1. In R4,findthedistanceofthevectory to the subspace W spanned by the orthogonal vectors x1, x2 and x3,where 1 2 1 1 0 1 1 3 x = 2 3, x = 2 3, x = 2 3 and y = 2 3. 1 1 2 3 3 0 3 617 6 17 6 17 6 47 6 7 6 7 6 7 6 7 4 5 4 5 4 5 4 5 (a) p15 (b) p12 (c) p14 (d) p11 (e) p17

Solution.Solution. Observe that x1, x2, x3 is an orthogonal basis for W . Therefore, the orthogonal projection of the vector y{ on to W}is 1 2 1 0 y x1 y x2 y x3 6 0 10 1 2 1 0 y =projWy = • x1 + • x2 + • x3 = 2 3 + 2 3 + 2 3 = 2 3. x1 x1 x2 x2 x3 x3 3 1 15 3 3 0 4 • • • 617 6 17 6 17 627 b 6 7 6 7 6 7 6 7 Since, 4 5 4 5 4 5 4 5 1 0 1 3 0 3 y y = 2 3 2 3 = 2 3 3 4 1 6 47 627 627 6 7 6 7 6 7 we have that the distance of the vectorb 4y to5 the4 subspace5 4 W5is equal to y y = ( 1)2 +32 +( 1)2 +22 = p15. k k p b 2. Find a least squares solution to the system 1 1 5 0 x 610 1 0 2 3 x = 2 3. 1 512 23 0 x 6401 7 3 697 6 74 5 6 7 4 5 4 5 Note that the columns a1, a2, a3 of the coecient matrix A form an orthogonal basis for Col A. 2/3 2/3 2/3 2/3 1 (a) 0 (b) 0 (c) 5/3 (d) 5/3 (e) 2 2 3 2 3 2 3 2 3 2 3 1/3 1/3 1/3 1/3 3 4 5 4 5 4 5 4 5 4 5 Solution.Solution. Since the columns a1, a2, a3 of the coecient matrix A form an orthogonal basis for W =ColA,wehavethattheorthogonalprojectionofthevector{ } b (the right-hand side of the system) on to W is 2/3 b a b a b a 36 9 b =proj b = • 1 a + • 2 a + • 3 a = a +0 a + a = A 0 . W a a 1 a a 2 a a 3 54 1 · 2 27 3 2 3 1 • 1 2 • 2 3 • 3 1/3 b 4 5 2/3 Therefore x = 0 is a least squares solution to the given system. 2 3 1/3 b 4 5 dy 3. Figure 1 shows the direction field for the di↵erential equation = f(y), where f(y)isa dt polynomial of third degree. y

10

8

6

4

2

0 t 4 8 12 16 20 -2

Figure 1 Which of the following statements is false? (a) The solution with initial value y(0) = 7.9isincreasingandbecomesequalto8infinite time. (b) The equilibrium solutions of this di↵erential equations are y =8,y =2andy =0. (c) y =8andy =0areasymtoticallystable solutions. (d) y =2isanunstable solution. (e) The solution with initial value y(0) = 1.9isdecreasingandgoingtozeroast goes to infinity.

Solution.Solution. Looking at this direction field we see that all statements are true, except the statement:“The solution with initial value y(0) = 7.9isincreasingandbecomesequalto8in finite time.” In fact, the solution with initial value y(0) = 7.9cannotmeettheequilibrium solution y(t)=8atsometimet0 > 0becausethenwewouldhavetwosolutionstothe dy equation = f(y)withinitialdatay(t )=8,contradictingthebasictheoremabout dt 0 existence and uniqueness of solution when f(y)iscontinuouslydi↵erentiable(seetextbok).

