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Question-154794




Question Number 154794 by Sobur2021 last updated on 21/Sep/21
Commented by tabata last updated on 21/Sep/21
(3m − 2 ) = 0 ⇒ m = (2/3) ⇒ f_1 (t) = c_1  e ^((2/3) t)     f_2 (t) = (1/(3 f^( ′)  (t ) − 2)) 4 e^(−t)  = (4/(−3 − 2))  e^(−t) = − (4/5) e^(−t)     f_3 (t) = a ⇒ f^′ (t) = 0 ⇒ 3 (0) − 2 (a) = 2 ⇒ a = −1    ∴ f_3 (t) = −1    ∴ f(t) = f_1 (t) + f_2 (t) + f_3 (t) = c_1  e^((2/3) t)  − (4/5) e^(− t)  − 1    ⟨ M . T  ⟩
$$\left(\mathrm{3}\boldsymbol{{m}}\:−\:\mathrm{2}\:\right)\:=\:\mathrm{0}\:\Rightarrow\:\boldsymbol{{m}}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:\Rightarrow\:\boldsymbol{{f}}_{\mathrm{1}} \left(\boldsymbol{{t}}\right)\:=\:\boldsymbol{{c}}_{\mathrm{1}} \:\boldsymbol{{e}}\overset{\frac{\mathrm{2}}{\mathrm{3}}\:\boldsymbol{{t}}} {\:} \\ $$$$ \\ $$$$\boldsymbol{{f}}_{\mathrm{2}} \left(\boldsymbol{{t}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}\:\boldsymbol{{f}}^{\:'} \:\left(\boldsymbol{{t}}\:\right)\:−\:\mathrm{2}}\:\mathrm{4}\:\boldsymbol{{e}}^{−\boldsymbol{{t}}} \:=\:\frac{\mathrm{4}}{−\mathrm{3}\:−\:\mathrm{2}}\:\:\boldsymbol{{e}}^{−\boldsymbol{{t}}} =\:−\:\frac{\mathrm{4}}{\mathrm{5}}\:\boldsymbol{{e}}^{−\boldsymbol{{t}}} \\ $$$$ \\ $$$$\boldsymbol{{f}}_{\mathrm{3}} \left(\boldsymbol{{t}}\right)\:=\:\boldsymbol{{a}}\:\Rightarrow\:\boldsymbol{{f}}\:^{'} \left(\boldsymbol{{t}}\right)\:=\:\mathrm{0}\:\Rightarrow\:\mathrm{3}\:\left(\mathrm{0}\right)\:−\:\mathrm{2}\:\left(\boldsymbol{{a}}\right)\:=\:\mathrm{2}\:\Rightarrow\:\boldsymbol{{a}}\:=\:−\mathrm{1} \\ $$$$ \\ $$$$\therefore\:\boldsymbol{{f}}_{\mathrm{3}} \left(\boldsymbol{{t}}\right)\:=\:−\mathrm{1} \\ $$$$ \\ $$$$\therefore\:\boldsymbol{{f}}\left(\boldsymbol{{t}}\right)\:=\:\boldsymbol{{f}}_{\mathrm{1}} \left(\boldsymbol{{t}}\right)\:+\:\boldsymbol{{f}}_{\mathrm{2}} \left(\boldsymbol{{t}}\right)\:+\:\boldsymbol{{f}}_{\mathrm{3}} \left(\boldsymbol{{t}}\right)\:=\:\boldsymbol{{c}}_{\mathrm{1}} \:\boldsymbol{{e}}^{\frac{\mathrm{2}}{\mathrm{3}}\:\boldsymbol{{t}}} \:−\:\frac{\mathrm{4}}{\mathrm{5}}\:\boldsymbol{{e}}^{−\:\boldsymbol{{t}}} \:−\:\mathrm{1} \\ $$$$ \\ $$$$\langle\:\boldsymbol{{M}}\:.\:\boldsymbol{{T}}\:\:\rangle \\ $$

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