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Question Number 182139 by Mastermind last updated on 04/Dec/22

Find constant a, b, so that  y(t)=(t+3)e^(2t)  is solution of IVP  y^′ =ay+e^(2t) ,           y(0)=b    .

$$\mathrm{Find}\:\mathrm{constant}\:\mathrm{a},\:\mathrm{b},\:\mathrm{so}\:\mathrm{that} \\ $$$$\mathrm{y}\left(\mathrm{t}\right)=\left(\mathrm{t}+\mathrm{3}\right)\mathrm{e}^{\mathrm{2t}} \:\mathrm{is}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{IVP} \\ $$$$\mathrm{y}^{'} =\mathrm{ay}+\mathrm{e}^{\mathrm{2t}} ,\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{b} \\ $$$$ \\ $$$$. \\ $$

Commented by mr W last updated on 06/Dec/22

do you never give a feedback when a  question of yours is answered?

$${do}\:{you}\:{never}\:{give}\:{a}\:{feedback}\:{when}\:{a} \\ $$$${question}\:{of}\:{yours}\:{is}\:{answered}? \\ $$

Answered by mr W last updated on 04/Dec/22

y′=e^(2t) +2(t+3)e^(2t)   e^(2t) +2(t+3)e^(2t) =a(t+3)e^(2t) +e^(2t)   ⇒2=a  y(0)=(0+3)e^0 =3=b

$${y}'={e}^{\mathrm{2}{t}} +\mathrm{2}\left({t}+\mathrm{3}\right){e}^{\mathrm{2}{t}} \\ $$$${e}^{\mathrm{2}{t}} +\mathrm{2}\left({t}+\mathrm{3}\right){e}^{\mathrm{2}{t}} ={a}\left({t}+\mathrm{3}\right){e}^{\mathrm{2}{t}} +{e}^{\mathrm{2}{t}} \\ $$$$\Rightarrow\mathrm{2}={a} \\ $$$${y}\left(\mathrm{0}\right)=\left(\mathrm{0}+\mathrm{3}\right){e}^{\mathrm{0}} =\mathrm{3}={b} \\ $$

Commented by Mastermind last updated on 22/Dec/22

Thank you boss

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{boss} \\ $$

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