4. Which of the follow di↵erential equations has the direction field shown in Figure 1 above? dy y y dy dy (a) = y(1 )(1 )(b) =(2 y)(y 8)(y 4) (c) = (2 y)2(8 y) dt 2 8 dt dt dy y y dy (d) = y(1 )(1 )(e)=(2 y)2(8 y) dt 2 8 dt Solution.Solution. Looking at the direction field in Figure 1, we see that only the di↵erential equation dy = y(1 0.5y)(1 0.125y)hastheequilibriumsolutionsy =0, 2, 8andtheslopesofthe dt arrows are consistent with the sign of the function f(y)= y(1 0.5y)(1 0.125y) 5. Atank,withacapacityof1000gallons,initiallycontains100gallonsofbrinewitha concentration of 0.2 lb salt per gallon. A solution containing 0.4 lb of salt per gallon is pumped into the tank at the rate of 5 gal/min, and the well-stirred mixture flows out of the tank at the rate of 2 gal/min. Write the initial value problem (i.e. a di↵erential equation and an initial condition) needed to find the amount S(t)ofsaltinthetankatanytimet (before the tank is full). DO NOT SOLVE. dS 2 dS 1 (a) + S =2,S(0) = 20. (b) + S =2,S(0) = 20. dt 100 + 3t dt 100 + 2t dS 2 dS 2 (c) + S =2,S(0) = 20. (d) + S =2,S(0) = 20. dt 100 + 5t dt 100 dS 2 (e) + S =2,S(0) = 100. dt 100 3t Solution.Solution. The quantity S(t)satisfiesthedi↵erentialequation dS S(t) dS 2 =5 (0.4) 2 or + S =2. dt · · 100 + (5 2)t dt 100 + 3t We also have S(0) = 20 .

2 6. If y(x)isthesolutiontothedi↵erentialequationxy0+5y = x , x>0, such that lim y(x)=0, x 0 then compute y(14). !

1 (a) 28 (b) 17 (c) 5 (d) 14 (e) 14 5 Solution.Solution. First, we write the DE in the form: y + y = x. Then, we compute the integrating 0 x 5 dx 5lnx 5 5 5 6 factor µ(x)=e x = e = x . Now, multiplying the last DE by x we obtain (x y)0 = x . R 1 c Furthermore, integrating it we obtain x5y = 1 x7 + c or y = x2 + . Finally, using the 7 7 x5 1 2 condition limx 0 y(x)=0wegetc =0andtherefore y = x . This gives y(14) = 28. ! 7

rt 7. For what values of r is y(t)=e asolutiontoy00 3y0 10y =0? (a) r =5orr = 2(b)r =5orr =2 (c) r =2orr = 3(d)r = 3only (e) r = 3orr =2 rt rt 2 rt Solution.Solution. If y = e ,theny0 = re and y00 = r e . Substituting into the ODE y00 3y0 10y =0givesr2ert 3rert 10ert =0i.e.(r2 3r 10)ert =0.Sowegetasolution ert when r2 3r 10 = 0 i.e. (r 5)(r +2)=0i.e.r =5or r = 2.

8. ArichdonorgiftsanamountA0 to ND for student scholarships. The ND investment oce invests this amount in an account earning annual interest of 5% compounded continuously and makes withdrawals continuously at the rate of 10 million dollars per year, without the amount in the account changing (i.e. it is always equal to A0). Find the initial amount A0. (a) 200 million dollars (b) This is impossible. (c) 100 million dollars (d) 50 million dollars (e) 500 million dollars Solution.Solution. If we denote by A(t) the amount (in millions of dollars) in the account at any time t,thenitsatisfiesthedi↵erentialequation dA =0.05A 10. dt dA 10 Since =0wemusthave0.05A 10 = 0 or A = = 200 millon dolars. dt 0.05

9. Find the interval of existence of the solution to the following initial value problem: dy x +(y +1)2 =0,x>0,y(1) = 4. dx

0.2 (a) (e , )(b)(0, )(c)(1, )(d)(ln4, )(e)(0.2, ) 1 1 1 1 1 Solution.Solution. Separating variables first and then integrating gives

dy dx 2 dx 1 = or (y +1) dy = or =lnx + c. (y +1)2 x x y +1 Z Z Letting x =1andy =4weget0.2=c, which gives the solution 1 y(x)= 1. ln x +0.2 Now, we observe that the denominator in the solution formula becomes zero when ln x = 0.2 0.2 or x = e . Then, the solution becomes infinity (blows up). Thus, the existence interval 0.2 of the solution to the given initial value problem is (e , ). 1

