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Exam II Review Solutions 1 Qualitative Behavior Consider the following 2nd order ODE’s (a) y 00 + 2y 0 + y = 0 Solution: The characteristic polynomial is r2 + 2r + 1 = 0, which has the root r = −1 with multiplicity 2. Thus, this is an example of free (since the equation is unforced) critically damped motion. Figure 1: A typical critically damped solution. (b) 3y 00 + √ 2y 0 + y = 0 √ 2 + 2r + 1 = 0, which Solution: The characteristic polynomial is 3r √ √ has the roots r = −6 2 ± 610 i. Thus, this is an example of free (since the equation is unforced) under-damped motion. 1 Figure 2: A typical under-damped solution. (c) y 00 + 1.1y = cos(t) Solution: Since the frequency of the homogeneous solution is approximately equal to the frequency of the forcing term, the solution to the equation will exhibit beats. Figure 3: A typical beats solution. (d) y 00 + 4y = sin(2t + π/2) Solution: Since the frequency of the homogeneous solution is equal to the frequency of the forcing term, the solution to the equation will 2 exhibit resonance. Figure 4: A typical resonance solution. Without solving, do the following for each equation given above: name the phenomena exhibited by the equation and give justification. Finally, sketch the qualitative behavior of the solution. 2 Solution Techniques For each of the equations listed below, state which method is the most appropriate solution technique (give justification). Then, solve the equation. (a) y 00 − y 0 − 2y = et , y(0) = y 0 (0) = 1 This is a linear second-order equation with constant coefficients for which there is a standard guess for the particular solution (see table 4.4.1 in the text). Thus, we may use Undetermined Coefficients. Solution: y(t) = e−t et − + e2t 2 2 (b) y 00 + y = sec2 (x) Hint: R sec(x)dx = ln | sec(x) + tan(x)| + c This is a linear second-order equation with constant coefficients, but with a forcing term for which we do not have a standard guess for the 3 particular solution (see table 4.4.1 in the text). Thus, we should try Variation of Parameters. Solution: y(t) = c1 cos(x) + c2 sin(x) − 1 + sin(x) ln | sec(x) + tan(x)| (c) y 00 + y = 1 + U t − π 2 [sin(t) − 1] , y(0) = 1, y 0 (0) = 0 This is a linear second-order equation with constant coefficients, but with a discontinuous forcing term. Thus, we should try Laplace Transform. Solution: y(t) = 1 − U(t − π2 ) 1 − sin(t) − 3 1 2 t− π 2 cos(t) Spring-Mass System A force of 400 Newtons stretches a spring 2 meters. A mass of 50 kilograms is attached to the end of the spring and it is initially released from the equilibrium position with an upward velocity of 10 m/s. Additionally, the mass is driven by an external force equal to f (t) = sin(2t). (a) Write the corresponding IVP. First, we must find the spring constant. From Newton’s laws we have, F = kl, thus k = 200N/m. This yields thw following IVP 50y 00 + 200y = sin(2t); y(0) = 0, y 0 (0) = −10 (b) Solve the IVP using any appropriate method. Name the behavior of the solution. Solution: y(t) = 4 −1 400 (1999 sin(2t) + 2t cos(2t)) ; Resonance Laplace Transform (a) For the following equation, find the Laplace transform using only the definition: f (t) = et cos(t) 4 Solution: F (s) s−1 (s − 1)2 − 1 (b) Find the Laplace transforms of the following functions using the table: (i) g(t) = (t + 1)3 Solution: G(s) = (ii) h(t) = 0 t < π/2 cos(3(t − π/2)) t ≥ π/2 Solution: H(s) = 5 6 + 6s + 3s2 + s3 s4 se−πs/2 s2 + 9 Inverse Laplace Transform Find the inverse Laplace transform of the following functions: (a) Y (s) = 2s+5 s2 +6s+34 Solution: y(t) = 15 e−3t (10 cos(5t) − sin(5t)) (b) Q(s) = e−s (2s−1) s2 (s+1)2 Solution: q(t) = U(t − 1) 4 − (t − 1) − 4e−(t−1) − 3(t − 1)e−(t−1) 6 One More Thing There are a few topics that are not included on this review, but that doesn’t mean they won’t be on the exam. You are still responsible for all material covered in lecture! 5