If x''+x'-6x=e^(-t), with x(0)=x'(0)=0, what is the value of X(s)?

Question

If $x^{″} + x^{'} - 6 x = e^{- t}, with x (0) = x^{'} (0) = 0$ , what is the value of X(s)?

Answer 1

Step 1
Given

x^{″} + x^{'} - 6 x = e^{- t}, with x (0) = x^{'} (0) = 0

To find X(s)
Step 2
Laplace transformation

L {x^{″}} = s^{2} X (s) - s x (0) - x^{'} (0)

L {x^{'}} = s X (s) - x (0)

L {x} = X (s)

L {e^{- t}} = \frac{1}{s + 1}

Step 3
Laplace transformation of given differential equation is

L {x^{″} + x^{'} - 6 x} = L {e^{- t}}

s^{2} X (s) - s x (0) - x^{'} (0) + s X (s) - x (0) - 6 X (s) = \frac{1}{s + 1}

s^{2} X (s) + s X (s) - 6 X (s) = \frac{1}{s + 1}

X (s) [s^{2} + s - 6] = \frac{1}{s + 1}

X (s) [(s + 3) (s - 2)] = \frac{1}{s + 1}

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

Expert Answer