Laplace Transformations can be used to solve initial value problems for differential equations. The Laplace Transform of a derivative can be expressed as an algebraic equation in terms of the initial values.
| Differential equation in terms of \(y(t)\) |
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| Take Laplace transformation |
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| Isolate \(Y(s)\) |
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| Take inverse Laplace to solve for \(y(t)\) |
Solve the DE \(\cfrac{d^2y}{dt^2} - \cfrac{dy}{dt} - 2y = 0\) using Laplace Transformations. Given \(y'(0) = 0\), \(y(0) = 1\).
First, we can 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 can 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 the following solutions:
\(A = \cfrac{1}{3}, B = \cfrac{2}{3}\)
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:
\( y(t) = \cfrac{1}{3} e^{2t} + \cfrac{2}{3} e^{-t} \)