96 lines
3.7 KiB
Markdown
96 lines
3.7 KiB
Markdown
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#start of lec 17
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summary of all the equations (separable linear exact, ...) available on eclass
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(homogenous (2 forms), Bernoulli equation, linear coefficients,)
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second order equations:
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put it in standard form
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solve the homogenous eq,
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with constant coefficients
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3 cases: overdamped critically damped under damped
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now find particular solution to get general solution $y=y_{h}+y_{p}$
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either use #mouc or #voparam
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we also have Cauchy-Euler equations (two methods, either guess y=x^r or use the general formula)
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reduction of order is a technique when you're given a solution and you have to find the other
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mechanical systems $my''+by'+ky=F_{o}\cos(\gamma t)$ (he recommends solving with mouc) that gives you the transient solution.
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the particular solution is the steady part.
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recall $\mu$=amplitude $\gamma_{r}$=resonance freq $\mu(\gamma_{_{r}})$=amplitude at resonance
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Laplace transforms: definition and 4 important properties
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That's all that will be covered on the midterm. 25 minutes spent covering what will be on the midterm!
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So, why did we learn all this stuff about laplace transforms? We will now see how its useful:
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#ex #LT #second_order_nonhomogenous
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$$y''+y=t^2+2 \qquad y(0)=1 \qquad y'(0)=-1$$
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LHS=RHS, so the Laplace transforms of each side must also be equal:
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$\mathcal{L}\{y''+y\}=\mathcal{L}\{t^2+2\}$
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applying linearity:
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$\mathcal{L}\{y''\}+\mathcal{L}\{y\}=\mathcal{L}\{t^2\}+\mathcal{L}\{2\}$
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using properties:
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$s^2Y(s)-s+1+Y(s)=\frac{2!}{s^3}+\frac{2}{s}$
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factor out Y(s)
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$Y(s)(s^2+1)=\frac{2(s^2+1)}{s^3}+s-1$
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$Y(s)=\frac{2}{s^3}+\frac{s}{s^2+1}-\frac{1}{s^2+1}$
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now we take the inverse LT to obtain y(s)
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$$y(t)=t^2+\cos(t)-\sin(t)$$
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Done!
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second example:
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#ex #LT #second_order_nonhomogenous
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$$y''-y=(t-2)e^{t-2} \qquad y(2)=0 \qquad y'(2)=0$$
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we need y at 0 but its as 2.
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so we make a substitution: $x=t-2$
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$\frac{d^2y}{dx^2}-y=xe^x$ <- notice the y here is not the same as the y above, lousy notation again :)
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where $y(0)=0 \qquad \frac{dy}{dx}(0)=0$
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hit it with the LT!
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notice 1/s^2 is LT of x so we have to shift it
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$s^2Y(s)-Y(s) =\frac{1}{(s-1)^2}$
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$Y(s)=\frac{1}{(s-1)^3(s+1)}=\frac{A}{s-1}+\frac{B}{(s-1)^2}+\frac{C}{(s-1)^3}+\frac{D}{s+1}$
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$\frac{{A(s-1)^2(s+1)+B(s-1)(s+1)+C(s+1)+D(s-1)^3}}{(s-1)^3(s+1)}$
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$\begin{matrix}A+D=0 \\-A-B-3D=0 \\ -A+c+3D=0 \\ A-B+C-D=1\end{matrix}$
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$A=\frac{1}{8}$
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$B=-\frac{1}{4}$
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$C=\frac{1}{2}$
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$D=-\frac{1}{8}$
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plug into expression then take inv LT to obtain y(t):
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final solution: $y(x)=\frac{1}{8}e^x-\frac{1}{4}xe^x+\frac{1}{4}x^2e^x-\frac{1}{8}e^-x$
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where x=t-2
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all done!
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#end of lec 17 #start of lec 18
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#ex
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$$y''+ty'-2y=2 \qquad y(0)=y'(0)=0$$
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hit it with the LT!
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$\mathcal{L}\{y''\}+\mathcal{L}\{ty'\}-2\mathcal{L}\{y\}=\frac{2}{s}$
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$s^2Y(s)-\frac{d}{ds}\mathcal{L}\{y'\}-2Y(s)=\frac{2}{s}$
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$s^2Y-\frac{d}{ds}(sY(s))-2Y=\frac{2}{s}$
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apply product rule:
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$s^2Y-Y-s\frac{dY}{ds}=\frac{2}{s}$
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^ Boooo! another differential equation! :(
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$\frac{dY}{ds}$ lies in the s "phase space"
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$-s\frac{dY}{ds}+s\left( s-\frac{3}{5} \right)=\frac{2}{s}$
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This is a linear equation!
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divide by -s to get it in standard form
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$\frac{dY}{ds}-\left( s-\frac{3}{5} \right)y=-\frac{2}{s^2}$
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compute integrating factor:
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$\mu(s)=e^{-\int (s-3/s) \, ds}=e^{-s^2/2}e^{\ln{s^3}}=s^3e^{-s^2/2}$
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$\frac{d}{ds}(s^3e^{-s^2/2}Y)=-2se^{-s^2/2}$
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$s^3e^{-s^2/2}Y=-2\int se^{-s^2/2} \, ds$
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use u sub.
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$u=\frac{s^2}{2}$
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$-2\int e^{-u} \, du$
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$=2e^{-s^2/2}+C$
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$Y(s)=2s^3+C \frac{e^{s^2/2}}{s^3}$
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is this even a legitimate thing to take an inverse of?
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the lim of the expression approaches inf as s approaches inf
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So what do we do? Well we have that C term. We have to set $C=0$
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then:
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$Y(s)=\frac{2}{s^3}$
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$$y(t)=t^2$$
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we just solved a new equation that hasn't fit into our previous equation types using LT. How cool is that!
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