Dear mister Tanmay y ′′ − y′ −6y = 2sin 3x find solution let y = e^(mx) ? or not


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Question Number 81586 by jagoll last updated on 14/Feb/20

$D e a r m i s t e r$ $T a n m a y$ $y^{″} - y^{'} - 6 y = 2 \sin 3 x$ $f i n d s o l u t i o n$ $l e t y = e^{m x} ? o r n o t$
Commented by jagoll last updated on 14/Feb/20

$y^{'} = {m e}^{m x} \Rightarrow y^{″} = m^{2} e^{m x}$ $m^{2} e^{m x} - {m e}^{x} - 6 e^{m x} = 0$ $e^{m x} (m^{2} - m - 6) = 0$ $(m - 3) (m + 2) = 0 \Rightarrow m = 3 \land m = - 2$ $y = {A e}^{3 x} + {B e}^{- 2 x}$
Commented by jagoll last updated on 14/Feb/20

$p a r t i c u l a r i n t e g r a l$ $y = \frac{2 \sin 3 x}{D^{2} - D - 6}$
Commented by TANMAY PANACEA last updated on 14/Feb/20

$y = 2 . \frac{s i n 3 x}{D^{2} - 6 - D}$ $= 2 . \frac{(D^{2} - 6 + D) s i n 3 x}{{(D^{2} - 6)}^{2} - D^{2}}$ $= 2 . \frac{(- 9 s i n 3 x - 6 s i n 3 x + 3 c o s 3 x)}{{(- 3^{2} - 6)}^{2} - (- 3^{2})}$ $= \frac{2}{225 + 9} \times \frac{3 c o s 3 x - 15 s i n 3 x}{1}$ $= \frac{2}{234} \times (3 c o s 3 x - 15 s i n 3 x)$
Commented by jagoll last updated on 14/Feb/20

$s i r h o w t o g e t l i n e 2 ?$
Commented by jagoll last updated on 14/Feb/20

$o o b y m u l t i p l y c o n j u g a t e$ $(D^{2} - 6) + D$
Commented by TANMAY PANACEA last updated on 14/Feb/20

$y e s . . . o b j e c t i v e t o m u l t i p l y b y c o n j u a g e t o g e t$ $d e n o m i n a t o r i n t e r m s o f D^{2}$
	Answered by Joel578 last updated on 14/Feb/20
	$• homogeneous solution with characteristic eq. λ^2 − λ − 6 = 0 ⇒ λ_1 = −2, λ_2 = 3 ⇒ y_h (x) = C_1 e^(−2x) + C_2 e^(3x) • particular solution y_p (x) = A cos 3x + B sin 3x y_p ′(x) = −3A sin 3x + 3B cos 3x y_p ′′(x) = −9A cos 3x − 9B sin 3x Substitute to the question (−15A − 3B)cos 3x + (−15B + 3A)sin 3x = 2 sin 3x ⇒ { ((−15A − 3B = 0)),((−15B + 3A = 2)) :} ⇒ { ((A = (1/(39)))),((B = −(5/(39)))) :} ⇒ y_p = (1/(39)) cos 3x − (5/(39)) sin 3x ∴ y(x) = y_h (x) + y_p (x) = C_1 e^(−2x) + C_2 e^(3x) + (1/(39)) cos 3x − (5/(39)) sin 3x$
	$∙ homogeneous solution with characteristic eq .$ $λ^{2} - λ - 6 = 0 \Rightarrow λ_{1} = - 2, λ_{2} = 3$ $\Rightarrow y_{h} (x) = C_{1} e^{- 2 x} + C_{2} e^{3 x}$ $∙ particular solution$ $y_{p} (x) = A \cos 3 x + B \sin 3 x$ $y_{p}^{'} (x) = - 3 A \sin 3 x + 3 B \cos 3 x$ $y_{p}^{″} (x) = - 9 A \cos 3 x - 9 B \sin 3 x$ $Substitute to the question$ $(- 15 A - 3 B) \cos 3 x + (- 15 B + 3 A) \sin 3 x = 2 \sin 3 x$ $\Rightarrow {\begin{cases} - 15 A - 3 B = 0 \\ - 15 B + 3 A = 2 \end{cases} \Rightarrow {\begin{cases} A = \frac{1}{39} \\ B = - \frac{5}{39} \end{cases}$ $\Rightarrow y_{p} = \frac{1}{39} \cos 3 x - \frac{5}{39} \sin 3 x$ $∴ y (x) = y_{h} (x) + y_{p} (x)$ $= C_{1} e^{- 2 x} + C_{2} e^{3 x} + \frac{1}{39} \cos 3 x - \frac{5}{39} \sin 3 x$
	Commented by jagoll last updated on 14/Feb/20

	$t h a n k y o u m i s t e r$
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