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

2022
Views:

359

Pages:

2
Author:

mylovaboyj9c

Selected Problems from Section 3.7 1. Problem 3: y 00 + 2y 0 + y = 3e−t Using Variation of Parameters: y1 = e−t y2 = te−t g(t) = 3e−t W = e−2t so −3te−t e−t 3 = −3t ⇒ u1 (t) = − t2 −2t e 2 −t −t e 3e u02 = = 3 ⇒ u2 (t) = 3t e−2t Therefore, the particular solution is yp = u1 y1 + u2 y2 : u01 = 3 3 yp (t) = − t2 e−t + 3t2 e−t = t2 e−t 2 2 Using the method of undetermined coefficients, we would have initially guessed that yp was a constant times the exponential function, but we have to multiply by t2 . Do that, and differentiate to substitute back into the ODE: yp = At2 e−t yp0 = A(2t − t2 )e−t yp00 = A(2 − 4t + t2 )e−t We end up with A = 23 , which is what we had using Variation of Parameters. 2. Problem 5: y 00 + y = tan(t) y1 = cos(t) y2 = sin(t) g(t) = tan(t) W = 1 Therefore, we get an integral that can be a bit tricky, but we can simplify the integrand first: sin2 (t) −1 + cos2 (t) = = − sec(t) + cos(t) u01 = − sin(t) tan(t) = − cos(t) cos(t) so that u1 = − ln | sec(t) + tan(t)| + sin(t). u02 = cos(t) tan(t) ⇒ u2 = Z sin(t) dt = − cos(t) Therefore, yp = (− ln | sec(t) + tan(t)| + sin(t)) cos(t) − cos(t) sin(t) = − cos(t) ln | sec(t) + tan(t)| 3. Problem 15: We won’t verify that y1 , y2 are solutions, but you should. Assuming that they are (remember to put the equation in standard form by dividing by t): y 1 = 1 + t y2 = e t −et te2t = −e2t tet For u2 , we’ll integrate by parts: u01 = u02 = g(t) = te2t ⇒ (1 + t)te2t = (1 + t)et t te 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 4. Problem 17 comes out nicely as well- Be sure to get standard form before going too far. Use u = ln(x) for the integrals.

Differential Equations Selected Problems from Section 3.7

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