Find Laplace transform forx"−3x′+2x=1−e2t

Step 1given equationx"−3x′+2x=1−e2tby assuming that initial condition is zero.x"=s2X(s)and x′=sX(s)and laplace transform of 1−e2t is 1s−1s−2=s−2−ss(s−2)=−2s2−2sStep 2put the calculated values in the above equations2X(s)−3sX(s)+2X(s)=−2s2−2sX(s)(s2−3s+2)=−2s2−2sX(s)=−2(s2−2s)(s2−3s+2)the X(s)=−2(s2−2s)(s2−3s+2) is the required laplace transform.

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e1s2kat26

Answered question

2021-02-01

Find Laplace transform for

x " - 3 x^{'} + 2 x = 1 - e^{2 t}

Answer & Explanation

2abehn

Skilled2021-02-02Added 88 answers

Step 1
given equation

x " - 3 x^{'} + 2 x = 1 - e^{2 t}

by assuming that initial condition is zero.

x " = s^{2} X (s)

and x^{'} = s X (s)

and laplace transform of

1 - e^{2 t} is \frac{1}{s} - \frac{1}{s - 2} = \frac{s - 2 - s}{s (s - 2)} = \frac{- 2}{s^{2} - 2 s}

Step 2
put the calculated values in the above equation

s^{2} X (s) - 3 s X (s) + 2 X (s) = \frac{- 2}{s^{2} - 2 s}

X (s) (s^{2} - 3 s + 2) = \frac{- 2}{s^{2} - 2 s}

X (s) = \frac{- 2}{(s^{2} - 2 s) (s^{2} - 3 s + 2)}

the X (s) = \frac{- 2}{(s^{2} - 2 s) (s^{2} - 3 s + 2)} is the required laplace transform.

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