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PSU MATH 251 - LECTURE NOTES

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Math 251 March 31, 2005 Second ExamANS KEY1. In Parts a and b determine the form of a particular solution yp= yp(t) having the leastnumber of unknown constants. DO NOT DETERMINE the unknown constantsappearing in your answers in Parts a and b.a. 2pt y00− 14y0+ 49y = 2t2e7tANS. t2(At2+ Bt + C)e7t7 is a double root of the characteristic polynomial.Take off 1pt more missing factor of t2.Take off 2pts for any other error.b. 2pt y00− 50y0+ 49y = 3tetANS. t(At + B)et1 is a simple root of the characteristic polynomial.Take off 1pt more missing factor of t.Take off 2pts for any other error.c. 7pt Without using Laplace transforms, find a particular solution to the followingODE:y00+ 3y = etsin 2t(In this part you need to determine the unknown constant(s) in the solution.ANS. Solve the complexified equation and then take imaginary party00+ 3y = e(1+2i)tPlug in yC= Ae(1+2i)tsince 1 + 2i is not a root of the characteristic polynomial. (2pts)y0C= A(1 + 2i)e(1+2i)tand y0C= A(−3 + 4i)e(1+2i)t.(1pt)Therefore, L[yC] = A(3 + (−3 + 4i))e(1+2i)t. (2pt)We conclude that A = 1/(4i) = −i/4 and hence the imaginary part of yCis yp=(−1/4) cos(2t) (2pt)Alternatively,plug in yp= et(A cos(2t) + B sin(2t)). (2pt)Taking the derivative: y0p= et((A + 2B) cos(2t) + (B − 2A) sin(2t)). (1pt)and again: y00p= et((−3A + 4B) cos(2t) + (−3B − 4A) sin(2t)). (2pt)Therefore, L[yp] = et(4B cos(2t) − 4A sin(2t)). (1pt)We conclude that 4B = 0 and −4A = 1, ie, yp= (−1/4) cos(2t). (1pt)2. Assume that acceleration due to gravity g is equal to 10 meter/sec2. An object withmass 2 kg stretches a spring 2.5 meters to the equilibrium position. Assume that thereis no damping device attached and also assume that at time t = 0 the object is released1 meter below its equilibrium position with a upward velocity of 4 meter/sec.a. 3pt Write down a differential equation with initial conditions for y(t) for thedisplacement of the object below its equilibrium position.ANS. mg = kL Therefore k = 8 (1pt)2y00+ 8y = 0 (1pt)y(0) = 1 y0(0) = −4 (1pt)b. 4pt Find a formula for y(t)ANS. y = c1cos 2t + c2sin 2t (2pt)y0= −2c1sin 2t + 2c2cos 2t (1pt)c1= 1 and c2= −2 (1pt)c. 2pt Find the maximum value of y(t).ANS. R =qc21+ c22=√5 (2pt)d. 2pt If a periodic external force equal to 3 cos ωt Newtons is applied, then for whatpositive value of ω does resonance occur?ANS. ω = natural frequency = 2 (2pt)3. a. 4pts For a spring-mass system system with mass e qual to 1 kg, spring contantequal to 25 Newton-sec/meter, which damping constant γ causes critical damping?ANS. Critical damping =√4mk =q(4)(1)(25) = 10b. 3pts If the damping constant γ in the above system is set to 2 Newton-sec/meter,then what can be said about the number of times does the object pass through itsequilibrium position?ANS. Since 2 is less than critical damping, infinitely many times.c. 4pts If the damping constant γ in the above system is set to 8 Newton-sec/meter,then what is the interval of time between the second time the object returns to itsequilibrium position and the third time it returns to its equilibrium position?ANS. y00+ 8y0+ 25y = 0 has characteristic polynomial r2+ 8r + 25 which