- #1
sourcandy
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Homework Statement
y=sinxcosx-cos^{2}x Differential equation= 2y+y'=2sin(2x)-1
Initial condition y(pi/4)=0
The Attempt at a Solution
2y=2sinxcosx-2cos^{2}x
y'=-sin^{2}x+cos^{2}x+2cosxsinx
ok... using double angle formulas... I get
2y=Sin2x-2cos^{2}x
y'=2cos^{2}x-1+2sinx
2y+y'
(Sin2x-2Cos^{2}x)+(2cos^{2}x-1+2sinx)
the 2cos mess cancells out leaves you with
Sin2x + (-1) +2Sinx
I don't understand how that equals 2sin2x-1
as the book says it does and the solution shows.
Stupid question I know... I struggle with the easy stuff.
Thanks
-Ed