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Math 54. Selected Solutions for Week 11 Section 4.6 (Page 435) 7. Find a general solution to the differential equation y 00 + 4y 0 + 4y = e−2t ln t using the method of variation of parameters. The characteristic polynomial is r2 + 4r + 4 = (r + 2)2 , so a fundamental solution set is y1 = e−2t , y2 = te−2t . One therefore needs to solve the system e−2t v10 + te−2t v20 = 0 ; −2e−2t v10 + (1 − 2t)e−2t v20 = e−2t ln t . Adding two times the first equation to the second gives e−2t v20 = e−2t ln t , so v20 = ln t , and (from the first equation) v10 = −t ln t . Integrating gives v1 = − t2 t2 ln t + 2 4 and v2 = t ln t − t . Therefore t2 t2 e−2t + (t ln t − t)te−2t yp = − ln t + 2 4 ln t 3 2 −2t = − t e . 2 4 The general solution is therefore ln t 3 2 −2t y= − t e + c1 e−2t + c2 te−2t . 2 4 Section 6.1 (Page 482) 2. Determine the largest interval (a, b) for which Theorem 1 guarantees the existence of a unique solution on (a, b) to the given initial value problem: √ y(π) = 0 , y 0 (π) = 11 , y 00 (π) = 3 . y 000 − xy = sin x ; Since √ x is only defined for x ≥ 0 , the largest possible interval is (0, ∞) . 1 2 4. Determine the largest interval (a, b) for which Theorem 1 guarantees the existence of a unique solution on (a, b) to the given initial value problem: x(x + 1)y 000 − 3xy 0 + y = 0 ; y(−1/2) = 1 , y 0 (−1/2) = y 00 (−1/2) = 0 . We have to divide the equation by x(x + 1) to get the coefficient of y 000 to equal 1 (which the theorem requires). This function is zero at x = 0 and x = −1 (and nowhere else), so the largest interval is (−1, 0) . 10. Determine whether the given functions are linearly dependent or linearly independent on the specified interval. Justify your decision. {sin x, cos x, tan x} on (−π/2, π/2) . They are linearly independent. To show this, we use the method of Example 3. Suppose that c1 sin x + c2 cos x + c3 tan x = 0 for all x ∈ (−π/2, π/2) . Plugging in x = 0 gives c2 = 0 , leaving c1 sin x+c3 tan x = 0 . Plugging in two other values gives 1/2 √ 1/ 2 are linearly independent. Since the matrix 12. c3 c1 +√ =0 (x = π/3) ; 2 3 c √1 + c3 = 0 (x = π/4) . 2 √ 1/ 3 is invertible, this forces c1 = c3 = 0 , so the functions 1 Determine whether the given functions are linearly dependent or linearly independent on the specified interval. Justify your decision. {cos 2x, cos2 x, sin2 x} on (−∞, ∞) . We have cos 2x = cos2 x − sin2 x on (−∞, ∞) , so the functions are linearly dependent. 34. Constructing Differential Equations. Given three functions f1 (x) , f2 (x) , f3 (x) that are each three times differentiable and whose Wronskian is never zero on (a, b) , show that the equation y f1 (x) f2 (x) f3 (x) 0 0 0 0 f1 (x) f2 (x) f3 (x) y 00 =0 f1 (x) f200 (x) f300 (x) y 00 000 f1 (x) f2000 (x) f3000 (x) y 000 3 is a third-order linear differential equation for which {f1 , f2 , f3 } is a fundamental solution set. What is the coefficient of y 000 in this equation? The function y = f1 satisfies the differential equation because the first and fourth columns of the matrix are equal, so the determinant is zero. Similarly, y = f2 and y = f3 are also solutions. This equation is a linear differential equation because you can expand about the last column to get an expression C44 y 000 + C34 y 00 + C24 y 0 + C14 y = 0 , where the cofactors Ci4 are functions of x not involving y . The coefficient of y 000 is just the Wronskian of f1 , f2 , f3 , and we are given that it is never zero, so the equation must be of third order (the term C44 y 000 does not disappear), and we can divide by the Wronskian and apply Theorem 3 to find that f1 , f2 , and f3 form a fundamental set of solutions of the differential equation. Section 6.2 (Page 488) 14. Find a general solution for the differential equation y (4) + 2y 000 + 10y 00 + 18y 0 + 9y = 0 with x as the independent variable. [Hint: y(x) = sin 3x is a solution.] The auxiliary polynomial is r4 + 2r3 + 10r2 + 18r + 9 . We are given that sin 3x is a solution, which suggests that ±3i are roots. In fact, dividing by r2 + 9 works out, and we have the factorization (r2 + 9)(r2 + 2r + 1) = (r2 + 9)(r + 1)2 . Therefore the general solution is y = c1 sin 3x + c2 cos 3x + c3 e−x + c4 xe−x . Section 9.1 (Page 503) 4. Express the system of differential equations in matrix notation: x01 = x1 − x2 + x3 − x4 , x02 = x1 + x4 , √ x03 = πx1 − x3 , x04 = 0 . 0 x1 1 x 2 1 = √ x3 π x4 0 −1 0 0 0 1 0 −1 0 −1 x1 1 x2 . 0 x3 0 x4 4 12. Express the given system of higher-order differential equations as a matrix system in normal form: x00 + 3x0 − y 0 + 2y = 0 , y 00 + x0 + 3y 0 + y = 0 . Letting x1 = x , x2 = x0 , x3 = y , and x4 = y 0 gives 0 x1 0 x 2 0 = x3 0 x4 0 1 −3 0 −1 0 −2 0 −1 0 x1 1 x2 . 1 x3 −3 x4