1. In R 4,findthedistanceofthevectory to the subspace W spanned by the orthogonal vectors x 1, x 2 and x 3,where and y = x 1 = 60

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1 . In R,finhedistanceofthevectory to the subsace W sanned by the orthogonal vectors x, x and x,where x = 0 5, x = 5, x = 05 and y = 5. (a) 5 (b) (c) (d) (e) Solution. Observe that {x, x, x } is an orthogonal basis for W. Therefore, the orthogonal rojection of the vector y on to W is 0 by =roj W y = y x x + y x x + y x x = x x x x x x Since, 0 y by = = 5 we have that the distance of the vector y to the subsace W is equal to ky byk = ( ) + +( ) + = = Find a least squares solution to the system 5 0 x x 5 = x 0 9 Note that the columns a, a, a of the coe cient matrix A form an orthogonal basis for Col A. (a) / 05 (b) / 05 (c) / 5/5 (d) / 5/5 (e) 5 / / / / Solution. Since the columns {a, a, a } of the coe cient matrix A form an orthogonal basis for W =ColA, wehavethattheorthogonalrojectionofthevectorb (the right-hand side of the system) on to W is bb =roj W b = b a a + b a a + b a a = a a a a a a 5 a +0 a + 9 a = A / 05. / Therefore bx = / 05 is a least squares solution to the given system. /

2 . Figure shows the direction field for the di erential equation = f(y), where f(y) isa olynomial of third degree. y t Figure Which of the following statements is false? (a) The solution with initial value y(0) =.9 isincreasingandbecomesequaltoinfinite time. (b) The equilibrium solutions of this di erential equations are y =,y =an =0. (c) y =an =0areasymtoticallystable solutions. (d) y =isanunstable solution. (e) The solution with initial value y(0) =.9 isdecreasingandgoingtozeroast goes to infinity. Solution. Looking at this direction field we see that all statements are true, excet the statement: The solution with initial value y(0) =.9 isincreasingandbecomesequaltoin finite time. In fact, the solution with initial value y(0) =.9 cannotmeettheequilibrium solution y(t) =atsometimet 0 > 0becausethenwewouldhavetwosolutionstothe equation = f(y) withinitialdatay(t 0) =, contradicting the basic theorem about existence and uniqueness of solution when f(y) iscontinuouslydi erentiable(seetextbok).. Which of the follow di erential equations has the direction field shown in Figure above? (a) (d) = y( y )( y = y( y )( y ) (b) =( y)(y )(y ) (c) = ( y) ( y) ) (e) =( y) ( y) Solution. Looking at the direction field in Figure, we see that only the di erential equation = y( 0.5y)( 0.5y) hastheequilibriumsolutionsy =0,, anhesloesofthe arrows are consistent with the sign of the function f(y) = y( 0.5y)( 0.5y)

3 5. Atank,withacaacityof000gallons,initiallycontains00gallonsofbrinewitha concentration of 0. lb salt er gallon. A solution containing 0. lb of salt er gallon is umed into the tank at the rate of 5 gal/min, and the well-stirred mixture flows out of the tank at the rate of gal/min. Write the initial value roblem (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. (a) (c) (e) t t + S =, S(0) = t S =, S(0) = 0. (b) + S =, S(0) = t S =, S(0) = 0. (d) + S =, S(0) = Solution. The quantity S(t) satisfiesthedi erentialequation We also have S(0) = 0. =5 (0.) S(t) 00 + (5 )t or t S =.. If y(x)isthesolutiontothedi erentialequationxy 0 +5y = x, x>0, such that lim x!0 y(x) =0, then comute y(). (a) (b) (c) 5 (d) (e) Solution. First, we write the DE in the form: y y = x. Then, we comute the integrating x factor µ(x) =e R 5 x dx = e 5lnx = x 5. Now, multilying the last DE by x 5 we obtain (x 5 y) 0 = x. Furthermore, integrating it we obtain x 5 y = x + c or y = x + c. Finally, using the x5 condition lim x!0 y(x) =0wegetc =0anherefore y = x. This gives y() =.. For what values of r is y(t) =e rt asolutiontoy 00 y 0 0y =0? (a) r =5orr = (b) r =5orr = (c) r =orr = (d) r = only (e) r = orr = Solution. If y = e rt,theny 0 = re rt and y 00 = r e rt. Substituting into the ODE y 00 y 0 0y =0givesr e rt re rt 0e rt =0i.e. (r r 0)e rt =0. Sowegetasolutione rt when r r 0 = 0 i.e. (r 5)(r +)=0i.e. r =5orr =.. ArichdonorgiftsanamountA 0 to ND for student scholarshis. The ND investment o ce invests this amount in an account earning annual interest of 5% comounded continuously and makes withdrawals continuously at the rate of 0 million dollars er year, without the amount in the account changing (i.e. it is always equal to A 0 ). Find the initial amount A 0. (a) 00 million dollars (b) This is imossible. (c) 00 million dollars (d) 50 million dollars (e) 500 million dollars

