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Question Number 59800 by necx1 last updated on 14/May/19
find the general solution y(t) of the ordinary differential equation y′′ + ω^2 y=cos ωt ,where w>0
$${find}\:{the}\:{general}\:{solution}\:{y}\left({t}\right)\:{of}\:{the} \\ $$$${ordinary}\:{differential}\:{equation} \\ $$$${y}''\:+\:\omega^{\mathrm{2}} {y}=\mathrm{cos}\:\omega{t}\:,{where}\:{w}>\mathrm{0} \\ $$
Answered by MJS last updated on 15/May/19
1^(st) attempt y=C_1 cos ωt +C_2 sin ωt y′′=−C_1 ω^2 cos ωt −C_2 ω^2 sin ωt y′′+ω^2 y=0 2^(nd) attempt y=C_1 cos ωt +C_2 sin ωt +C_3 tsin wt y′′=−C_1 ω^2 cos ωt −C_2 ω^2 sin ωt +2C_3 ωcos ωt −C_3 ω^2 tsin ωt y′′+ω^2 y=2C_3 ωcos ωt ⇒ C_3 =(1/(2ω)) ⇒ y(t)=C_1 cos ωt +((t/(2ω))+C_2 )sin ωt
$$\mathrm{1}^{\mathrm{st}} \:\mathrm{attempt} \\ $$$$\:\:\:\:\:{y}={C}_{\mathrm{1}} \mathrm{cos}\:\omega{t}\:+{C}_{\mathrm{2}} \mathrm{sin}\:\omega{t} \\ $$$$\:\:\:\:\:{y}''=−{C}_{\mathrm{1}} \omega^{\mathrm{2}} \mathrm{cos}\:\omega{t}\:−{C}_{\mathrm{2}} \omega^{\mathrm{2}} \mathrm{sin}\:\omega{t} \\ $$$$\:\:\:\:\:{y}''+\omega^{\mathrm{2}} {y}=\mathrm{0} \\ $$$$\mathrm{2}^{\mathrm{nd}} \:\mathrm{attempt} \\ $$$$\:\:\:\:\:{y}={C}_{\mathrm{1}} \mathrm{cos}\:\omega{t}\:+{C}_{\mathrm{2}} \mathrm{sin}\:\omega{t}\:+{C}_{\mathrm{3}} {t}\mathrm{sin}\:{wt} \\ $$$$\:\:\:\:\:{y}''=−{C}_{\mathrm{1}} \omega^{\mathrm{2}} \mathrm{cos}\:\omega{t}\:−{C}_{\mathrm{2}} \omega^{\mathrm{2}} \mathrm{sin}\:\omega{t}\:+\mathrm{2}{C}_{\mathrm{3}} \omega\mathrm{cos}\:\omega{t}\:−{C}_{\mathrm{3}} \omega^{\mathrm{2}} {t}\mathrm{sin}\:\omega{t} \\ $$$$\:\:\:\:\:{y}''+\omega^{\mathrm{2}} {y}=\mathrm{2}{C}_{\mathrm{3}} \omega\mathrm{cos}\:\omega{t} \\ $$$$\:\:\:\:\:\Rightarrow\:{C}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}\omega} \\ $$$$\:\:\:\:\:\Rightarrow\:{y}\left({t}\right)={C}_{\mathrm{1}} \mathrm{cos}\:\omega{t}\:+\left(\frac{{t}}{\mathrm{2}\omega}+{C}_{\mathrm{2}} \right)\mathrm{sin}\:\omega{t} \\ $$
Answered by tanmay last updated on 15/May/19
(d^2 y/dt^2 )+w^2 y=coswt (D^2 +w^2 )y=coswt C.F determination (d^2 y/dt^2 )+w^2 y=0 let y=Ae^(αt) is a solution (dy/dt)=Aαe^(αt) (d^2 y/dt^2 )=Aα^2 e^(αt) Aα^2 e^(αt) +w^2 Ae^(αt) =0 Ae^(αt) (α^2 +w^2 )=0 Ae^(αt) ≠0 α^2 +w^2 =0 α=±iw C.F y=C_1 e^(iwt) +C_2 e^(−iwt) Particular intregal y=((coswt)/(D^2 +w^2 )) p=((coswt)/(D^2 +w^2 )) q=((sinwt)/(D^2 +w^2 )) p+iq=(e^(iwt) /((D+iw)(D−iw))) p+iq=(e^(iwt) /((iw+iw)))×(1/(D+iw−iw)) p+iq=(e^(iwt) /(2iw))×x [(1/D)×1=x] p+iq=((x(coswt+isinwt))/(2iw)) =((xcoswt)/(2w))×(i/(−1))+((xsinwt)/(2w)) p+iq=(((xsinwt)/(2w)))+i(((−xcoswt)/(2w))) p=((−xcoswt)/(2w)) →(taking the real part) complte solution y=C_1 e^(iwt) +C_2 e^(−iwt) +(((−xcoswt)/(2w))) pls check the answer
