Asked by abc
y"-y'=sin x
solve in undetermined coefficients
solve in undetermined coefficients
Answers
Answered by
Damon
probably sinusoidal
y = a sin x + b cos x
y' = a cos x - b sin x
y" = -a sin x - b cos x = -y
-a sx -b cx -a cx + b sx = 1 sx
b = -a
-a + b = 1
-2a = 1
a = -1/2
b = +1/2
so I get y = -(1/2) (sin x - cos x)
y = a sin x + b cos x
y' = a cos x - b sin x
y" = -a sin x - b cos x = -y
-a sx -b cx -a cx + b sx = 1 sx
b = -a
-a + b = 1
-2a = 1
a = -1/2
b = +1/2
so I get y = -(1/2) (sin x - cos x)
Answered by
Steve
That pesky y' gets in the way. The solution is really
y = c1 e^x + c2 - 1/2 (sinx - cosx)
y = c1 e^x + c2 - 1/2 (sinx - cosx)
Answered by
Damon
Whoops, yes, use Steve's
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