286practicefinalsolns - 286-E1 Spring 2010Practice Final...

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Unformatted text preview: 286-E1, Spring 2010Practice Final Exam - Solutions11 (a) Find alla∈Rso that the solution of the initial value problemx=x4-1, x(0) =asatisfieslimt→∞x(t) =-1.Solution.Sincex4-1 = (x-1)(x+ 1)(x2+ 1), the autonomous equationx=x4-1 has equilib-rium solutionsx(t) =-1 andx(t) = 1.x4-1 is negative only on (-1,1), so we conclude thatx(t) =-1is a stable equilibrium, whilex(t) = 1 is unstable. (Sketch the phase diagram!) Therefore we have thatlimt→∞x(t) =-1 precisely whenx(0) =a∈(-∞,1).(b) Classify the type of origin as a critical point of the system:ddt~x=-kk~xwherek6= 0is an arbitrary nonzero real number. Describe geometrically the trajectories ofthis system.Solution.The characteristic polynomial is:p(λ) = det-λ-kk-λ=λ2+k2= (λ+ik)(λ-ik)Since the egeinvaluesλ=±ikare purely imaginary, (0,0) is a center. Solution trajectories are circlescentered at the origin, since for (x1(t),x2(t)) satisfying the system we get:ddtx1(t)x2(t)=-kx2(t)kx1(t)⇒ddt(x21+x22) = 2x1x1+ 2x2x2=-2kx1x2+ 2kx1x2= 02. Find the general solution of the equation:(x2+ 1)dydx+xy=xSolution.The equation is linear, so we use the integrating factor method.dydx+xx2+ 1y=xx2+ 1r(x) = expZxx2+ 1dx= exp12Zduu= exp12ln(x2+ 1)= (x2+ 1)12ddx(x2+ 1)12y=x(x2+ 1)12(x2+ 1)12y=Zx(x2+ 1)12dx(x2+ 1)12y=Zdv2v12=v12+C(x2+ 1)12y= (x2+ 1)12+Cy= 1 +C(x2+ 1)-12286-E1, Spring 2010Practice Final Exam - Solutions23. Find the general solution of the equation:(2xsinycosy)y= 4x2+ sin2ySolution.After substitutionv= siny, the equation becomes homogeneous:2xvdvdx= 4x2+v2dvdx=4x2+v22xv= 2xv+12vxu+xdudx=2u+u2whereu=vx=sinyxxdudx=2u-u22udu4-u2=dxx-ln|4-u2|= ln|x|+C4-u2=Kx4-sin2yx2=Kx4. Compute the Fourier series of the4π-periodic extension of the function:f(t) =(,≤t≤2π2,-2π≤t≤Solution.Fourier series with half-periodL= 2πare of the form:f(t) =a2+∞Xn=1ancosnt2+bnsinnt2a=12πZ2π-2πf(t)dt=12πZ-2π2dt= 2an=12πZ2π-2πf(t)cosnt2dt=12πZ-2π2cosnt2dt=2nπsinnt2-2π= 0bn=12πZ2π-2πf(t)sinnt2dt=12πZ-2π2sinnt2dt=-2nπcosnt2-2π=-2nπ(1-(-1)n)f(t) = 1 +∞Xn=1-2nπ(1-(-1)n)cosnt2286-E1, Spring 2010Practice Final Exam - Solutions35. Suppose that a massm= 2kgis attached to a spring with spring constantk= 32N/mwhose other end is fixed to a wall.(a) Find the value of damping coefficientcfor which this system is critically damped. Whatis the general solution for the position functionx(t)of the mass at timetin this case?Solution.We want to findcso that the characteristic polynomial 2r2+cr+ 32 of the equation2x00+cx+ 32x= 0 has a repeated root. The condition for that isc2= 4·2·32 = 162soc= 16corresponds to the critically damped case. In this situation, the characteristic polynomial:2r2+ 16r+ 32 = 2(r2+ 8r+ 16) = 2(r+ 4)2has a double rootr=-4 , so the general solution isx(t) =e-4t(c1t+c2)....
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This note was uploaded on 08/25/2010 for the course ECE 210 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.

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286practicefinalsolns - 286-E1 Spring 2010Practice Final...

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