Laplace Transform of a 2nd order Differential Equation

**craig** · October 13th 2011, 12:59 AM

Another Laplace question, this time a second order DE.

We have $\ddot{x} + 3\dot{x} + 2x = 1$ , with zero initial conditions. The Laplace Transform gives:

$s^2\mathcal{L}\{x\} -sx_0 - \dot{x}_0 + 3(s\mathcal{L}\{x\} - x_0) + 2\mathcal{L}\{x\} = \frac{1}{s}$

$(s^2 + 3s + 2)\mathcal{L}\{x\} = \frac{1}{s}$

$\mathcal{L}\{x\} = \frac{1}{s(s+1)(s + 2)} = \frac{1}{2s} - \frac{1}{s+1} + \frac{1}{2s + 4}$ by partial fractions.

So using the inverse Laplace:

$\mathcal{L}\{x\} = \frac{1}{2} \frac{1}{s} - \frac{1}{s-(-1)} + \frac{1}{2} \frac{1}{s -(-2)}$

$x(t) = \frac{1}{2} - e^{-t} + \frac{e^{-2t}}{2}$

Does this seem about right?

**FernandoRevilla** · October 13th 2011, 01:34 AM

Originally Posted by craig

$x(t) = \frac{1}{2} - e^{-t} + \frac{e^{-2t}}{2}$ Does this seem about right?

Right.

**craig** · October 13th 2011, 02:18 AM

Thankyou!

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