Question

Find Laplace transform for x"-3x'+2x=1-e^{2t}

Laplace transform
ANSWERED
asked 2021-02-01
Find Laplace transform for
\(x"-3x'+2x=1-e^{2t}\)

Answers (1)

2021-02-02
Step 1
given equation
\(x"-3x'+2x=1-e^{2t}\)
by assuming that initial condition is zero.
\(x"=s^{2}X(s)\)
\(\text{and}\ x'=sX(s)\)
and laplace transform of \(1-e^{2t} \text{ is } \frac{1}{s}-\frac{1}{s-2}=\frac{s-2-s}{s(s-2)}=\frac{-2}{s^{2}-2s}\)
Step 2
put the calculated values in the above equation
\(s^{2}X(s)-3sX(s)+2X(s)=\frac{-2}{s^{2}-2s}\)
\(X(s)(s^{2}-3s+2)=\frac{-2}{s^{2}-2s}\)
\(X(s)=\frac{-2}{(s^{2}-2s)(s^2-3s+2)}\)
\(\text{the } X(s)=\frac{-2}{(s^{2}-2s)(s^2-3s+2)} \text{ is the required laplace transform.}\)
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