(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 .


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

$\frac{d^{3} x}{{d t}^{3}} + 27 \frac{d^{2} x}{{d t}^{2}} + 243 \frac{d x}{d t} + 729 x = t \cdot e^{- 9 t}$ $x (0) = x^{'} (0) = x^{^{″}} (0) = 0$ $U s e L a p l a c e T r a n s f o r m a t i o n t o s o l v e i t .$
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