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

Find Laplace transform for
$x"-3{x}^{\prime }+2x=1-{e}^{2t}$
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2abehn
Step 1
given equation
$x"-3{x}^{\prime }+2x=1-{e}^{2t}$
by assuming that initial condition is zero.
$x"={s}^{2}X\left(s\right)$

and laplace transform of
Step 2
put the calculated values in the above equation
${s}^{2}X\left(s\right)-3sX\left(s\right)+2X\left(s\right)=\frac{-2}{{s}^{2}-2s}$
$X\left(s\right)\left({s}^{2}-3s+2\right)=\frac{-2}{{s}^{2}-2s}$
$X\left(s\right)=\frac{-2}{\left({s}^{2}-2s\right)\left({s}^{2}-3s+2\right)}$