(2x-3)

dy/dx=-3e^(2x)sin(2x-3)+cos(2x-3)6e^(2x)

d^2y/dx^2

now i did

lndy/dx=-ln3e^(2x)+lnsin(2x-3)+lncos(2x-3)+ln6e^(2x)

d^2y/dx^2.1/dy/dx=-6e^(2x)/(3e^2(x)+2cos(2x-3)/sin(2x-3)-2sin(2x-3)/cos(2x-3)+12e^(2x)

so whats up any 1 see it too much

sir steve my best man check

2 answers

y=3e^(2x)cos(2x-3)
y' = 6e^(2x) (cos(2x-3)-sin(2x-3))
y" = -24e^(2x)sin(2x-3)

y" - 4y' + 8y
= -24e^(2x)sin(2x-3) - 24e^(2x)(cos(2x-3)-sin(2x-3)) + 24e^(2x)cos(2x-3)
= 0

Note that the DE

y" - 4y' + 8y = 0
has the solution
y = e^(2x)(a cos2x + b sin2x)
If we let
a = 3cos3
b = 3sin3
y= 3e^(2x)cos(2x-3)
ln(ab+cd) is NOT

lna +lnb + lnc +lnd