# Laplace Transform of a 2nd order Differential Equation

• Oct 13th 2011, 12:59 AM
craig
Laplace Transform of a 2nd order Differential Equation
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?
• Oct 13th 2011, 01:34 AM
FernandoRevilla
Re: Laplace Transform of a 2nd order Differential Equation
Quote:

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

Right.
• Oct 13th 2011, 02:18 AM
craig
Re: Laplace Transform of a 2nd order Differential Equation
Thankyou!