Math 244 Homework Set 1 Solutions 1 2008

Homework Set 1 Solutions 1. Exercise 22, p 363 (Section 7.1) Construct the DE’s by looking at “Rate In - Rate Out”, and make sure your units are matching up. Define Q1 (t) and Q2 (t) to be the ounces of salt in Tanks 1 and 2 (respectively). Before we start, we take note of the rates in and out of each tank- The total amounts of water in tanks 1 and 2 (30 gallons and 20 gallons, respectively) does not change (important for the rates in and out). Filling in the quantities from the figure on p. 364, we have: 1.5 gal 1 oz 1.5 gal Q2 oz 3 gal Q1 oz dQ1 = · + · − · dt min gal min 20 gal min 30 gal dQ2 1 gal 3 oz 3 gal Q1 oz 4 gal Q2 oz = · + · − · dt min gal min 30 gal min 20 gal Simplifying, and putting them in order: 1 3 3 dQ1 = − Q1 + Q2 + dt 10 40 2 1 1 dQ2 = Q1 − Q2 + 3 dt 10 5 The system is currently non-homogeneous because of the constants 3/2 and 3. The equilibria are found by solving for where the derivatives are zero. We simplify to make Cramer’s Rule easier to apply: 1 3 − 10 Q1 + 40 Q2 + 32 1 1 10 Q1 − 5 Q2 + 3 =0 =0 ⇒ −4Q1 + 3Q2 Q1 − 2Q2 = −60 = −30 Therefore, using the determinants, the equilibrium solution is: Q1 = 210 = 42 5 Q2 = 180 5 The last part of the question shows that a change of variables results in a homogeneous differential equation. That is, if x1 = Q1 − 42 and x2 = Q2 − 36, then substitution into the system of DE’s: x01 = Q01 x02 = Q02 and 1 3 3 1 3 (x1 + 42) + (x2 + 36) + = − x1 + x2 10 40 2 10 40 Similarly, substitution into the second equation and substituting: − 1 1 1 1 (x1 + 42) − (x2 + 36) + 3 = x1 − x2 10 5 10 5 Notice that our change of coordinates simply shifted the coordinate system so that the equilibrium is now at the origin. To get the initial conditions, make the last substitution: x01 x02 3 1 x1 + 40 x2 = − 10 1 1 = 10 x1 − 5 x2 x1 (0) = −17, 1 x2 (0) = −21 2. Short Answer: (a) Is every second order linear homogeneous differential equation (with constant coefficients) equivalent to a system of first order equations? Yes. (b) Can every 2×2 system of DEs be converted into an equivalent second order system? (Hint: To do our technique, what must be true?) No. In particular, one of the variables must be solvable in terms of the other. Example where this does not work: x01 x02 = 3x1 = −x2 3. Give the solution to each system. If it has an infinite number of solutions, give your answer in vector form: 3x + 2y 2x − y 3x + 2y 6x + 4y 3x + 2y 6x + 4y =1 =2 ⇒ =1 =3 ⇒ =1 =3 3 1 y = − x+ 2 2 x = 1, y = −1 No Solution.  ⇒ x y   =x 1 −3/2    0 + 1/2 4. Write each of the previous systems in matrix-vector form. Verify that the determinant of the first matrix is not zero, but is zero for the second and third. Straightforward. 5. Write each system of differential equations in matrix-vector form or write the system from the matrix-vector form:   x01 = 3x1 − x2 −1 2 0 x x = 2 1 x02 = 9x1 − 3x2 Straightforward. 6. Find the equilibrium solutions to the previous autonomous linear differential equations. The determinant of the first is zero, which means we have an infinite number of solutions, 3x1 − x2 = 0, or x2 = 3x1 . In vector form, the solutions are:   1 x1 3 In the second DE, the determinant of the coefficients is not zero, so the only equilibrium is the origin. For each of the problems below, define     1 2 −1 A= b= −1 1 1 and I is the identity matrix: 2  B= 3 1 1 3  7. Compute (B − 4I)b    3 1 4 − 1 3 0 0 4   b= −1 1 1 −1  −1 1  7 5   = −2 2  8. Compute det(B − 4I) = 0 9. Are AB and BA the same? No:   5 7 AB = −2 2 10. Compute A −1 1 b= 3  1 1 −2 1   2 −2 −1 0  BA = −1 1   = 11. Verify: Ab − 3b = (A − 3I)b 12. Compute B T