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

2021
Views:

439

Pages:

2
Author:

Janiya Drake

Selected Solutions: 3.2 1. Problem 16, 3.2: We are told that y = sin(t2 ) is a solution- Substitute it into the DE to determine p(t) and q(t): y 0 = 2t cos(t2 ) y = sin(t2 ) y 00 = 2 cos(t2 ) − 4t2 sin(t2 ) Substituting into y 00 + p(t)y 0 + q(t)y = 0, we get: y 00 p(t)y 0 q(t)y 0 = 2 cos(t2 ) −4t2 sin(t2 ) 2 = p(t)2t cos(t ) = q(t) sin(t2 ) = (2 + 2tp(t)) cos(t2 ) +(q(t) − 4t2 ) sin(t2 ) Therefore, if sin(t2 ) was a solution, p(t) = make p(t) not continuous at t = 0. −1 t and q(t) = 4t2 . However, this would 2. Problem 18, Sect 3.2: We’re given the Wronskian and a function f (t) = t. Find the function g: W (f, g)(t) = t g(t) = tg 0 (t) − g(t) = t2 et 1 g 0 (t) That is a linear differential equation in g: 1 g 0 (t) − g(t) = tet t The integrating factor: e− ( g(t) 0 ) = et t R (1/t) dt ⇒ = 1t , so: g(t)/t = et + C ⇒ g(t) = t(et + C) 3. Problem 22, Section 3.2: Find the fundamental set (Theorem 3.2.5) if y 00 +4y 0 +3y = 0. We know the general solution is C1 e−t + C2 e−3t . In Theorem 3.2.5, we construct two solutions that are guaranteed to form a fundamental set: • y1 solves the ODE with initial conditions y(1) = 1, y 0 (1) = 0: 3 1 C1 e−1 + C2 e−3 = 1 ⇒ C1 = e, C2 = − e3 −1 −3 −C1 e − 3C2 e =0 2 2 Therefore, y1 (t) = 3e −t e 2 − e3 −3t e . 2 • Similarly, y2 solves the ODE with I.C.s: y(1) = 0, y 0 (1) = 1: e e3 C1 e−t + C2 e−3 = 0 ⇒ C1 = , C2 = − −1 −3 −C1 e − 3C2 e =1 2 2 Therefore, y2 (t) = 2e e−t − e3 −3t e 2 Even though y1 and y2 look a lot alike, Theorem 3.2.5 guarantees that they are linearly independent, and that they form a fundamental set. Note that Theorem 3.2.5 is more of a formal result than something we would actually compute with- However, it does give conditions on which we can always guarantee that we can find a fundamental set of solutions. 4. Problem 26, Section 3.2: The verification is straightforward. Before we consider the question of whether we have a fundamental set, look at where the solutions would be valid. We will have a discontinuity of p, q, g at the x−values where: 1 − x cot(x) = 0 ⇒ x cot(x) = 1 ⇒ x cos(x) =1 sin(x) ⇒ x cos(x) = sin(x) Therefore, existence and uniqueness is only guaranteed on intervals which avoid these points. Going to the original question, does x and sin(x) constitute a fundamental set? Look at the Wronskian: x sin(x) = x cos(x) − sin(x) W (x, sin(x)) = 1 cos(x) This looks very familiar! The Wronskian will be non-zero for intervals which also satisfy the existence and uniqueness theorem. Finally, is the given interval one such example? The expression will not be zero on 0 < x < π (check this graphically). 5. Problem 27, Section 3.2: Just a couple of notes here. You should find that y1 , y3 do form a fundamental set; y2 , y3 do NOT form a fundamental set. To show that y1 , y4 do form a fundamental set, notice that, since y1 , y2 do form a fundamental set, y1 y20 − y10 y2 6= 0 at t0 Now form the Wronskian between y1 and y4 : W (y1 , y4 ) = y1 y1 + 2y2 y10 y10 + 2y20 = y1 y10 + 2y1 y20 − y1 y10 − 2y10 y2 = 2W (y1 , y2 ) 6= 0 The last set, y4 , y5 does NOT form a fundamental set. You can show that y4 = y1 +2y2 , and y5 = 2y4 .

Differential Equations Selected Solutions 3.2

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