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

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

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

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

[…]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 […]Let \(f\) be a piecewise continuous function on the interval \([-T,T]\). The Fourier Series of \(f\) is the trigonometric series

\[f(x)\sim\frac{a_0}{2}+\sum_{n=1}^\infty \left\{a_n\cos\left(\frac{n\pi x}{T}\right)+b_n\sin\left(\frac{n\pi x}{T}\right)\right\}\]

where the \(a_n\)'s and \(b_n\)'s are given by the formulas\[a_n=\frac{1}{T}\int_{-T}^Tf(x)\cos\left(\frac{n\pi x}{T}\right)dx\; , \; n=0,1,2,...\]

\[b_n=\frac{1}{T}\int_{-T}^Tf(x)\sin\left(\frac{n\pi x}{T}\right)dx\; , \; n=1,2,3,...\]

For example, let \(f(x)\) be defined by $$!f(x)=\begin{cases} 0 & , […]\[\ddot{x}+x=0.5\cos(0.8x)\]

The homogeneous solution is\[x_h(t)=c_1\cos(t)+c_2\sin(t)\]

Using variation of parameters we get a nonhomogeneous solution of\[x_p(t)=\left(\frac{5}{4}\cos(0.2t)+\frac{5}{36}\cos(1.8t)\right)\cos(t)+\left(\frac{5}{4}\sin(0.2t)+\frac{5}{36}\sin(1.8t)\right)\sin(t)\]

Distributing and regrouping\[x_p(t)=\frac{5}{4}\cos(t)\cos(0.2t)+\frac{5}{4}\sin(t)\sin(0.2t)+\frac{5}{36}\cos(1.8t)\cos(t)+\frac{5}{36}\sin(1.8t)\sin(t)\]

Reversing the cosine angle difference formula\[x_p(t)=\frac{5}{4}\cos(0.8t)+\frac{5}{36}\cos(0.8t)=\frac{25}{18}\cos(0.8t)\]

Using the initial conditions \(x(0)=0\), \(\dot{x}(0)=0\), we get \(c_1=\frac{-25}{18}\) and \(c_2=0\) So a specific solution is\[x(t)=\frac{-25}{18}\cos(t)+\frac{25}{18}\cos(0.8t)\]

\[x(t)=\frac{25}{18}\left(\cos(0.8t)-\cos(t)\right)\]

Regrouping and applying a difference to product formula\[x(t)=\frac{25}{18}\left(2\sin\left(\frac{1-0.8}{2}\right)\sin\left(\frac{1+0.8}{2}\right)\right)\]

[…]Bernoulli equations are of the form:

\[\frac{dy}{dx}+P(x)y=Q(x)y^n\]

Dividing by \(y^n\):\[\frac{dy}{dx}y^{-n}+P(x)y^{1-n}=Q(x)\]

If \(n=0\) or \(n=1\) then the equation is linear. Otherwise a substitution of \(u=y^{1-n}\) will give us a linear equation. If \(u=y^{1-n}\), then \(\frac{du}{dx}u=(1-n)\frac{dy}{dx}y^{-n}\). Substituting these into the second equation from above:\[\frac{1}{1-n}\frac{du}{dx}+P(x)u=Q(x)\]

For example,\[y'+2xy=xy^4\]

Dividing by \(y^4\), we get \(y'y^{-4}+2xy^{-3}=x\) Substituting \(u=y^{-3}\) […]There are two ways to view "homogeneous" first-order ordinary differential equations. 1. \(\frac{dy}{dx}=f(x,y)\) is a general first order ordinary differential equation. If the function on the right, \(f(x,y)\), can be rewritten as a function of \(\frac{y}{x}\), e.g. \(f(x,y)=\frac{y-2x}{x}\) can be rewritten as \(g\left(\frac{y}{x}\right)=\frac{y}{x}-2\). If we let \(v=\frac{y}{x}\), then we would have \(g(v)=v-2\). In general, the […]

Here is a link to a short, 15 min, introduction to differential equations: http://youtu.be/kIKp8rLjsHY Here is a long video recorded in Blackboard Collaborate: https://sas.elluminate.com/p.jnlp?psid=2012-06-27.1134.M.415E25ABC3437BC9A4A0FBE3CC971C.vcr&sid=837