Find-the-particular-solution-to-the-differential-equation-2y-5y-2y-0-subject-to-the-initial-conditions-y-0-2y-y-0-1-

Question Number 161462 by ZiYangLee last updated on 18/Dec/21

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{particular}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{the}\:\mathrm{differential} \\ $$$$\mathrm{equation}\:\mathrm{2}{y}''+\mathrm{5}{y}'+\mathrm{2}{y}=\mathrm{0}\:\mathrm{subject}\:\mathrm{to}\:\mathrm{the}\:\mathrm{initial} \\ $$$$\mathrm{conditions}\:\:{y}\left(\mathrm{0}\right)=\mathrm{2}{y}\:,\:\:{y}'\left(\mathrm{0}\right)=\mathrm{1}\:. \\ $$

Answered by TheSupreme last updated on 18/Dec/21

2λ^2 +5λ+2=0 λ_(1,2) =((−5±(√(25−16)))/4)=((−5±3)/4)=−2, −(1/2) y_0 =c_1 e^(−2t) +c_2 e^(−(t/2)) y(0)=c_1 +c_2 =2 y′(0)=−2c_1 −(1/2)c_2 =1 −2c_2 +4−(1/2)c_2 =1 −(5/2)c_2 =−3 c_2 =(6/5) →c_1 =(4/5) y(t)=(4/5)e^(−2t) +(6/5)e^(−(t/2))

$$\mathrm{2}\lambda^{\mathrm{2}} +\mathrm{5}\lambda+\mathrm{2}=\mathrm{0} \\ $$$$\lambda_{\mathrm{1},\mathrm{2}} =\frac{−\mathrm{5}\pm\sqrt{\mathrm{25}−\mathrm{16}}}{\mathrm{4}}=\frac{−\mathrm{5}\pm\mathrm{3}}{\mathrm{4}}=−\mathrm{2},\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${y}_{\mathrm{0}} ={c}_{\mathrm{1}} {e}^{−\mathrm{2}{t}} +{c}_{\mathrm{2}} {e}^{−\frac{{t}}{\mathrm{2}}} \\ $$$${y}\left(\mathrm{0}\right)={c}_{\mathrm{1}} +{c}_{\mathrm{2}} =\mathrm{2} \\ $$$${y}'\left(\mathrm{0}\right)=−\mathrm{2}{c}_{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}{c}_{\mathrm{2}} =\mathrm{1} \\ $$$$−\mathrm{2}{c}_{\mathrm{2}} +\mathrm{4}−\frac{\mathrm{1}}{\mathrm{2}}{c}_{\mathrm{2}} =\mathrm{1} \\ $$$$−\frac{\mathrm{5}}{\mathrm{2}}{c}_{\mathrm{2}} =−\mathrm{3} \\ $$$${c}_{\mathrm{2}} =\frac{\mathrm{6}}{\mathrm{5}}\:\rightarrow{c}_{\mathrm{1}} =\frac{\mathrm{4}}{\mathrm{5}} \\ $$$${y}\left({t}\right)=\frac{\mathrm{4}}{\mathrm{5}}{e}^{−\mathrm{2}{t}} +\frac{\mathrm{6}}{\mathrm{5}}{e}^{−\frac{{t}}{\mathrm{2}}} \\ $$

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