MIT differential equation---9,10
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周一回家了。这次图省钱坐的阿联酋航空,路上一共花了将近35小时了吧。这两天休息的差不多了,开始实行暑假计划~
Linear second-order ODE with constant coefficients
y'' + Ay' + By = 0 homogenous
solution in the form of y = c1y1 + c2y2
spring mass dashpot system:
mx'' = -kx^2 + cx'
basic method:
try y = e^rt
plug in and you can get a quadratic equation, also known as characteristic equation
case 1 : r1 != r2, both are real(overdamped)
ex: y'' + 4y' + 3y = 0
one can immediately write out the characteristic equation:
r^2 + 4r + 3 = 0. r1 = -3; r2 = -1
given by condition where y(0) = 1 & y'(0) = 0
one can solve y = c1y1 + c2y2
case 2: roots are complex(underdamped)
if u+iv is the solution to the characteristic equation, then u, v are two different solutions to the ODE
y = e^(at + ibt) => y = e^at (cosbt + i sinbt) => y = e^at (c1 cosbt + c2 sinbt)
ex:y''+4y'+5=0
y = e^-2(c1cost+c2sint)
physically, there's gonna be oscillation only when there's complex roots
case 3: r1 = r2 = -a&& real(critical damped)
y = e^-at *u
backward-euler formula
sinx = (e^xt + e^-xt)/2
cosx = (e^xt - e^-xt)/2i
change the name of spring mass dashpot system formula to
y'' + 2p y' + (wo^2) y = 0
2p = c / m; wo^2 = k/m
solution can be written in the form of y = Ae^p cos(w1t + phi)
p and w1 depend on ODE
A, phi depend on initial condition
case 1 : p = 0; c = 0; oscillation undamped
pure oscillation wo is the circular frequency
case 2: p^2 < wo^2; oscillation damping
p < wo; p is the half damping
w1 = sqrt(-p^2 + w0^2)
T = pi/w1 ; w1 is the circular pseudofrequency
今天上了两课。上午看了一节,晚上7点睡觉十点醒来再看了一集。。这时差倒的。。。
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