First apply the Laplace Transformation on both sides of the equation.
\( \mathcal L \{\cfrac{d^2y}{dt^2} - \cfrac{dy}{dt} - 2y \}= \mathcal L \{0\} \)
\( (s^2 Y - s y(0) - y'(0)) - (sY - y(0) - 2Y = 0\)
\( s^2 Y - s y(0) - y'(0) - sY + y(0) - 2Y = 0\)
Next, isolate for \(Y\) subtituting the initial value conditions:
\( s^2 Y - sY - 2Y = s y(0) + y'(0) - y(0)\)
\( Y(s^2 - s - 2) = s - 1\)
\( Y = \cfrac{s - 1}{s^2 - s - 2}\)
Now we need to evaluate the Laplace inverse. It is helpful to factor the demoninator and then express the quotient
as a partial fraction:
\( Y = \cfrac{s - 1}{(s - 2)(s+1)}\)
To express the right hand side as a partial fraction, we want:
\( \cfrac{s - 1}{(s - 2)(s+1)} = \cfrac{A}{s-2} + \cfrac{B}{s+1} = \cfrac{A(s+1) + B(s-2)}{(s - 2)(s+1)} \)
Now we set the numerators equal and solve for \(A\) and \(B\). Note that the numerator should be one degree smaller than the denominator.
For a denominator \(s^2 + 3\) we would choose a numerator \(Ax+B\).
\( s - 1 = A(s+1) + B(s-2) = As + A + Bs - 2B \)
We set up a system of equations for each exponent on \(s\):
\( \bbox[5px,border:1px solid red] {s} \bbox[5px,border:1px solid blue]{- 1} = \bbox[5px,border:1px solid red]{As} + \bbox[5px,border:1px solid blue]{A} + \bbox[5px,border:1px solid red]{Bs} \bbox[5px,border:1px solid blue]{- 2B} \)
\( s = As + Bs \Rightarrow 1 = A + B \)
\( - 1 = A - 2B \)
Solving the system yields:
\(A = \cfrac{1}{3}, B = \cfrac{2}{3}\)
Now we can rewrite the equation and take the Lapalce inverse:
\( Y(s) = \cfrac{1}{3}\cfrac{1}{s-2} + \cfrac{2}{3}\cfrac{1}{s+1} \)
\( \mathcal L^{-1} \{Y(s)\} = \mathcal L^{-1} \{ \cfrac{1}{3}\cfrac{1}{s-2} + \cfrac{2}{3}\cfrac{1}{s+1} \}\)
Using the table from the previous lesson, we find that:
\( y(t) = \cfrac{1}{3} e^{2t} + \cfrac{2}{3} e^{-t} \)