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Form a differential equation represent family of curve given by $y= (c_1+c_2) -\sin (x+c_3)-c_4 e^{x+c_5}$

$(a)\;y''-2y'''-3y'+y=0 \\ (b)\;y'''-y''-y+y'=0\\ (c)\;y''-y'''-y'+y=0\\ (d)\;y'-y+y'''+y''=0 $

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$y= A \sin x \cos c_3+ A \cos x \sin C_3 -c_4 . e^{c_5}.e^x$
$y= c \sin x + D \cos x-B e^x$
$y'= -c \cos x -D \sin x -Be^x$
$y''= -c \sin x -D \cos x -B e^x$
$y'''= -y'-2 Be^x$
So, $y''-y'''-y'+y=0$
Hence c is the correct answer.
answered Feb 7, 2014 by meena.p
edited Mar 20, 2014 by balaji.thirumalai

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