MA 262 Exam 1 Fall 2011 Instructor: Raphael Hora Name: Max Possible Student ID#: 1234567890 1. No books or notes are allowed. 2. You CAN NOT USE calculators or any electronic devices. 3. Show all work to receive full credit. 4. Boa Sorte! (Good Luck in portuguese) Problem Max. Possible Points 1 10 10 2 15 15 3 10 10 4 10 10 5 10 10 6 10 10 7 10 10 8 10 10 9 15 15 Total 100 100 1
1. Find the general solution of (10 points) First divide both sides by x Now compute the integrating factor x dy dx 2y = x3 e x, x > 0 dy dx 2 x y = x2 e x I(x) = e 2 x dx = e 2 ln x = x 2 So, multiplying by the integrating factor d dx (x 2 y) = e x x 2 y = e x dx + C y = x 2 e x + Cx 2 2. Verify that y(x) = 1 solves dy dx = 2xy + 2 +2 and compute other solutions x 1 to this problem. Where are these solutions valid? (15 points) First, if y = 1, then dy dx = 0 and 2xy + 2 x 1 + 2 = 2x + 2 2(x 1) + 2 = + 2 = 2 + 2 = 0 x 1 x 1 Now dy dx = 2xy + 2 2xy + 2x + 2 = = x 1 x 1 This is a separable equation, so 1 y + 1 dy = 2x x 1 dx 2x(y + 1) x 1 Now you either integrate the left-hand side as ( 2x 2(x 1) x 1 dx = x 1 + 2 ) dx = 2x + 2 ln x 1 + C x 1 or make the substitution u = x 1, du = dx, x = u + 1. We have then ln y + 1 = 2x + ln(x 1) 2 + C Then exponentiating both sides and making e C = k, we get y + 1 = k(x 1) 2 e 2x 2xy + 2x Note that f(x, y) = and f x 1 y = 2x are both discontinuous at x 1 x = 1, so the soltuions above are unique given an initial condition y(b) = a, a, b R and b 1. 2
3. Solve the following initial value problem: (10 points) dy dx = 4x y x + 2y, y(0) = 1 You can either rewrite this differential equation as (x + 2y)dy + (y 4x)dx = 0 and see that this is an exact differential equation, or use the substitution v = y/x. Using the first idea, we have that for N(x, y) = x+2y and M(x, y) = y 4x, M y = 1 = N x, so this equation is exact. Then integrating M(x, y) with respect to x one gets φ(x, y) = (y 4x)dx + h(y) = xy 2x 2 + h(y). Then using that φ y = N, we take the partial derivative of φ with respect to y x + h (y) = x + 2y h (y) = 2y h(y) = y 2 Hence the solution is Since y(0) = 1, we get xy 2x 2 + y 2 = C xy 2x 2 + y 2 = 1 Using the second idea, we see that this is a homogeneous equation and make v = y/x, so dy dx = 4 y/x 1 + 2y/x v + xdv dx = 4 v 1 + 2v xdv dx This is a separable differential equation 1 + 2v 2(2 v v 2 ) dv = 1 x dx 1 2 = 4 2v 2v2 1 + 2v 1 + 2v 1 v 2 + v 2 dv = x dx + C The integral in v is computed by using the substitution u = v 2 + v 2, du = (2v + 1)dv, so we get 1 2 ln v2 + v 2 = ln x + C ln v 2 + v 2 = 2 ln x 2C Exponentiating both sides and making e 2C = k, we get v 2 + v 2 = kx 2 y 2 x + y 2 x 2 = kx 2 Multiplying by x 2 both sides y 2 + xy 2x 2 = k Since y(0) = 1, we get y 2 + xy 2x 2 = 1 3
4. Solve the following initial value problem: (10 points) Let v = y, then v = y, so y = y x 3x(y ) 2, y(1) = 1, y (1) = 1 v = v x 3xv2 v v x = 3xv2 This is a Bernoulli equation, so divide both sides by v 2 v 2 v v 1 x = 3x and make u = v 1, so u = v 2 v and then u 1 x u = 3x u + 1 x u = 3x This is a first-order diff. eq., so we compute the integrating factor I(x) = e 1 x dx = e ln x = x and then multiply both sides by x to obtain d dx (xu) = 3x2 xu = 3x 2 dx + C 1 xu = x 3 + C 1 x v = x3 + C 1 Since y (1) = v(1) = 1, we get C 1 = 0. Thus y = v = x 2 y = x 2 dx + C 0 = 1 x + C 0 Since y(1) = 1, we get C 0 = 0. Hence y = 1 x 4
