Question

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

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

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