B, Tr(A), and Tr(B)   10 6 BT B = 6 10 Tr(A) = 2 Tr(B) = 6 R1 13. If x is as defined below, compute x0 (t), and 0 x(t) dt:   t2 − 3 x(t) = 3et − 2e3t 0  x = 2t 3et − 6e3t  1 Z 0 1 x(t) dt = = 3  −8 9e − 2e3 − 7  14. Solve the system of equations given by first converting it into a second order linear ODE (then use Chapter 3 methods):   −2 1 0 x (a) x = 1 −2 From the first equation, we get x2 = x01 + 2x1 . Substitute this into the second equation to get: x001 + 2x01 = x1 − 2(x01 + 2x1 ) ⇒ x001 + 4x01 + 3x1 = 0 We get a characteristic equation: r2 + 4r + 3 = 0, and: x1 = C1 e−t + C2 e−3t x2 = x01 + 2x1 = C1 e−t − C2 e−3t Recall that we can verify our solution (not necessary, but good for practice):  0    x1 −C1 e−t − 3C2 e−3t = x02 −C1 e−t + 3C2 e−3t And, using the matrix, we get the same vector:       −2 1 x1 −2 1 C1 e−t + C2 e−3t = 1 −2 x2 1 −2 C1 e−t − C2 e−3t 3  0 2 (b) x = x −2 0 Following our usual procedure, use one equation to solve for one of the variables. Using the first equation, we have: x2 = 12 x01 . Substitute this into the second to get a DE in terms of x1 alone: 0  x02 = −2x1 ⇒ 1 0 x = −2x2 2 1 ⇒ x001 + 4x1 = 0 The characteristic equation gives r = ±2i, so x1 = C1 cos(2t) + C2 sin(2t) and x2 = 21 x01 = −C1 sin(2t) + C2 cos(2t). 15. The four graphs in the attached figure show the direction fields for the four systems of differential equations below. Try to reason out which direction field goes with which system.   3 −1 (a) x = x Upper right graph 4 −2   −2 1 x Lower left graph (b) x = 1 −2   0 2 (c) x = x Upper left graph −2 0   −1 −4 (d) x = x Lower right graph. 1 −1 16. Convert the following second order differential equations into a system of autonomous, first order equations. Using methods from Chapter 3, give the solution to the system. An example follows before the exercises: y 00 + 3y 0 + 2y = 0 SOLUTION: We’ll get the homogenous solution first. The roots to the characteristic equation are −1, −2. The general solution is: y = C1 e−t + C2 e−2t To get an equivalent system, let x1 = y and x2 = y 0 . Then x01 = y 0 = x2 x02 = y 00 = −2y − 3y 0 = −2x1 − 3x2 so the system is (in matrix-vector form):   0 1 x= x −2 −3 Since x1 = y, then x1 = C1 e−t + C2 e−2t . Since x2 = y 0 , then x2 = −C1 e−t − 2C2 e−2t . In vector form, this means our solution is:     1 1 x = C1 e−t + C2 e−2t −1 −2 Here we go: 4 (a) y 00 + 4y 0 + 3y = 0 The solution is: y = C1 e−t + C2 e−3t . Using our usual substitution x1 = y, x2 = y 0 , we have:  0     x01 = y 0 = x2 x1 0 1 x1 ⇒ = x02 = y 00 = −3y − 4y 0 = −3x1 − 4x2 x02 −3 −4 x2 And we can write the solution as:         x1 C1 e−t + C2 e−3t 1 1 −t −3t = = C1 e + C2 e x2 −C1 e−t − 3C2 e−3t −1 −3 (b) y 00 + 5y 0 = 0 The solution is: y = C1 + C2 e−5t . The system, with the usual substitutions, is:  0     x1 0 1 x1 = x02 0 −5 x2 The solution can be written as:         1 1 x1 C1 + C2 e−5t −5t e = C + C = 2 1 0 −5 −5C2 e−5t x2 (c) y 00 + 4y = 0 We’ve seen one like this already in 14(b): The solution is y = C1 cos(2t) + C2 sin(2t). Using the substitutions x1 = y, x2 = y 0 , we have:  0     x1 0 1 x1 x01 = x2 x02 = −4y = −4x1 ⇒ = x02 −4 0 x2 and we can write the solution as a vector (but not as neatly as before:       sin(2t) cos(2t) x1 + C2 = C1 2 cos(2t) −2 sin(2t) x2 (d) y 00 − 2y 0 + y = 0 The characteristic equation gives r = 1, 1, so y(t) = C1 et + C2 tet Using the usual substitutions, the matrix-vector form of the first order system is:  0     x1 0 1 x1 = x02 −1 2 x2 Writing the solution as a system, we have:     x1 C1 et + C2 tet = x2 C1 et + C2 et + C2 tet For future reference, notice that this can be written as:        1 1 0 t t C1 e + C2 e t + 1 1 1 5