5. A tank contains 64L (liters) of water in which is dissolved 650g (grams) of (10 points) salt. Seawater (saline water) containing 10g/L of salt flows into the tank at a rate of 9L/min, and the well-stirred saline water flows out at a rate of 3L/min. What is the amount of salt after 6 minutes? (a) 748 (b) 812 (c) 998 (d) 1008 (THIS IS THE ANSWER) (e) 1112 We have dv dt = 9 3 = 6 V (t) = 6t + V 0, since V (0) = 64, we get V (t) = 6t + 64. So da dt = (9)(10) 3 6t + 64 A da dt + 3 6t + 64 A = 90 Computing the integrating factor I(t) = e 3 6t+64 dt = e 3 6 ln 6t+64 = (6t + 64) 1/2 Hence (6t+64) 1/2 A(t) = 90(6t+64) 1/2 dt+c = (90) ( ) 2 1 3 6 (6t+64)3/2 +C = 10(6t+64) 3/2 +C Dividing both sides by (6t + 64) 1/2 Since A(0) = 650 we have A(t) = 10(6t + 64) + C 6t + 64 650 = 10(64) + C 64 C = 80 A(t) = 10(6t + 64) + 80 6t + 64 Thus A(6) = 10(36 + 64) + 80 = 1000 + 80 = 1008 36 + 64 100 5
6. For what values of a, b and c is it true that (10 points) [ ] [ ] [ ] [ ] 1 2 a b a b 1 2 =? 3 4 0 c 0 c 3 4 Multiplying the given matrices, one obtains [ ] [ a b + 2c a + 3b 2a + 4b = 3a 3b + 4c 3c 4c ] So a = a + 3b b = 0 and then one concludes from the any other entry that a = c. Thus this is true for any a R, b = 0, and c = a. 7. Compute the rank of the following 4 X 4 matrix 3 0 2 0 1 3 0 3 2 3 2 2 4 6 4 1 (10 points) (a) 0 (b) 1 (c) 2 (d) 3 (THIS IS THE ANSWER) (e) 4 Switch row-1 and row-2: 1 3 0 3 3 0 2 0 2 3 2 2 4 6 4 1 Replace row-2 by (row-2)-3(row-1), row-3 by (row-3)-2(row-1) and row-4 by (row-4)-4(row-1) 1 3 0 3 0 9 2 9 0 9 2 4 0 18 4 13 Replace row-3 by (row-3)-(row-2) and row-4 by (row-4)-2(row-2) 1 3 0 3 1 3 0 3 0 9 2 9 0 0 0 5 0 9 2 9 0 0 0 5, 0 0 0 5 0 0 0 0 where in the last step we replace row-4 by (row-4)-(row-3). The rank is 3. 6
8. The entry b 23 of the matrix A 1 = (b ij ) inverse to A = equal to 1 2 3 2 2 3 1 0 1 is (10 points) (a) 1 2 (b) 2 3 (c) 3 2 (THIS IS THE ANSWER) (d) 1 2 (e) 1 We have 1 2 3 2 2 3 1 0 1 0 1 0 0 0 1 1 2 3 0 2 3 0 2 2 2 1 0 1 0 1, where we replaced row-2 by (row-2)-2(row-1) and row-3 by (row-3)-(row-1). Then replace row-1 by (row-1)+(row-2) and row-3 by (row-3)-(row-2) 1 1 0 0 2 3 2 1 0 0 0 1 1 1 1 Now replace row-2 by (row-2)+3(row-1) to get 1 1 0 0 2 0 1 2 3 0 0 1 1 1 1 Finally divide row-2 by 2 to get 0 1 0 0 0 1 1 1 0 1/2 1 3/2 1 1 1, Thus A 1 = 1 1 0 1/2 1 3/2 1 1 1 then b 23 = 3/2. 7
9. Find all values of a such that the following system has an infinite number (15 points) of solutions. x 1 + 2x 2 + 3x 3 = 1 2x 1 3x 2 8x 3 = 7 2x 1 + 6x 2 + (a 2 + 1)x 3 = 8 We have 1 2 3 2 3 8 2 6 a 2 + 1 Replace row-2 by (row-2)+2(row-1) and row-3 by (row-3)-2(row-1) to get 1 2 3 1 0 1 2 5 0 2 a 2 5 10 1 7 8 Now replace row-3 by (row-3)-2(row-2) to get 1 2 3 1 0 1 2 5 0 0 a 2 1 0 Thus the system has infinite many solutions if and only if a 2 1 = 0 a = ±1 8