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MATH 54 − HINTS TO HOMEWORK 11 PEYAM TABRIZIAN Here are a couple of hints to Homework 11. Make sure to attempt the problems before you check out those hints. Enjoy! S ECTION 4.6: VARIATION OF PARAMETERS The easiest way to do the problems in this section is to look at my differential equations handout! The formula is: Let y1 (t) and y2 (t) be the solutions to the homogeneous equation, and suppose yp (t) = v1 (t)y1 (t) + v2 (t)y2 (t). Let: y (t) y (t) 1 2 f (t) = 0 W y1 (t) y20 (t) And solve: 0 0 v1 (t) f = W (t) 0 v2 (t) f (t) where f (t) is the inhomogeneous term. S ECTION 6.1: BASIC THEORY OF LINEAR DIFFERENTIAL EQUATIONS 6.1.2, 6.1.4. First of all, make sure that the coefficient of y 000 is equal to 1. Then look at the domain of each term, including the inhomogeneous term (more precisely, the part of the domain which contains the initial condition). Then the answer is just the intersection of the domains you found! 6.1.10. Use the Wronskian with x = 0 6.1.12. cos2 (x) − sin2 (x) = cos(2x) 6.1.17. Verify that the three functions solve the differential equations, then show they’re linearly independent (by using the Wronskian at x = 1) Date: Tuesday, April 14th, 2015. 1 2 PEYAM TABRIZIAN 6.1.21. Use the Wronskian to check that the functions are linearly independent (try to simplify first). Then the general solution is y(t) = Ax + Bx ln(x) + Cx (ln(x))2 + ln(x), and then plug in the initial condition. 6.1.34. It’s third-order because it involves y 000 . In fact the coefficient of y 000 is 1, by expanding the determinant along the fourth row. What happens when you plug in y(t) = f1 (t), is the determinant in fact 0 or not? Is f1 (t) a solution or not? S ECTION 6.2: H OMOGENEOUS LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 6.2.3, 6.2.9, 6.2.19, 6.2.21. The following fact might be useful: Rational roots theorem: If a polynomial p has a zero of the form r = ab , then a divides the constant term of p and b divides the leading coefficient of p. This helps you ‘guess’ a zero of p. Then use long division to factor out p. 6.2.14. Since sin(3x) is a solution, this means that one of the roots of the auxiliary equation is r = ±3i, which tells you that the auxiliary equation has the form (r2 + 9)× something. Use long division to figure out what that something is, by dividing r4 + 2r3 + 10r2 + 18r + 9 by r2 + 9. 6.2.17. The reason this is written out in such a weird way is because the auxiliary polynomial is easy to figure out! Here, the auxiliary polynomial is (r + 4)(r − 3)(r + 2)3 (r2 + 4r + 5)2 r5 = 0. S ECTION 9.1: I NTRODUCTION k 9.1.7. Let z = y 0 , then z 0 = y 00 = − mb y 0 − m y = − mb z − ( y0 = z k y − mb z z0 = − m k y, m then we get: MATH 54 − HINTS TO HOMEWORK 11 3 9.1.11, 9.1.12. For 9.1.11, let: x1 = x x = x 0 = x 0 2 1 x = y 3 x4 = y 0 = x03 Now calculate x0i and put them in a system, using the above relations. In particular x02 = x00 = −3x − 3y = −3x1 − 2x3 .