$$\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }+{w}^{\mathrm{2}} {y}={coswt} \\ $$$$\left({D}^{\mathrm{2}} +{w}^{\mathrm{2}} \right){y}={coswt} \\ $$$${C}.{F}\:\:{determination} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }+{w}^{\mathrm{2}} {y}=\mathrm{0} \\ $$$${let}\:{y}={Ae}^{\alpha{t}} \:{is}\:{a}\:{solution} \\ $$$$\frac{{dy}}{{dt}}={A}\alpha{e}^{\alpha{t}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }={A}\alpha^{\mathrm{2}} {e}^{\alpha{t}} \\ $$$${A}\alpha^{\mathrm{2}} {e}^{\alpha{t}} +{w}^{\mathrm{2}} {Ae}^{\alpha{t}} =\mathrm{0} \\ $$$${Ae}^{\alpha{t}} \left(\alpha^{\mathrm{2}} +{w}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${Ae}^{\alpha{t}} \neq\mathrm{0} \\ $$$$\alpha^{\mathrm{2}} +{w}^{\mathrm{2}} =\mathrm{0} \\ $$$$\alpha=\pm{iw} \\ $$$${C}.{F} \\ $$$${y}={C}_{\mathrm{1}} {e}^{{iwt}} +{C}_{\mathrm{2}} {e}^{−{iwt}} \\ $$$${Particular}\:{intregal} \\ $$$${y}=\frac{{coswt}}{{D}^{\mathrm{2}} +{w}^{\mathrm{2}} } \\ $$$${p}=\frac{{coswt}}{{D}^{\mathrm{2}} +{w}^{\mathrm{2}} } \\ $$$${q}=\frac{{sinwt}}{{D}^{\mathrm{2}} +{w}^{\mathrm{2}} } \\ $$$${p}+{iq}=\frac{{e}^{{iwt}} }{\left({D}+{iw}\right)\left({D}−{iw}\right)} \\ $$$${p}+{iq}=\frac{{e}^{{iwt}} }{\left({iw}+{iw}\right)}×\frac{\mathrm{1}}{{D}+{iw}−{iw}} \\ $$$${p}+{iq}=\frac{{e}^{{iwt}} }{\mathrm{2}{iw}}×{x}\:\:\left[\frac{\mathrm{1}}{{D}}×\mathrm{1}={x}\right] \\ $$$${p}+{iq}=\frac{{x}\left({coswt}+{isinwt}\right)}{\mathrm{2}{iw}} \\ $$$$=\frac{{xcoswt}}{\mathrm{2}{w}}×\frac{{i}}{−\mathrm{1}}+\frac{{xsinwt}}{\mathrm{2}{w}} \\ $$$${p}+{iq}=\left(\frac{{xsinwt}}{\mathrm{2}{w}}\right)+{i}\left(\frac{−{xcoswt}}{\mathrm{2}{w}}\right) \\ $$$${p}=\frac{−{xcoswt}}{\mathrm{2}{w}}\:\rightarrow\left({taking}\:{the}\:{real}\:{part}\right) \\ $$$${complte}\:{solution} \\ $$$${y}={C}_{\mathrm{1}} {e}^{{iwt}} +{C}_{\mathrm{2}} {e}^{−{iwt}} +\left(\frac{−{xcoswt}}{\mathrm{2}{w}}\right) \\ $$$${pls}\:{check}\:{the}\:{answer} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by necx1 last updated on 15/May/19
hmmm....I think I need to study this more
$${hmmm}….{I}\:{think}\:{I}\:{need}\:{to}\:{study}\:{this}\: \\ $$$${more} \\ $$

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