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Differential Calculus

Scorching goIITian

 Joined: 20 Jul 2011 Post: 256
25 May 2012 14:26:16 IST
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3
312
Higher Order derivatives
Engineering Entrance , JEE Main , JEE Main & Advanced , Mathematics , Differential Calculus

Find A and B so that y = Asin3x + Bcos3x satisfies the equation y2 + 4y1 + 3y  = 10cos3x ( where y2 and y1 are the second order and first order derivatives respectively)....

Blazing goIITian

Joined: 19 Jan 2008
Posts: 1089
27 May 2012 12:36:47 IST
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$\hspace{-16}Here \bf{y=A\sin(3x)+B\cos(3x)}\\\\\\ So \bf{\quad y'=3A\cos(3x)-3B\sin(3x)}\\\\\\ and \bf{\quad y''=-9\times\quad y'}\\\\\\\ So \bf{-9.\quad y'+4.\quad y'+3y=10\cos(3x)}\\\\\\ \bf{-5.\quad y'+3y=10\cos(3x)}\\\\\\ \bf{\left(-15A+3B\right)\cos(3x)+\left(3A+15B\right)\sin(3x)=10\cos(3x)+0.\sin(3x)}\\\\\\ \bf{A=\frac{25}{36}} and \bf{y=\frac{5}{36}}$

Scorching goIITian

Joined: 20 Jul 2011
Posts: 256
28 May 2012 13:46:53 IST
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That's not the correct answers Jagdish.......The answers will be A = 2/3 and B = -1/3 .

Just check your third step. There'll be y instead of y' .

Scorching goIITian

Joined: 20 Jul 2011
Posts: 256
28 May 2012 20:34:48 IST
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Thanks for the answer bhaiya..I've got the correct one....

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