Home » Ask & Discuss » Mathematics
« Back to Discussion

Differential Calculus

Scorching goIITian

Joined:	20 Jul 2011
Post:	256

25 May 2012 14:26:16 IST

0 People liked this

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)....

Comments (3)

jagdish singh

Blazing goIITian

Joined: 19 Jan 2008

Posts: 1089

27 May 2012 12:36:47 IST

Like 0 people liked this

$\hspace{-16}$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}}$$

Sumiran agrawal

Scorching goIITian

Joined: 20 Jul 2011

Posts: 256

28 May 2012 13:46:53 IST

Like 0 people liked this

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' .

Sumiran agrawal

Scorching goIITian

Joined: 20 Jul 2011

Posts: 256

28 May 2012 20:34:48 IST

Like 0 people liked this

Thanks for the answer bhaiya..I've got the correct one....

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team