Solve the laplace transforms x˙−2x¨+x=et⋅t fiven t=0 and x˙=0 and x=1

Name: Solve the laplace transforms x˙−2x¨+x=et⋅t fiven t=0 and x˙=0 and x=1
Uploaded: 2022-02-23T17:22:57+00:00
Description: Solve the laplace transforms x˙−2x¨+x=et⋅t fiven t=0 and x˙=0 and x=1

Solution: x˙−2x¨+x=et⋅t,x˙(0)=0,x(0)=1 Take both sides Laplace Transformation L{x˙}−2L{x¨(t)}+L{x(t)}=L{et⋅t} ⇒{sX(s)−x(0)}−2{s2X(s)−sX(0)−x˙(0)}+x(s)=−dds(1s−1)=1(s−1)2 ⇒(sX(s)−1)−2(s2X(s)−s−0)+x(s)=1(s−1)2 ⇒(−2s2+s+1)x(s)=1(s−1)2−2s+1 ⇒−(2s+1)(s−1)X(s)=1(s−1)2−2s+1 ⇒x(s)=−1(s−1)3(2s+1)+2s−1(2s+1)(s−1) ⇒x(t)=L−1{−1(s−1)3(s+1)}+L−1{2s−1(2s+1)(s−1)} =L−1{−4271(s−1)+291(s−1)2−161(s−1)3+82712s+1}+L−1{43⋅12s+1+131(s−1)} =L−1

Expert answers to "Solve the laplace transforms x˙−2x¨+x=et⋅t fiven t=0 and..."

Ernstfalld

Answered question

2021-09-14

Solve the laplace transforms
$\dot{x} - 2 \ddot{x} + x = e^{t} \cdot t$
fiven $t = 0$ and $\dot{x} = 0$ and $x = 1$

Answer & Explanation

Arham Warner

Skilled2021-09-15Added 102 answers

Solution:
$\dot{x} - 2 \ddot{x} + x = e^{t} \cdot t, \dot{x} (0) = 0, x (0) = 1$
Take both sides Laplace Transformation
$L {\dot{x}} - 2 L {\ddot{x} (t)} + L {x (t)} = L {e^{t} \cdot t}$
$\Rightarrow {s X (s) - x (0)} - 2 {s^{2} X (s) - s X (0) - \dot{x} (0)} + x (s) = \frac{- d}{d s} (\frac{1}{s - 1}) = \frac{1}{(s - 1)^{2}}$
$\Rightarrow (s X (s) - 1) - 2 (s^{2} X (s) - s - 0) + x (s) = \frac{1}{{(s - 1)}^{2}}$
$\Rightarrow (- 2 s^{2} + s + 1) x (s) = \frac{1}{{(s - 1)}^{2}} - 2 s + 1$
$\Rightarrow - (2 s + 1) (s - 1) X (s) = \frac{1}{{(s - 1)}^{2}} - 2 s + 1$
$\Rightarrow x (s) = \frac{- 1}{{(s - 1)}^{3} (2 s + 1)} + \frac{2 s - 1}{(2 s + 1) (s - 1)}$
$\Rightarrow x (t) = L^{- 1} {\frac{- 1}{(s - 1)^{3} (s + 1)}} + L^{- 1} {\frac{2 s - 1}{(2 s + 1) (s - 1)}}$
$= L^{- 1} {\frac{- 4}{27} \frac{1}{(s - 1)} + \frac{2}{9} \frac{1}{(s - 1)^{2}} - \frac{1}{6} \frac{1}{(s - 1)^{3}} + \frac{8}{27} \frac{1}{2 s + 1}} + L^{- 1} {\frac{4}{3} \cdot \frac{1}{2 s + 1} + \frac{1}{3} \frac{1}{(s - 1)}}$
$= L^{- 1}$

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