EXAM 2 MARCH 17, 2004

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Gabriel Douglas
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1 8.034 EXAM MARCH 7, 004 Name: Problem : /30 Problem : /0 Problem 3: /5 Problem 4: /5 Total: /00 Instructions: Please write your name at the top of every page of the exam. The exam is closed book, closed notes, and calculators are not allowed. You will have approximately 50 minutes for this exam. The point value of each problem is written next to the problem use your time wisely. Please show all work, unless instructed otherwise. Partial credit will be given only for work shown. You may use either pencil or ink. If you have a question, need extra paper, need to use the restroom, etc., raise your hand. Date : Spring 004.

2 Name: Problem : /30 Problem (30 points) A driven, damped harmonic oscillator satisfies the following ODE, where b, and are positive real numbers. y + by + y = cos(t), (a)(0 points) Using the method of undetermined coefficients, find a solution of the y(t) form = C cos(t) + C sin(t). Solution A particular solution is the real part of the complex valued solution of, y + bỹ + ỹ = e it. The complex number i is not a root of the characteristic polynomial. Therefore, we guess that y = e it A, for some complex number A. Substituting in gives, it. e it ( + ib + )A = e Therefore A = i/b. This gives, y(t) = sin(t) i cos(t). b b So a particular solution is the real part, y d (t) = sin(t). b (b)(0 points) Let R be a positive real number, R /(b) (this guarantees that y d (t) = R for some t > 0). There is a multi valued function T = T () for the set of positive numbers where y(t ) = R. This can be made into a single valued function T th n by specifying that T n is the n smallest positive number such that y(t ) = R. So T is the smallest positive number such that y(t ) = R, T is the smallest positive number greater than T such that y(t ) = R, etc. or at least one choice n of > 0, find a formula for T n (). Solution The solution below is only valid if is strictly less than /(br). The functions T n, n, are discontinuous at = /(br). Therefore, assume that < 0 < /(br). The positive number T equals T n for some n if sin(t ) = R. b Solving for T gives, T () = sin (br/ ). The different values T, T, etc. correspond to the different positive branches of sin (br/ ). In particular, defining sin (θ) to be the usual branch, T () = sin (br/ ), 0 < T () < π, the formula for T n is, T n () = ((n )π + sin (br/ )), ((n )π sin (br/ )), if n is odd if n is even

3 (c)(0 points) Suppose that b, and R are fixed, but is allowed to vary, 0 <. br Write T = T n for some positive integer n. Let be a critical point of T (). Prove there is an equation, T () = α, where α is a real number independent b, of and R. Moreover, find an equation that α satisfies. (Remark You will see there are many choices α. for You are not responsible for matching choices of α to choices of n; just write down an equation that α satisfies). Solution Implicitly differentiating the following relation with respect, to br sin(t ) =, gives the following, ( cos(t )(T + T ) sin(t )) = 0. Equivalently, this is, T + T = tan(t). At a critical point, T () = 0, giving the equation, T = tan(t ). So T () = α, where α is a positive solution of the equation, α = tan(α). Observe that the corresponding values of and T are, br = sin(α), T = α csc(α). br Therefore, to be perfectly accurate, T = α only gives a critical point Tof n if n = m + is odd, in which case the corresponding value α of is the unique solution, ( ) α = tan(α), mπ < α m + π. (But this is more detail than you were asked to give). Remark I was asked why the functions T n do not have a critical point n if is even. The answer is related to the discontinuity of T n at /(br). or n even, the function T n does have a continuous extension to the interval (0, br ] (although it does not agree with the definition given above). The only extremal point of T n is the global minimum of T n, which occurs at = /(br). Extra credit(5 points) Prove the solutions of your equation α for give local minima of T (). Solution The implicit differentiation above gives, T + T = tan(t). Implicitly differentiating once again gives, T + T = ( sec (T )(T + T ) tan(t )) = ) sin(t ) cos(t )). cos ((T + T (T ) Plugging in T = α and T = 0 gives, T = 3 cos (α sin(α) cos(α)). (α)

4 Because α = tan(α), sin(α) cos(α) = α cos (α). So the equation simplifies to, T = α tan (α). In particular, this is positive. So gives a local minimum of T (). 3

