e1s2kat26

2021-02-01

Find Laplace transform for

$x"-3{x}^{\prime}+2x=1-{e}^{2t}$

2abehn

Skilled2021-02-02Added 88 answers

Step 1

given equation

$x"-3{x}^{\prime}+2x=1-{e}^{2t}$

by assuming that initial condition is zero.

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

$\text{and}\text{}{x}^{\prime}=sX(s)$

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

by assuming that initial condition is zero.

and laplace transform of

Step 2

put the calculated values in the above equation

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$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{({s}^{2}-{a}^{2})(s-2b)}$

My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$

$K=0$

$A=F(s)\ast (s+a)$