find particular solution of (D^3 +4D) y = sin 2x by using inverse operation ?


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Question Number 118574 by bramlexs22 last updated on 18/Oct/20

$f i n d p a r t i c u l a r s o l u t i o n o f$ $(D^{3} + 4 D) y = \sin 2 x b y u s i n g$ $i n v e r s e o p e r a t i o n ?$
	Answered by TANMAY PANACEA last updated on 18/Oct/20
	$y=((sin2x)/(D(D^2 +4))) y=(e^(i2x) /(D(D^2 +4))) =(e^(i2x) /((i2){(D+i2)^2 +4})) =(e^(i2x) /((i2)(D^2 +4iD))) =(e^(i2x) /((i2)))×(1/(D×4i×(1+(D/(4i))))) =(1/(−8))×(1/((1+((i2)/(4i)))))×(e^(i2x) /1)×x =(1/(−8))×(2/3)×((cos2x+isin2x)/1)×x =(x/(−12))(cos2x)+i×(x/(−12))(sin2x) P.I=(x/(−12))×sin2x→y=((xsin2x)/(−12)) pls check my answer$
	$y = \frac{s i n 2 x}{D (D^{2} + 4)}$ $y = \frac{e^{i 2 x}}{D (D^{2} + 4)}$ $= \frac{e^{i 2 x}}{(i 2) {{(D + i 2)}^{2} + 4}}$ $= \frac{e^{i 2 x}}{(i 2) (D^{2} + 4 i D)}$ $= \frac{e^{i 2 x}}{(i 2)} \times \frac{1}{D \times 4 i \times (1 + \frac{D}{4 i})}$ $= \frac{1}{- 8} \times \frac{1}{(1 + \frac{i 2}{4 i})} \times \frac{e^{i 2 x}}{1} \times x$ $= \frac{1}{- 8} \times \frac{2}{3} \times \frac{c o s 2 x + i s i n 2 x}{1} \times x$ $= \frac{x}{- 12} (c o s 2 x) + i \times \frac{x}{- 12} (s i n 2 x)$ $P . I = \frac{x}{- 12} \times s i n 2 x \to y = \frac{x s i n 2 x}{- 12}$ $p l s c h e c k m y a n s w e r$
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