Solve the Laplace transforms x˙−2x¨+x=ett given t=0 and x=0 and x=1

Name: Solve the Laplace transforms x˙−2x¨+x=ett given t=0 and x=0 and x=1
Uploaded: 2022-02-23T17:22:57+00:00
Description: Solve the Laplace transforms x˙−2x¨+x=ett given t=0 and x=0 and x=1

Given :The given Laplace Transform−2x¨+x˙+x=ettwith the given conditions x(0)=1,x˙(0)=0Step 2Solution :The given Laplace Transform−2x¨+x˙+x=ett…(1)with the given conditionsx(0)=1,x˙(0)=0Taking Laplace transform of both sides in equation (1) , we have−2[s2x¯−sx(0)−x′(0)]+[sx¯−x(0)]+x¯=1(s−1)2−2[s2x¯−s]+[sx¯−1]+x¯=1(s−1)2[−2s2+s+1]x¯+[2s−1]=1(s−1)2[−2s2+2s−s+1]x¯=1(s−1)2−[2s−1][−2s(s−1)−(s−1)]x¯=1(s−1)2−[2s−1][(−2s−1)(s−1)]x¯=1(s−1)2−[2s−1]x¯=1(s−1)2[(−2s−1)(s−1)]+[2s−1][(2s+1)(s−1)]x¯=1(s−1)3(−2s−1)+[2s−1][(2s+1)(s−1)]Solving By Partial fractions ,[2s−1][(2s+1)(s−1)]=A(s−1)+B(2s+1)[2s−1][(2s+1)(s−1)]=A(2s+1)+B(s−1)(s−1)(2s+1)2A+B=2,A−B=−13A=1⇒A=1313−B=−1⇒B=43[2s−1][(2s+1)(s−1)]=13(s−1)+43(2s+1)Now , another term ,−1(s−1)3(2s+1)=A(2s+1)+B(s−1)+C(s−1)2+D(s−1)3

Solved: "Solve the Laplace transforms x˙−2x¨+x=ett given t=0"

pedzenekO

Answered question

2020-11-01

Solve the Laplace transforms

\dot{x} - 2 \ddot{x} + x = e^{t} t

given t = 0 and x = 0 and x = 1

Answer & Explanation

Ayesha Gomez

Skilled2020-11-02Added 104 answers

Given :
The given Laplace Transform

- 2 \ddot{x} + \dot{x} + x = e^{t} t

with the given conditions

x (0) = 1, \dot{x} (0) = 0

Step 2
Solution :
The given Laplace Transform

- 2 \ddot{x} + \dot{x} + x = e^{t} t \dots (1)

with the given conditions

x (0) = 1, \dot{x} (0) = 0

Taking Laplace transform of both sides in equation (1) , we have

- 2 [s^{2} \bar{x} - s x (0) - x^{'} (0)] + [s \bar{x} - x (0)] + \bar{x} = \frac{1}{(s - 1)^{2}}

- 2 [s^{2} \bar{x} - s] + [s \bar{x} - 1] + \bar{x} = \frac{1}{(s - 1)^{2}}

[- 2 s^{2} + s + 1] \bar{x} + [2 s - 1] = \frac{1}{(s - 1)^{2}}

[- 2 s^{2} + 2 s - s + 1] \bar{x} = \frac{1}{(s - 1)^{2}} - [2 s - 1]

[- 2 s (s - 1) - (s - 1)] \bar{x} = \frac{1}{(s - 1)^{2}} - [2 s - 1]

[(- 2 s - 1) (s - 1)] \bar{x} = \frac{1}{(s - 1)^{2}} - [2 s - 1]

\bar{x} = \frac{1}{(s - 1)^{2} [(- 2 s - 1) (s - 1)]} + \frac{[2 s - 1]}{[(2 s + 1) (s - 1)]}

\bar{x} = \frac{1}{(s - 1)^{3} (- 2 s - 1)} + \frac{[2 s - 1]}{[(2 s + 1) (s - 1)]}

Solving By Partial fractions ,

\frac{[2 s - 1]}{[(2 s + 1) (s - 1)]} = \frac{A}{(s - 1)} + \frac{B}{(2 s + 1)}

\frac{[2 s - 1]}{[(2 s + 1) (s - 1)]} = \frac{A (2 s + 1) + B (s - 1)}{(s - 1) (2 s + 1)}

2 A + B = 2, A - B = - 1

3 A = 1 \Rightarrow A = \frac{1}{3}

\frac{1}{3} - B = - 1 \Rightarrow B = \frac{4}{3}

\frac{[2 s - 1]}{[(2 s + 1) (s - 1)]} = \frac{1}{3 (s - 1)} + \frac{4}{3 (2 s + 1)}

Now , another term ,

- \frac{1}{(s - 1)^{3} (2 s + 1)} = \frac{A}{(2 s + 1)} + \frac{B}{(s - 1)} + \frac{C}{(s - 1)^{2}} + \frac{D}{(s - 1)^{3}}

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