University:
Whitman College
Course:
MATH-244 | Differential Equations
Academic year:

2022
Views:

313

Pages:

3
Author:

Mike Pollard

Selected Problems from Section 3.4 1. Problem 25. Let y 00 + 2y 0 + 6y = 0 with y(0) = 2 and y 0 (0) = α ≥ 0. 2 (a) Solve it: The √ characteristic equation is r + 2r + 6 = 0. Therefore, r = −1 ± 5 i and the general solution is: √ √ y(t) = e−t C1 cos( 5 t) + C2 sin( 5 t) √ , so that With the initial conditions, find that C1 = 2 and C2 = α+2 5 the solution to the IVP is: √ √ α+2 y(t) = e−t 2 cos( 5 t) + √ sin( 5 t) 5 (b) Find α so that y(1) = 0: Algebraically, we get: √ √ cos( 5) √ −2=α −2 5 sin( 5) (c) Find the smallest value of t > 0 so that y(t) = 0. We’ll write our answer as a function of α. Algebraically, solving this: √ √ α+2 −t 0=e 2 cos( 5 t) + √ sin( 5 t) 5 will get us to: √ −2 cos( 5 t) α+2 √ = √ sin( 5 t) 5 We want to solve this for t, remembering that we want the smallest t > 0. I think it is easiest to work with the tangent rather than the cotangent, √ √ −2 5 tan( 5 t) = α+2 √ Now consider the graph of tan( 5 t). It √ looks much like tangent, except that it is periodic with period π/ 5. √ If we consider the graph of tan( 5 t), as shown in Maple in Figure 1, we see that solving directly for t: √ ! 1 −2 5 −1 t = √ tan α+2 5 √ will put us one period (π/ 5) short. Therefore, the full solution is: √ ! 1 −2 5 π −1 t = √ tan +√ α+2 5 5 1 √ √ Figure 1: Plot of tan( 5t) and a sample value of −2 5/(α + 2) marked as A. Using the standard restriction (so that the tangent is invertible), we see that we are one period too short- Add one period to get the actual solution. √ Taking the limit as α → ∞, we get that t = π/ 5. NOTE: The inverse tangent is an odd function, so one could write the solution as: √ ! π 1 2 5 −1 t = √ − √ tan α+2 5 5 2. Problem 29: Using Euler’s Formula and noting that e−iθ = cos(−θ) + i sin(−θ) = cos(θ) − i sin(θ) we can simplify the following: eit + e−it = cos(t) + i sin(t) + cos(t) − i sin(t) = 2 cos(t) and eit − e−it = cos(t) + i sin(t) − cos(t) + i sin(t) = 2i sin(t) 3. Problem 32: Differentiate expressions in i as if i is a constant (it is a constant): If φ = u + iv, then φ0 = u0 + iv 0 and φ00 = u00 + iv 00 . Therefore, substituting this into the differential equation, we get: (u00 + iv 00 ) + p(t)(u0 + iv 0 ) + q(t)(u + iv) = 0 2 Separate into real and imaginary components: (u00 + p(t)u0 + q(t)u) + i(v 00 + p(t)v 0 + q(t)v) = 0 We note that if a + ib = 0, then a = 0 and b = 0. Apply this principle to the above equation so that u00 + p(t)u0 + q(t)u = 0 and v 00 + p(t)v 0 + q(t)v = 0 Therefore, u(t) and v(t) must separately solve the homogeneous second order linear differential equation. 3

Selected Problems from Section 3.4 Answer Key

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