Determiner l′original de laplace F(x)=(1/((x^2 +x+1)^2 ))


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 147711 by lapache last updated on 22/Jul/21

$D e t e r m i n e r l^{'} o r i g i n a l d e l a p l a c e$ $F (x) = \frac{1}{{(x^{2} + x + 1)}^{2}}$
	Answered by Olaf_Thorendsen last updated on 23/Jul/21
	$G(x) = F(x−(1/2)) = (1/((x^2 +(3/4))^2 )) H(x) = G(((√3)/2)x) = ((16)/9).(1/((x^2 +1)^2 )) H(x) = ((16)/9)[−(1/(4(x+i)^2 ))−(1/(4(x−i)^2 ))+(i/(4(x+i)))−(i/(4(x−i)))] H(x) = (4/9)[−(1/((x+i)^2 ))−(1/((x−i)^2 ))+(i/(x+i))−(i/(x−i))] L^(−1) ((1/(x−a))) = e^(at) et L^(−1) ((1/((x−a)^2 ))) = te^(at) L^(−1) {H}(t) = (4/9)[−te^(−it) −te^(it) +ie^(−it) −ie^(it) ] L^(−1) {H}(t) = (4/9)[−2tcost+2sint] L^(−1) {H}(t) = (8/9)(sint−tcost) L^(−1) {G(x)}(t) = L^(−1) {H((x/((√3)/2)))}(t) L^(−1) {G(x)}(t) = ((√3)/2)((1/((√3)/2))L^(−1) {H((x/((√3)/2)))}(t) ) L^(−1) {G(x)}(t) = ((√3)/2)(L^(−1) {H}(((√3)/2)t)) L^(−1) {G}(t) = (4/(3(√3)))(sin(((√3)/2)t)−((√3)/2)tcos(((√3)/2)t)) L^(−1) {F(x)}(t) = L^(−1) {G(x+(1/2))}(t) L^(−1) {F(x)}(t) = e^(−(1/2)t) L^(−1) {G(x)}(t) L^(−1) {F}(t) = (4/(3(√3)))e^(−(t/2)) (sin(((√3)/2)t)−((√3)/2)tcos(((√3)/2)t))$
	$G (x) = F (x - \frac{1}{2}) = \frac{1}{{(x^{2} + \frac{3}{4})}^{2}}$ $H (x) = G (\frac{\sqrt{3}}{2} x) = \frac{16}{9} . \frac{1}{{(x^{2} + 1)}^{2}}$ $H (x) = \frac{16}{9} [- \frac{1}{4 {(x + i)}^{2}} - \frac{1}{4 {(x - i)}^{2}} + \frac{i}{4 (x + i)} - \frac{i}{4 (x - i)}]$ $H (x) = \frac{4}{9} [- \frac{1}{{(x + i)}^{2}} - \frac{1}{{(x - i)}^{2}} + \frac{i}{x + i} - \frac{i}{x - i}]$ $L^{- 1} (\frac{1}{x - a}) = e^{a t} et L^{- 1} (\frac{1}{{(x - a)}^{2}}) = {t e}^{a t}$ $L^{- 1} {H} (t) = \frac{4}{9} [- {t e}^{- i t} - {t e}^{i t} + {i e}^{- i t} - {i e}^{i t}]$ $L^{- 1} {H} (t) = \frac{4}{9} [- 2 t \cos t + 2 \sin t]$ $L^{- 1} {H} (t) = \frac{8}{9} (\sin t - t \cos t)$ $L^{- 1} {G (x)} (t) = L^{- 1} {H (\frac{x}{\frac{\sqrt{3}}{2}})} (t)$ $L^{- 1} {G (x)} (t) = \frac{\sqrt{3}}{2} (\frac{1}{\frac{\sqrt{3}}{2}} L^{- 1} {H (\frac{x}{\frac{\sqrt{3}}{2}})} (t))$ $L^{- 1} {G (x)} (t) = \frac{\sqrt{3}}{2} (L^{- 1} {H} (\frac{\sqrt{3}}{2} t))$ $L^{- 1} {G} (t) = \frac{4}{3 \sqrt{3}} (\sin (\frac{\sqrt{3}}{2} t) - \frac{\sqrt{3}}{2} t \cos (\frac{\sqrt{3}}{2} t))$ $L^{- 1} {F (x)} (t) = L^{- 1} {G (x + \frac{1}{2})} (t)$ $L^{- 1} {F (x)} (t) = e^{- \frac{1}{2} t} L^{- 1} {G (x)} (t)$ $L^{- 1} {F} (t) = \frac{4}{3 \sqrt{3}} e^{- \frac{t}{2}} (\sin (\frac{\sqrt{3}}{2} t) - \frac{\sqrt{3}}{2} t \cos (\frac{\sqrt{3}}{2} t))$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum