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

Question

the differential equation be solved using the Laplace Transform? A=3

x^{I I} (t) - 3 x^{I} (t) + 2 x (t) = A e^{2 t}

x (0) = - 3

x^{I} (0) = 5

Answer 1

Step 1
Some Laplace Transforms,

1. L (x^{″} (t)) = s^{2} x (s) - s x (0) - x^{'} (0)

2. L (x^{'} (t)) = s x (s) - x (0)

3. L (x (t)) = x (s)

\Rightarrow L^{- 1} (x (s)) = x (t)

4. L (e^{a t}) = \frac{1}{(s - a)} then L (e^{2 t}) = \frac{1}{(s - 2)}

\Rightarrow L^{- 1} (\frac{1}{(s - a)}) = e^{a t}

5. L (t e^{a t}) = \frac{1}{(s - a)^{2}} \Rightarrow L^{- 1} (\frac{1}{(s - a)^{2}}) = t e^{a t}

Step 2
Given differential equation is

x^{″} (t) - 3 x^{'} (t) + 2 x (t) = 3 e^{2 t} (as given A = 3) and x (0) = - 3, x^{'} (0) = 5

Take Laplace Transform of every term of given differential equation, we get

L (x^{″} (t)) - 3 L (x^{'} (t)) + 2 L (x (t)) = 3 L (e^{2 t})

using step 1, we get

\Rightarrow s^{2} x (s) - s x (0) - x^{'} (0) - 3 (s x (s) - x (0)) + 2 x (s) = 3 (\frac{1}{(s - 2)})

as given initial conditions are

x (0) = - 3, x^{'} (0) = 5 then

\Rightarrow s^{2} x (s) - s (- 3) - 5 - 3 (s x (s) - (- 3)) + 2 x (s) = 3 (\frac{1}{(s - 2)})

\Rightarrow s^{2} x (s) + 3 s - 5 - 3 s x (s) - 9 + 2 x (s) = 3 (\frac{1}{(s - 2)})

\Rightarrow s^{2} x (s) - 3 s x (s) + 2 x (s) + 3 s - 14 = \frac{3}{(s - 2)}

\Rightarrow (s^{2} - 3 s + 2) x (s) + 3 s - 14 = \frac{3}{(s - 2)}

\Rightarrow (s - 2) (s - 1) x (s) = \frac{3}{(s - 2)} - 3 s + 14

\Rightarrow x (s) = \frac{3}{(s - 2)^{2} (s - 1)} - \frac{(3 s - 14)}{(s - 2) (s - 1)}

Step 3
Partial fractional decomposition

\frac{(3 s - 14)}{(s - 2) (s - 1)} = \frac{11}{(s - 1)} - \frac{8}{(s - 2)}

\frac{3}{(s - 2)^{2} (s - 1)} = \frac{3}{(s - 1)} - \frac{3}{(s - 2)} + \frac{3}{(s - 2)^{2}}

then equation (1) becomes

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