10. (11 pts) Apply the Gram-Schmidt process to find an orthogonal basis for the column space of the following matrix 3 78 15 2 2 3 111 6 3 517 6 7 You need not normalize your basis.4 5

Solution.Solution. Denoting by x1, x2, x3 the three column vectors of the matrix and applying the Gram-Schmidt process we obtain the following orthogonal vectors u1, u2 and u3: 3 7 3 1 1 x2 u1 5 40 1 3 u1 = x1 = 2 3, u2 = x2 • u1 = 2 3 2 3 = 2 3, 1 u1 u1 1 20 1 3 6 37 • 6 57 6 37 6 17 6 7 6 7 6 7 6 7 and 4 5 4 5 4 5 4 5 8 3 1 3 x3 u1 x3 u2 2 30 1 10 3 1 u3 = x3 • u1 • u2 = 2 3 2 3 2 3 = 2 3. u1 u1 u2 u2 1 20 1 20 3 1 • • 6 17 6 37 6 17 6 37 6 7 6 7 6 7 6 7 4 5 4 5 4 5 4 5 11. (12 pts) (1) Applying the fundamental ODE theorem for linear first order di↵erential equations, one concludes that the following initial value problem

4 dy 3 t2 (1 + t ) +4t y =6t 2te ,y(0) = 9, dt has a unique solution for

Solution.Solution. d 4 t2 (1) The di↵erential equation can be written as [(1 + t )y]=6t 2te .Therefore, dt 2 t2 2 3t + e + c (1 + t4)y =3t2 + e t + c or y(t)= . Since 9 = y(0) = 1 + c we obtain 1+t4 3t2 + e t2 +8 that y(t)= . 1+t4 1/3 (2) First, we observe that y1 = 0 is a solution. Then, separating variables gives y dy = 3 2/3 dt.Furthermore,integratingweget2 y dy = t + c.Lettingt =0andy =0tothis 2 3/2 formula gives c = 0. Finally, solving for y we obtain y = t , which gives the ± 3 following two solutions ⇣ ⌘ 2 3/2 2 3/2 3 t ,t 0, 3 t ,t 0, y2(t)= and y3(t)= ,t<, ,t<. ( 0 0 ( 0 0 dy Thus, we found three! di↵erent solutions to the the initial value problem = y1/3,y(0) = 0. dt In fact, it has infinitely many solutions (see textbook). 12. (14 pts) Assume that a region’s population p(t)(inbillions)atanytimet (in years) is modeled by the di↵erential equation dp =0.02p(1 p). dt (1) Solve this di↵erential equation with initial value p(0) = p0 > 0tofindanexplicit formula for p(t). (2) Use the formula in (1) to compute limt p(t). (3) Sketch the solution curves that correspond!1 to the initial values p(0) = 0.25, p(0) = 1, and p(0) = 1.5, clearly showing where they are increasing/decreasing.

p

1.5

1.25

1

0.75

0.5

0.25 t 0 50 100 150 200 250 dp 1 Solution.Solution. (1) Separating variables we have =0.02dt.Then,writing = p(1 p) p(1 p) 1 1 p + ,andintegratinggiveslnp ln 1 p =0.02t + c or ln | | =0.02t + c or p 1 p | | | | 1 p p | | = ece0.02t. Finally, letting C = ec we get the following general solution in implicit 1 p ± ± p p form = Ce0.02t.Lettingp = p and t =0,thisformulagives 0 = C.Substituting 1 p 0 1 p p p 0 this C into the general solution we get the formula = 0 e0.02t,whichsolvedforp 1 p 1 p0 gives the desired solution to the given initial value problem p p(t)= 0 . p +(1 p )e 0.02t 0 0 (2) From this formula in (1) we see that lim p(t)=1. t !1 (3) The solution curve with p(0) = 1 is the equilibrium solution p = 1. Using the direction fields we see that the solution with p(0) = 1.5isdecreasingtowardstheequilibriumsolution p = 1, while the solution with p(0) = 0.25 is increasing towards the equilibrium solution p = 1. Below, are the graphs of these solution curves and the direction field of our ODE. p

1.25

1

0.75

0.5

0.25

t 50 100 150 200 250