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Question Number 49358 by peter frank last updated on 06/Dec/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18

	$1) 9 \frac{d^{2} x}{{d t}^{2}} + 6 \frac{d x}{d t} + x = - 50 s i n t$ $s t e p 1$ $9 \frac{d^{2} x}{{d t}^{2}} + 6 \frac{d x}{d t} + x = 0$ $x = e^{α t} i s a s o l u t i i n$ $9 (α^{2} e^{α t}) + 6 (α e^{α t}) + e^{α t} = 0$ $e^{α t} \neq 0$ $9 α^{2} + 6 α + 1 = 0$ ${(3 α + 1)}^{2} = 0$ $α = \frac{- 1}{3}$ $C . F = {A t e}^{\frac{- t}{3}} + {B e}^{\frac{- t}{3}} = e^{\frac{- t}{3}} (A t + B)$ $P . I c a l c u l a t i o n . . .$ $D = \frac{d}{d t} D^{2} = \frac{d^{2}}{{d t}^{2}}$ $(9 D^{2} + 6 D + 1) x = - 50 s i n t$ $x = \frac{- 50}{9 D^{2} + 6 D + 1} \times s i n t$ $= - 50 \times \frac{9 D^{2} + 1 - 6 D}{{(9 D^{2} + 1)}^{2} - 36 D^{2}} \times s i n t$ $D (s i n t) = c o s t D^{2} (s i n t) = - s i n t$ $= - 50 \times \frac{- 9 s i n t + s i n t - 6 c o s t}{{(9 \times - 1^{2} + 1)}^{2} - 36 (- 1^{2})}$ $= - 50 \times \frac{- 8 s i n t - 6 c o s t}{64 + 36}$ $= \frac{1}{2} \times (8 s i n t + 6 c o s t) = 4 s i n t + 3 c o s t$ $5 (\frac{4}{5} s i n t + \frac{3}{5} c o s t) [\frac{4}{5} = s i n α \frac{3}{5} = c o s α]$ $= 5 c o s (t - α)$ $c o m p l e t e s o l u t i o n i s$ $x = e^{\frac{- t}{3}} (A t + B) + 5 c o s (t - α)$
	Commented by peter frank last updated on 07/Dec/18

	$thank you sir$
	Commented by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18

	$m o s t w e l c o m e . . .$
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