y-2y-5y-3sin-4t-

Question Number 91287 by jagoll last updated on 29/Apr/20

$${y}''+\mathrm{2}{y}'+\mathrm{5}{y}=\mathrm{3sin}\:\mathrm{4}{t} \\ $$

Answered by niroj last updated on 29/Apr/20

y^(′′) +2y′+5y= 3sin 4t (D^2 +2D+5)y= 3sin 4t A.E., m^2 +2m+5=0 m= −1+^− 2i CF = e^(−t) (Acos2t+Bsin2t) PI= ((3sin4t)/(D^2 +2D+5))= ((3sin4t)/(−16+2D+5)) = ((3sin4t)/(2D−11))= (((2D+11)3sin4t)/(4D^2 −121)) = (((2D+11)3sin4t)/(−64−121))= −(1/(185))(24cos4t+33sin4t) y=CF+PI y= e^(−t) (Acos2t+Bsin2t)− (1/(185))( 24cos4t+ 33 sin4t)

$$\:\:\:\mathrm{y}^{''} +\mathrm{2y}'+\mathrm{5y}=\:\mathrm{3sin}\:\mathrm{4t} \\ $$$$\:\:\left(\mathrm{D}^{\mathrm{2}} +\mathrm{2D}+\mathrm{5}\right)\mathrm{y}=\:\mathrm{3sin}\:\mathrm{4t}\:\: \\ $$$$\:\:\mathrm{A}.\mathrm{E}.,\:\mathrm{m}^{\mathrm{2}} +\mathrm{2m}+\mathrm{5}=\mathrm{0} \\ $$$$\:\:\:\:\mathrm{m}=\:−\mathrm{1}\overset{−} {+}\mathrm{2i} \\ $$$$\:\:\mathrm{CF}\:=\:\:\mathrm{e}^{−\mathrm{t}} \left(\mathrm{Acos2t}+\mathrm{Bsin2t}\right) \\ $$$$\:\:\mathrm{PI}=\:\:\frac{\mathrm{3sin4t}}{\mathrm{D}^{\mathrm{2}} +\mathrm{2D}+\mathrm{5}}=\:\frac{\mathrm{3sin4t}}{−\mathrm{16}+\mathrm{2D}+\mathrm{5}} \\ $$$$\:\:\:\:=\:\frac{\mathrm{3sin4t}}{\mathrm{2D}−\mathrm{11}}=\:\frac{\left(\mathrm{2D}+\mathrm{11}\right)\mathrm{3sin4t}}{\mathrm{4D}^{\mathrm{2}} −\mathrm{121}} \\ $$$$\:\:\:=\:\:\frac{\left(\mathrm{2D}+\mathrm{11}\right)\mathrm{3sin4t}}{−\mathrm{64}−\mathrm{121}}=\:−\frac{\mathrm{1}}{\mathrm{185}}\left(\mathrm{24cos4t}+\mathrm{33sin4t}\right) \\ $$$$\:\mathrm{y}=\mathrm{CF}+\mathrm{PI} \\ $$$$\:\mathrm{y}=\:\mathrm{e}^{−\mathrm{t}} \left(\mathrm{Acos2t}+\mathrm{Bsin2t}\right)−\:\frac{\mathrm{1}}{\mathrm{185}}\left(\:\mathrm{24cos4t}+\:\mathrm{33}\:\mathrm{sin4t}\right) \\ $$$$ \\ $$

Commented by jagoll last updated on 29/Apr/20

how to get ((3sin 4t)/(D^2 +2D+5)) = ((3sin 4t)/(−16+2D+5)) ?

$${how}\:{to}\:{get}\:\frac{\mathrm{3sin}\:\mathrm{4}{t}}{{D}^{\mathrm{2}} +\mathrm{2}{D}+\mathrm{5}}\:=\:\frac{\mathrm{3sin}\:\mathrm{4}{t}}{−\mathrm{16}+\mathrm{2}{D}+\mathrm{5}}\:? \\ $$

Commented by niroj last updated on 29/Apr/20

The case of (1/(F(D^2 ))) sin ax and (1/(F(D^2 )))cos ax when f(−a^2 )=0. ((sin ax)/(f(D^2 )))= (1/(f(−a^2 )))sinax.

$$ \\ $$$$\:\:\mathrm{The}\:\:\mathrm{case}\:\:\mathrm{of}\:\:\frac{\mathrm{1}}{\mathrm{F}\left(\mathrm{D}^{\mathrm{2}} \right)}\:\mathrm{sin}\:\boldsymbol{\mathrm{a}}\mathrm{x}\:\mathrm{and}\:\frac{\mathrm{1}}{\mathrm{F}\left(\mathrm{D}^{\mathrm{2}} \right)}\mathrm{cos}\:\boldsymbol{\mathrm{a}}\mathrm{x} \\ $$$$\:\mathrm{when}\:\mathrm{f}\left(−\mathrm{a}^{\mathrm{2}} \right)=\mathrm{0}. \\ $$$$\:\:\frac{\mathrm{sin}\:\mathrm{ax}}{\mathrm{f}\left(\mathrm{D}^{\mathrm{2}} \right)}=\:\frac{\mathrm{1}}{\mathrm{f}\left(−\mathrm{a}^{\mathrm{2}} \right)}\mathrm{sinax}. \\ $$$$ \\ $$$$\: \\ $$

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