Friday Review

Here are a few problems to review from Unit 2. We discussed the ERF, undetermined coefficients, and variation of parameter, mainly focusing on the ERF.

  1. \(y''+7y'+12y=3\cos(2x)\)
  2. \(\ddot{x}+x=3\sin(2t)+3\cos(2t)\)
  3. \(\ddot{x}+\dot{x}+2x=5\sin(2t)+\cos(2t)\)

From Unit 1 we discussed:

\[m\dot{v}=-mg+kv \; , \; \mathrm{with} \; v(0)=v_0\]

versus

\[m\dot{v}=-mg-kv \; , \; \mathrm{with} \; v(t_m)=0\]

where \(t_0\) is the time the object reaches its maximum height (when \(v=0\)) and upward is positive.

Last, I introduced the Laplace transform:

\[\mathop{\mathscr{L}}\{f(t)\}=\int_0^{\infty}f(t)\cdot e^{-st}dt=\hat{f}(s)\]

For example

\[\mathop{\mathscr{L}}\{1\}=\int_0^{\infty}1\cdot e^{-st}dt=\frac{1}{s}\]

\[\mathop{\mathscr{L}}\{e^{at}\}=\int_0^{\infty}e^{at}\cdot e^{-st}dt=\frac{1}{s-a}\]

and finally (this one is miraculous):

\[\mathop{\mathscr{L}}\{f'(t)\}=s\hat{f}(s)-f(0)\]

2 Responses to “Friday Review”

  1. gary says:

    Wow. That last formula is interesting. So if \(f(t)=t\) then \(f'(t)=1\) so

    \[\mathop{\mathscr{L}}\left\{\frac{d}{dt}[t]\right\}=s\hat{f}(s)-f(0)\]


    OK. \(\mathop{\mathscr{L}}\left\{\frac{d}{dt}[t]\right\}=\mathop{\mathscr{L}}\{1\}=\frac{1}{s}\) and \(f(0)=0\), so

    \[\frac{1}{s}=s\hat{f}(s)\]


    \[\mathop{\mathscr{L}}\{t\}=\frac{1}{s^2}\]

  2. andrew says:

    What is f hat? And what is s?

    -Andrew

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