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Question Number 205172 by shunmisaki007 last updated on 12/Mar/24
x(t)=c+tx′(t) ∣ c is constant. (1) Find x(t). (2) Find x(t) when x(0)=x_0 .
$${x}\left({t}\right)={c}+{tx}'\left({t}\right)\:\mid\:{c}\:\mathrm{is}\:\mathrm{constant}. \\ $$$$\left(\mathrm{1}\right)\:\mathrm{Find}\:{x}\left({t}\right). \\ $$$$\left(\mathrm{2}\right)\:\mathrm{Find}\:{x}\left({t}\right)\:\mathrm{when}\:{x}\left(\mathrm{0}\right)={x}_{\mathrm{0}} . \\ $$
Commented by lepuissantcedricjunior last updated on 12/Mar/24
x(t)=c+tx′(t) c∈R on a tx′(t)−x(t)=−c x_k =e^(∫(1/t)dt) =ke^(lnt) =kt faisons varier la constance k(t) x(t)=k(t)t x′(t)=k′(t)t+k(t) tx′(t)−x(t)=−c t^2 k′(t)+k(t)t−k(t)t=−c k′(t)=−(c/t^2 )=>k(t)=(c/t)+m ou m∈R x_m (t)=mt+c x(o)=x_0 =>x_m (0)=m(0)+c=x_0 =>c=x_0 x_m (t)=mt+x_0 ou m∈R x_m (t)=mt+x_0 ou m∈R .............le puissant Dr......................
$$\boldsymbol{{x}}\left(\boldsymbol{{t}}\right)=\boldsymbol{{c}}+\boldsymbol{{tx}}'\left(\boldsymbol{{t}}\right)\:\boldsymbol{{c}}\in\mathbb{R} \\ $$$$\boldsymbol{{on}}\:\boldsymbol{{a}}\: \\ $$$$\boldsymbol{{tx}}'\left(\boldsymbol{{t}}\right)−\boldsymbol{{x}}\left(\boldsymbol{{t}}\right)=−\boldsymbol{{c}} \\ $$$$\boldsymbol{{x}}_{\boldsymbol{{k}}} =\boldsymbol{{e}}^{\int\frac{\mathrm{1}}{\boldsymbol{{t}}}\boldsymbol{{dt}}} =\boldsymbol{{ke}}^{\boldsymbol{{lnt}}} =\boldsymbol{{kt}} \\ $$$$\boldsymbol{{faisons}}\:\boldsymbol{{varier}}\:\boldsymbol{{la}}\:\boldsymbol{{constance}}\:\boldsymbol{{k}}\left(\boldsymbol{{t}}\right) \\ $$$$\boldsymbol{{x}}\left(\boldsymbol{{t}}\right)=\boldsymbol{{k}}\left(\boldsymbol{{t}}\right)\boldsymbol{{t}} \\ $$$$\boldsymbol{{x}}'\left(\boldsymbol{{t}}\right)=\boldsymbol{{k}}'\left(\boldsymbol{{t}}\right)\boldsymbol{{t}}+\boldsymbol{{k}}\left(\boldsymbol{{t}}\right) \\ $$$$\boldsymbol{{tx}}'\left(\boldsymbol{{t}}\right)−\boldsymbol{{x}}\left(\boldsymbol{{t}}\right)=−\boldsymbol{{c}} \\ $$$$\boldsymbol{{t}}^{\mathrm{2}} \boldsymbol{{k}}'\left(\boldsymbol{{t}}\right)+\boldsymbol{{k}}\left(\boldsymbol{{t}}\right)\boldsymbol{{t}}−\boldsymbol{{k}}\left(\boldsymbol{{t}}\right)\boldsymbol{{t}}=−\boldsymbol{{c}} \\ $$$$\boldsymbol{{k}}'\left(\boldsymbol{{t}}\right)=−\frac{\boldsymbol{{c}}}{\boldsymbol{{t}}^{\mathrm{2}} }=>\boldsymbol{{k}}\left(\boldsymbol{{t}}\right)=\frac{\boldsymbol{{c}}}{\boldsymbol{{t}}}+\boldsymbol{{m}}\:\boldsymbol{{ou}}\:\boldsymbol{{m}}\in\mathbb{R} \\ $$$$\boldsymbol{{x}}_{\boldsymbol{{m}}} \left(\boldsymbol{{t}}\right)=\boldsymbol{{mt}}+\boldsymbol{{c}} \\ $$$$\boldsymbol{{x}}\left(\boldsymbol{{o}}\right)=\boldsymbol{{x}}_{\mathrm{0}} \\ $$$$=>\boldsymbol{{x}}_{\boldsymbol{{m}}} \left(\mathrm{0}\right)=\boldsymbol{{m}}\left(\mathrm{0}\right)+\boldsymbol{{c}}=\boldsymbol{{x}}_{\mathrm{0}} \\ $$$$=>\boldsymbol{{c}}=\boldsymbol{{x}}_{\mathrm{0}} \\ $$$$\boldsymbol{{x}}_{\boldsymbol{{m}}} \left(\boldsymbol{{t}}\right)=\boldsymbol{{mt}}+\boldsymbol{{x}}_{\mathrm{0}} \:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{ou}}\:\:\:\:\:\:\:\boldsymbol{{m}}\in\mathbb{R} \\ $$$$\boldsymbol{{x}}_{\boldsymbol{{m}}} \left(\boldsymbol{{t}}\right)=\boldsymbol{{mt}}+\boldsymbol{{x}}_{\mathrm{0}} \:\:\:\:\:\boldsymbol{{ou}}\:\boldsymbol{{m}}\in\mathbb{R} \\ $$$$………….{le}\:{puissant}\:\boldsymbol{{D}}{r}…………………. \\ $$$$ \\ $$

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