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Question Number 79626 by M±th+et£s last updated on 26/Jan/20
find tbe general solution of the D.E  y^(′′) −5y^′ +4y=e^t cos(t)
$${find}\:{tbe}\:{general}\:{solution}\:{of}\:{the}\:{D}.{E} \\ $$$${y}^{''} −\mathrm{5}{y}^{'} +\mathrm{4}{y}={e}^{{t}} {cos}\left({t}\right) \\ $$
Answered by mind is power last updated on 27/Jan/20
y′′−5y′+4y=0  ⇒y=ae^t  +be^(4t)   f(t)=e^t (acos(t)+bsin(t))  ⇒f′(t)=e^t (−asin(t)+bcos(t)+acos(t)+bsin(t))  f′′(t)=e^t (−2asin(t)+2bcos(t))  e^t (3asin(t)−3bcos(t)−acos(t)−bsin(t))=e^t cos(t)  3a−b=0  −(3b+a)=1⇒3b+a=−1  ⇒a=−(1/(10)),b=−(3/(10))  f=−((e^t (cos(t)+3sin(t)))/(10))  general solution ae^t +be^(4t) −((e^t (cos(t)+3sin(t)))/(10))
$${y}''−\mathrm{5}{y}'+\mathrm{4}{y}=\mathrm{0} \\ $$$$\Rightarrow{y}={ae}^{{t}} \:+{be}^{\mathrm{4}{t}} \\ $$$${f}\left({t}\right)={e}^{{t}} \left({acos}\left({t}\right)+{bsin}\left({t}\right)\right) \\ $$$$\Rightarrow{f}'\left({t}\right)={e}^{{t}} \left(−{asin}\left({t}\right)+{bcos}\left({t}\right)+{acos}\left({t}\right)+{bsin}\left({t}\right)\right) \\ $$$${f}''\left({t}\right)={e}^{{t}} \left(−\mathrm{2}{asin}\left({t}\right)+\mathrm{2}{bcos}\left({t}\right)\right) \\ $$$${e}^{{t}} \left(\mathrm{3}{asin}\left({t}\right)−\mathrm{3}{bcos}\left({t}\right)−{acos}\left({t}\right)−{bsin}\left({t}\right)\right)={e}^{{t}} {cos}\left({t}\right) \\ $$$$\mathrm{3}{a}−{b}=\mathrm{0} \\ $$$$−\left(\mathrm{3}{b}+{a}\right)=\mathrm{1}\Rightarrow\mathrm{3}{b}+{a}=−\mathrm{1} \\ $$$$\Rightarrow{a}=−\frac{\mathrm{1}}{\mathrm{10}},{b}=−\frac{\mathrm{3}}{\mathrm{10}} \\ $$$${f}=−\frac{{e}^{{t}} \left({cos}\left({t}\right)+\mathrm{3}{sin}\left({t}\right)\right)}{\mathrm{10}} \\ $$$${general}\:{solution}\:{ae}^{{t}} +{be}^{\mathrm{4}{t}} −\frac{{e}^{{t}} \left({cos}\left({t}\right)+\mathrm{3}{sin}\left({t}\right)\right)}{\mathrm{10}} \\ $$

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