5 Name: Problem : /0 Problem (0 points) In each case below, y (t), y (t) is a pair of solutions of a real, constant coefficient, linear homogeneous ODE in normal form. Determine the least degree of this ODE, and write down the ODE of this degree that the pair satisfies. :(Hint In each case, write down each nonzero solution as the real or imaginary part e λt of g(t) where g(t) is a polynomial. What does the degree of g(t), and the vanishing/nonvanishing of the imaginary part λ of tell you about the characteristic equation of the ODE? Remember, the ODE real is a ODE.) (a)(5 points)y t (t) = 0, y (t) = e. Solution The first equation, 0, is a solution of ANY homogeneous linear ODE. So this imposes no condition. The second condition imposes that is a root of the characteristic polynomial. Therefore the minimal degree is, and the corresponding ODE is, y y = 0. (b)(5 points) y (t) = e t, y (t) = e. Solution rom the two equations, both and are roots of the characteristic polynomial. Therefore the characteristic polynomial is divisible by z + ( )(z + ) = z + 3z +. So the minimal degree is, and the corresponding ODE is, y + 3y + y = 0. (c)(5 points) y t (t) = t, y (t) = e. Solution rom the first solution, 0 is a double root of the characteristic polynomial. rom the second solution, is a root of the characteristic polynomial. Therefore the characteristic polynomial is divisible by z (z ) = z 3 z. So the minimal degree is 3, and the corresponding ODE is, y y = 0. (d)(5 points) y (t) = sin(), y (t) = cos(3t). Solution rom the first solution, i is a root of the characteristic polynomial. Therefore the complex conjugate, i is also a root. rom the second solution, and 3i 3i are roots. Therefore the characteristic polynomial is divisible by (z + 4)(z + 9). So the minimal degree is 4, and the corresponding ODE is, y + 3y + 36y = 0.

6 Name: Problem 3: /5 Problem 3(5 points) or a certain linear ODE in normal Ly form a basic solution set of Ly = 0 is given by, y (t) = e, y (t) = + +. (a)(0 points) Compute the Wronskian of this basic solution pair. Is your answer consistent with Abel s theorem? Solution The derivatives of the functions are, y (t) = e, y (t) = + +, y (t) = e, y (t) = 4t + Therefore the Wronskian is W [y, y ](t) = (4t + )e ( + + )e = 4t e. or t > 0, the Wronskian is nonzero. Therefore the computation of the Wronskian is consistent with Abel s theorem. (b)(5 points) Using the method of variation of parameters, find a particular solution of the inhomogeneous ODE, Ly = t e. Solution By the method of variation of parameters, a particular solution is t y d (t) = K(s, t)f(s)ds, where K(s, t) is the Green s kernel, K(s, t) = (y (s)y (t) y (t)y (s)). W (s) In this case, y (s)y (t) y (t)y (s) = e s ( + + ) e (s + s + ). Multiplying the Green s kernel by s e s gives, K(s, t)f(s) = (e s ( + + ) e (s + s + )). 4 t 0 Antidifferentiating gives the particular solution, ( ( )) y d (t) = (e )( + + ) e t 3 + t + t. 4 3 Simplifying, this gives, y d (t) = t 3 e te + ( + + ) Of course the last term is a solution of the homogeneous equation. So a particular solution is, 3 y d (t) = (4t 3)e. 4

7 Name: Problem 4: /5 Problem 4(5 points) One solution of the linear ODE, is the equation y (t) = e. y + a(t)y + b(t)y = ( ) y + t y 4 + t y = 0, (a)(0 points) Let y (t) be a second solution. Define W (t) to be the Wronskian of y (t), y (t). This satisfies the differential equation W = a(t)w. ind a solution W (t) of this differential equation. Solution The differential equation is separable, ( ) dw = + dt. W t So, or equivalently, Taking A = 4 gives, ln(w ) = + ln(t) + C, W (t) = At e. W (t) = 4t e. (b)(5 points) or your solution W (t), solve the first order linear ODE, y (t)v y (t)v = W (t). Plug in y (t) = v to find a basic solution pair of the ODE, Ly = 0. Solution The ODE is, e v e v = 4t e. Simplifying, this is, v v = 4t. Because 0 is not a root of the characteristic polynomial, by the method of undetermined coefficients we guess, v(t) = a t + a t + a 0. Substituting in gives, a t + (a a )t + (a a 0 ) = 4t. Therefore a =, a = a =, and a 0 = a / =. So the solution is, v(t) = + +. So, by Wronskian reduction of order, a basic solution pair of the ODE is, y (t) = e, y (t) = + +. This is the same solution pair as in the previous problem! In particular, now that we know the operator Ly, it is straightforward to check that the particular solution y d (t) of Ly = t e actually does give a solution.

Non-homogeneous equations (Sect. 3.6).

Non-homogeneous equations (Sect. 3.6). We study: y + p(t) y + q(t) y = f (t). Method of variation of parameters. Using the method in an example. The proof of the variation of parameter method. Using the