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August 2012

Systems recordings

Here is a recoding of a solution to a system of linear, first-order differential equations with a slope field and solution curves near eigenvectors. or get it on You Tube, http://youtu.be/A-RwYm2SLiA Here is a recording of a similar system but with complex eigenvalues: http://youtu.be/q9nCdpdlqE4  

Exam 3

Here is the third test.  I have one more comprehensive part to post: coming soon. For now enjoy part 1 of the last test: mth256_exam3_summer12

Laplace Transform, impulse response

Suppose a differential equation, \(a\ddot{w}+b\dot{w}+cw=\delta(t)\), has the following unit impulse response:

\[w(t)=e^{-t}-e^{-3t}\]

What would its response to the unit step function be?

\[a\ddot{v}+b\dot{v}+cv=u(t)\]

Using convolution we get

\[v(t)=\int_0^t w(t-\tau)*u(\tau)d\tau=\int_0^t w(\tau)*u(t-\tau)d\tau=\int_0^te^{-t}-e^{-3t} d\tau\]

So, \(v(t)=\left[-e^{-\tau}+\frac{1}{3}e^{-3\tau}\right]_0^t=-e^{-t}+1+\frac{1}{3}e^{-3t}-\frac{1}{3}=\frac{2}{3}-e^{-t}+\frac{1}{3}e^{-3t}\) Alternatively, we know that \(w(t)=\mathop{\mathscr{L^{-1}}}\left\{\frac{1}{as^2+bs+c}\right\}\). Comparing that to \(\mathop{\mathscr{L}}\left\{w(t)\right\}=\mathop{\mathscr{L}}\left\{e^{-t}-e^{-3t}\right\}=\frac{1}{s+1}-\frac{1}{s+3}\).

\[\frac{1}{s+1}-\frac{1}{s+3}=\frac{2}{(s+1)(s+3)}=\frac{1}{\frac{1}{2}(s^2+4s+3)}=\frac{1}{\frac{1}{2}s^2+2s+\frac{3}{2}}\]

Cool! \(a=\frac{1}{2}\) , \(b=2\) , and \(a=\frac{3}{2}\). For fun we could solve […]

Laplace Transform Example

\[\dot{x}-4x=e^t \; ,\; x(0)=f(o)=1\]

Applying the Laplace transform operator to both sides:

\[\mathop{\mathscr{L}}\{\dot{x}-4x\}=\mathop{\mathscr{L}}\{e^t\}\]

\[\mathop{\mathscr{L}}\{\dot{x}\}-4\mathop{\mathscr{L}}\{x\}=\frac{1}{s-1}\]

\[s\hat{f}(s)-f(0)-4\hat{f}(s)=\frac{1}{s-1}\]

\[(s-4)\hat{f}(s)-1=\frac{1}{s-1}\]

\[\hat{f}(s)=\frac{1}{(s-1)(s-4)}+\frac{1}{s-4}\]

And now for the inverse Laplace transform:

\[x(t)=\mathop{\mathscr{L}^{-1}}\{\hat{f}(s)\}=\mathop{\mathscr{L}^{-1}}\left\{\frac{1}{(s-1)(s-4)}+\frac{1}{s-4}\right\}\]

\[x(t)=\mathop{\mathscr{L}^{-1}}\left\{\frac{1}{(s-1)(s-4)}\right\}+\mathop{\mathscr{L}^{-1}}\left\{\frac{1}{s-4}\right\}\]

The second term is easy:

\[\mathop{\mathscr{L}^{-1}}\left\{\frac{1}{s-4}\right\}=e^{4t}\]

The first term requires partial fraction decomposition or completing the square and involving complex numbers.  I am going to show the latter.

\[(s-1)(s-4)=s^2-5s+4=\left(s-\frac{5}{2}\right)^2-\frac{9}{4}=\left(s-\frac{5}{2}\right)^2+\left(\frac{3i}{2}\right)^2\]

[…]

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