Math 244 Selected Problems from Section 3.6 (Variation of Params)

Selected Problems from Section 3.6 (Variation of Params) 1. Solve: y 00 − 5y 0 + 6y = 2et NOTE: If we were not specifically told to use Variation of Parameters, we should choose Method of Undetermined Coefficients to solve this problem. For this exercise, we’re asked to do both. The homogeneous part has y1 = e2t and y2 = e3t for its fundamental set. Therefore, • Compute the Wronskian: W (e2t , e3t ) = e5t . Therefore, u01 = −2e4t = −2e−t 5t e u02 = 2e3t = 2e−2t e5t u1 = 2e−t ⇒ ⇒ u1 = −e−2t Therefore, yp = u1 y1 + u2 y2 = 2et − et = et • Verify the previous answer using Method of Undetermined Coefficients, with yp = Aet . Note that y = y 0 = y 00 , so substitution yields: et (A − 5A + 6A) = 2et ⇒ A=1 so yp (t) = et . 2. Same idea as Exercise 1. If we take y2 = e−t y1 = e2t Then: W (y1 , y2 ) = −3et , and 2 u01 = e−3t 3 ⇒ and 2 ⇒ 3 Substitution into yp = u1 y1 + u2 y2 gives u02 = − 2 u1 = − e−3t 9 2 u2 = − t 3 2 2 yp (t) = − e−t − te−t 9 3 Since e−t is part of yh , we can simplify yp : 2 yp (t) = − te−t 3 And, using Method of Undetermined Coefficients, we would initially guess yp = Ae−t . Since that is part of yh , we multiply by t so that yp = Ate−t And you should be able to verify that substitution into the DE yields the same answer as before. 6. The solution to the characteristic equation is r = ±3i, so the solution to the homogeneous part of the equation is yh = C1 cos(3t) + C2 sin(3t) The Wronskian is 3. Setting up the formulas and integrals, we get: Z u1 = −3 sin(3t) dt cos2 (3t) And use u, du substitution to get u1 (t) = −csc(3t). For the other function, u2 = 3 Z 1 dt cos(3t) The integral of the secant is a bit tricky, so I would not give you that integral on an in-class exam. The integral simplifies to: u2 = ln | sec(3t) + tan(3t)|. Finish up by constructing yp , then put everything together. 9. If we take y1 = cos(t/2) y2 = sin(t/2) then W = 1/2 and (writing the DE in standard form): g(t) = sec(t/2)/2. Setting up the integrals, we get: u1 = − Z sin(t/2) dt = 2 ln | cos(t/2)| cos(t/2) (which we integrated using u, du substitution with (u = cos(t/2))), and u2 = Z 1 dt = t Therefore, the particular part of the solution is: yp = 2 cos(t/2) ln | cos(t/2)| + t sin(t/2) To finish, write down yh + yp . 12. If we take y1 = cos(2t) and y2 = sin(2t), then W = 2, and u1 = − 1Z g(t) sin(2t) dt 2 1 u2 = g(t) cos(2t) dt 2 We can then write the particular part of the solution as u1 y1 + u2 y2 . The textbook uses a trig identity to get the answer in the back of the text (this simplification would not be required on an exam/quiz). 13. Taking p = 0, q = −2/t2 , and g(t) = 3 − 1/t2 , we get a Wronskian that is −3 which leads us to: Z −3t−1 + t−3 1 dt = ln |t| + t−2 u1 = −3 6 and Z 3t2 − 1 1 1 u2 = dt = − t3 + t −3 3 3 Then form yp = u1 y1 + u2 y2 . 15. We won’t verify that y1 , y2 are solutions to the homogeneous equation, but you should. Assuming that they are (remember to put the equation in standard form by dividing by t), we compute the Wronskian and use the formulas from Variation of Parameters: y1 = 1 + t y2 = et g(t) = te2t −et te2t = −e2t tet For u2 , we’ll integrate by parts: u01 = u02 = ⇒ (1 + t)te2t = (1 + t)et tet W = tet 1 u1 (t) = − e2t 2 ⇒ u2 (t) = tet The particular solution is u1 y1 + u2 y2 which simplifies to 1 yp = e2t (t − 1) 2 29. This problem is good for practice using Reduction of Order: The DE in standard form is: 2 2 y 00 − y 0 + 2 y = 4 t t so that y2 = tv(t) = tv, and substitution yields Equation (ii) in Exercise 28: tv 00 = 4 v 00 = 4 t ⇒ v 0 = 4 ln(t) + C1 Therefore, v(t) = 4 Z ln(t) dt + C1 t + C2 The integral of ln(t) is done by “integration by parts” with u = ln(t) and dv = dt (I would either not give you this integral or give you a hint if this were on an exam). Our final version of v is then: v = 4t ln(t) − 4t + C1 t + C2 Since y1 = t, we can take vy1 = 4t2 ln(t) + C3 t2 , so that y2 = t2 and yp = 4t2 ln(t).