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Question Number 92524 by naka3546 last updated on 07/May/20

(d^3 x/dt^3 ) + 27 (d^2 x/dt^2 ) + 243 (dx/dt) + 729x = t∙ e^(−9t)         x(0) = x′(0) = x^(′′) (0) = 0  Use  Laplace  Transformation  to  solve  it .

$$\frac{{d}^{\mathrm{3}} {x}}{{dt}^{\mathrm{3}} }\:+\:\mathrm{27}\:\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }\:+\:\mathrm{243}\:\frac{{dx}}{{dt}}\:+\:\mathrm{729}{x}\:=\:{t}\centerdot\:{e}^{−\mathrm{9}{t}} \\ $$$$\:\:\:\:\:\:{x}\left(\mathrm{0}\right)\:=\:{x}'\left(\mathrm{0}\right)\:=\:{x}^{''} \left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$$${Use}\:\:{Laplace}\:\:{Transformation}\:\:{to}\:\:{solve}\:\:{it}\:. \\ $$

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