has rootsr = −4 ± 3i. (2pt)The object returns to equilibrium when c1cos 3t + c2sin 3t is zero. (1pt)This happens at intervals of c onstant length π/3. (1pt)4. a. 2pts What is the definition of the Laplace transform L{e3t}?ANS. L{e3t} =R∞0e−ste3tdt (2pt)No credit for any errors.b. 4pts Use the answer to Part a to calculate L{e3t}. (Be sure to explain why thisexists only when s > 3).ANS.R∞0e−ste3tdt =R∞0e−(s−3)tdt (1pt)R∞0e−(s−3)tdt = limA→∞−1s − 3e−(s−3)A− 1(2pt)The limit on the right hand side is−1(s − 3)whenever −(s − 3) is negative, or equiva-lently, when s > 3. (1pt)c. 2pts Suppose that the Laplace transform of y is Y . If y(0) = 2 and y0(0) = −3,then find the Laplace transform of y00.ANS. L{y0} = sY −2 (1pt)L{y00} = s(sY −2) −(−3) = s2Y −2s + 3 (1pt)d. 3pts Find the function f(t) whose Laplace transform is equal toss2+ 2s + 5ANS.ss2+ 2s + 5=s(s + 1)2+ 22(1pt)ss2+ 2s + 5=s + 1(s + 1)2+ 22−1(s + 1)2+ 22(1pt)= L{e−t(cos(2t) − (1/2) sin(2t))} (1pt)5. a. 5pt Find f(t) so that L{f(t)} =e−3s(s − 1)(s + 3)ANS.1(s − 1)(s + 3)=1/4s − 1−1/4s + 3(2pt)1/4s − 1−1/4s + 3= L{u(t)((1/4)et) − (1/4)e−3t} (1pt)e−3s(s − 1)(s + 3)= L{u(t − 3)((1/4)e(t−3)− (1/4)e−3(t−3))} (2pt)b. 6pt Assume that acceleration due to gravity g is equal to 10 meter/sec2. An objectwith mass 2 kg stretches a spring 4 m to equilibrium. At time t = 0 it is released 2meters below its equilibrium position with an upward velocity of 3 meters/sec . Attime t = 6 it is struck with a hammer and as a result its momentum is decreased by7 kg-meters/sec at that moment in time. At time t = 8 a constant external force of 9Newtons is added and at t = 10 it is removed. Write down a differential equation withinitial conditions for the displacement y(t) of the object below its equilibrium position.DO NOT SOLVE THIS EQUATIONANS. 2y00+ 5y = −7δ(t − 6) + 9(u(t − 8) − u(t − 10)) y(0) = 2, y(0) = −3(1pt) for the lhs of the ODE.(2pt) for each of the two terms on the rhs of the ODE.(1pt) for the initial conditions.6. a. 3pt Consider the functionf(t) =(t, if t < 22, if 2 ≤ tSketch a graph of this function and find a formula for f(t) in terms of unit step functionsu(t − c), for appropriate values of c. (Note that u(t − c) and uc(t) denote the samefunction.)ANS. f(t) = t((u(t) − u(t − 2)) + 2u(t − 2)(1pt) for the graph of f(t)(1pt) for each of the two terms in the above expression for f(t).b. 4pt Determine the Laplace transform of f(t) in Part aANS. Note that f(t) = tu(t) − (t − 2)u(t − 2)) (1pt)L{tu(t)} = 1/s2(1pt)L{(t − 2)u(t − 2)} = e−2s/s2(2pt)c. 4pt Find the Laplace transform of u(t − π) sin(t).ANS. Note that cos(t − π) = sin(t) (2pt)L{u(t − π) cos(t − π)} = e−πsss2+ 1(2pt)7. 11pt Solve the following IVP:y0+ 3y = δ(t − 1) + u(t − 2) y(0) = −4ANS.Take the Laplace transform of both sides: sY + 4 + 3Y = e−s+e−2ss(1pt) for the lhs(1pt) for Laplace of delta(1pt) for Laplace of unit step fcn.Hence, solving for Y Y =−4s + 3+ e−s1s + 3+ e−2s1(s + 3)s(1pt) for solving for Y1(s + 3)s=1/3s−1/3s + 3(1pt)Therefore,y(t) = −4e−3t+ u(t − 1)e−3(t−1)+131 − e−3(t−2)u(t − 2)(1pt) for the first term in y(t)(2pt) for the second term in y(t)(3pt) for the third


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PSU MATH 251 - LECTURE NOTES

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