4 Solution. If we denote by A(t) the amount (in millions of dollars) in the account at any time t, thenitsatisfiesthedi erentialequation Since da da =0.05A 0. 0 =0wemusthave0.05A 0 = 0 or A = 0.05 = 00 millon dolars. 9. Find the interval of existence of the solution to the following initial value roblem: x dx +(y +) =0,x>0, y() =. (a) (e 0., ) (b) (0, ) (c) (, ) (d) (ln, ) (e) (0., ) Solution. Searating variables first and then integrating gives (y +) = dx Z Z dx x or (y +) = x or y + Letting x =an =weget0. =c, which gives the solution y(x) = ln x +0.. =lnx + c. Now, we observe that the denominator in the solution formula becomes zero when ln x = 0. or x = e 0.. Then, the solution becomes infinity (blows u). Thus, the existence interval of the solution to the given initial value roblem is (e 0., ). 0. ( ts) Aly the Gram-Schmi rocess to find an orthogonal basis for the column sace of the following matrix You need not normalize your basis. Solution. Denoting by x, x, x the three column vectors of the matrix and alying the Gram-Schmi rocess we obtain the following orthogonal vectors u, u and u : and u = x = 5, u x u = x u = u u x u x u u = x u u = u u u u = 5 = 5, 5.

5 . ( ts) () Alying the fundamental ODE theorem for linear first order di erential equations, one concludes that the following initial value roblem ( + t ) +t y =t te t,y(0) = 9, has a unique solution for <t<. Find this solution. () While the initial value roblem () has a unique solution, show that the following initial value roblem = y/,y(0) = 0, has more than one solution by finding at least two of them. Solution. () The di erential equation can be written as d [( + t )y] =t te t. Therefore, ( + t )y =t + e t + c or y(t) = t + e t + c +t. Since 9 = y(0) = + c we obtain that y(t) = t + e t + +t. () First, we observe that y = 0 is a solution. Then, searating variables gives y / =. Furthermore,integratingweget y/ = t + c. Lettingt =0an =0tothis formula gives c = 0. Finally, solving for y we obtain y = ± t /, which gives the following two solutions ( y (t) = t /, t 0, 0, t < 0, and y (t) = ( t /, t 0, 0, t < 0. Thus, we found three! di erent solutions to the the initial value roblem = y/,y(0) = 0. In fact, it has infinitely many solutions (see textbook).

6 . ( ts) Assume that a region s oulation (t) (inbillions)atanytimet(in years) is modeled by the di erential equation d =0.0( ). () Solve this di erential equation with initial value (0) = 0 > 0tofindanexlicit formula for (t). () Use the formula in () to comute lim t! (t). () Sketch the solution curves that corresond to the initial values (0) = 0.5, (0) =, and (0) =.5, clearly showing where they are increasing/decreasing t

7 d Solution. () Searating variables we have ( ) =0.0. Then, writing ( ) = +,andintegratinggivesln ln =0.0t + c or ln =0.0t + c or = ±ec e 0.0t. Finally, letting C = ±e c we get the following general solution in imlicit form = 0 Ce0.0t.Letting = 0 and t =0,thisformulagives = C. Substituting 0 this C into the general solution we get the formula = 0 e 0.0t,whichsolvedfor 0 gives the desired solution to the given initial value roblem 0 (t) = 0 +( 0 )e. 0.0t () From this formula in () we see that lim t! (t) =. () The solution curve with (0) = is the equilibrium solution =. Using the direction fields we see that the solution with (0) =.5 isdecreasingtowardstheequilibriumsolution =, while the solution with (0) = 0.5 is increasing towards the equilibrium solution =. Below, are the grahs of these solution curves and the direction field of our ODE t