# the differential equation be solved using the Laplace Transform? A=3 x^{II}(t)-3x^{I}(t)+2x(t)=A e^{2t} x(0)=-3 x^{I}(0)=5

the differential equation be solved using the Laplace Transform? A=3
${x}^{II}\left(t\right)-3{x}^{I}\left(t\right)+2x\left(t\right)=A{e}^{2t}$
$x\left(0\right)=-3$
${x}^{I}\left(0\right)=5$
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averes8
Step 1
Some Laplace Transforms,
$1.L\left({x}^{″}\left(t\right)\right)={s}^{2}x\left(s\right)-sx\left(0\right)-{x}^{\prime }\left(0\right)$
$2.L\left({x}^{\prime }\left(t\right)\right)=sx\left(s\right)-x\left(0\right)$
$3.L\left(x\left(t\right)\right)=x\left(s\right)$
$⇒{L}^{-1}\left(x\left(s\right)\right)=x\left(t\right)$

$⇒{L}^{-1}\left(\frac{1}{\left(s-a\right)}\right)={e}^{at}$
$5.L\left(t{e}^{at}\right)=\frac{1}{\left(s-a{\right)}^{2}}⇒{L}^{-1}\left(\frac{1}{\left(s-a{\right)}^{2}}\right)=t{e}^{at}$
Step 2
Given differential equation is

Take Laplace Transform of every term of given differential equation, we get
$L\left({x}^{″}\left(t\right)\right)-3L\left({x}^{\prime }\left(t\right)\right)+2L\left(x\left(t\right)\right)=3L\left({e}^{2t}\right)$
using step 1, we get
$⇒{s}^{2}x\left(s\right)-sx\left(0\right)-{x}^{\prime }\left(0\right)-3\left(sx\left(s\right)-x\left(0\right)\right)+2x\left(s\right)=3\left(\frac{1}{\left(s-2\right)}\right)$
as given initial conditions are

$⇒{s}^{2}x\left(s\right)-s\left(-3\right)-5-3\left(sx\left(s\right)-\left(-3\right)\right)+2x\left(s\right)=3\left(\frac{1}{\left(s-2\right)}\right)$
$⇒{s}^{2}x\left(s\right)+3s-5-3sx\left(s\right)-9+2x\left(s\right)=3\left(\frac{1}{\left(s-2\right)}\right)$
$⇒{s}^{2}x\left(s\right)-3sx\left(s\right)+2x\left(s\right)+3s-14=\frac{3}{\left(s-2\right)}$
$⇒\left({s}^{2}-3s+2\right)x\left(s\right)+3s-14=\frac{3}{\left(s-2\right)}$
$⇒\left(s-2\right)\left(s-1\right)x\left(s\right)=\frac{3}{\left(s-2\right)}-3s+14$
$⇒x\left(s\right)=\frac{3}{\left(s-2{\right)}^{2}\left(s-1\right)}-\frac{\left(3s-14\right)}{\left(s-2\right)\left(s-1\right)}$
Step 3
Partial fractional decomposition
$\frac{\left(3s-14\right)}{\left(s-2\right)\left(s-1\right)}=\frac{11}{\left(s-1\right)}-\frac{8}{\left(s-2\right)}$
$\frac{3}{\left(s-2{\right)}^{2}\left(s-1\right)}=\frac{3}{\left(s-1\right)}-\frac{3}{\left(s-2\right)}+\frac{3}{\left(s-2{\right)}^{2}}$
then equation (1) becomes