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

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

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

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

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

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

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

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

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

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