Third In-Class Exam Solutions Math 246, Professor David Levermore Thursday, 3 December 2009 (1) [6] Given that 2 is an eigenvalue of the matrix

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1 Third In-Class Exam Solutions Math 26, Professor David Levermore Thursday, December 2009 ) [6] Given that 2 is an eigenvalue of the matrix A 2, 0 find all the eigenvectors of A associated with 2. Solution. The eigenvectors of A associated with 2 are all nonzero vectors v that satisfy Av 2v. Equivalently, they are all nonzero vectors v that satisfy A 2I)v 0, which is 0 v v The entries of v thereby satisfy the homogeneous linear algebraic system v v 2 + v 0, v v 2 v 0, v 2 + v 0. You may solve this system either by elimination or by row reduction. By either method you find that its general solution is v 2α, v 2 α, v α, for any constant α. The eigenvectors of A associated with therefore have the form α 2 for any nonzero constant α. 2) [] A matrix has eigenpairs, 2, 2, 2,, 0. a) Give an invertible matrix V and a diagonal matrix D such that A VDV. You do not have to compute V!) Solution. V 2 2 0, D

2 2 b) Give a general solution to the system dx dt Ax. Solution. xt) c e t 2 + c 2 e 2t 2 + c e t 0. ) [6] Transform the equation d w w dt td dt + dw w cost) into a first-order system dt of ordinary differential equations. Solution: Because the equation is fourth order, the first order system must have dimension four. The simplest such first order system is x x 2 x w d x 2 dt x x x, where x 2 x w w. cost) + x x 2 + tx x w x ) [0] Consider the vector-valued functions x t) a) Compute the Wronskian W[x,x 2 ]t). Solution. t t 2, x 2 t) + t 6. t W[x,x 2 ]t) det t 2 + t 6 + t 6 t 6. b) Find At) such that x, x 2 is a fundamental set of solutions to the system dx dt At)x wherever W[x,x 2 ]t) 0. t Solution. Let Ψt) t 2 + t 6. Because Ψt) At)Ψt), one has dt At) Ψt) 0 t t dt Ψt) 2t 6t t 2 + t 6 0 t + t 6 t 2t 6t t 2 t t 6t t 7 t. c) Give a general solution to the system you found in part b). Solution. Because x t), x 2 t) is a fundamental set of solutions to the system found in part b), a general solution is given by t xt) c x t) + c 2 x 2 t) c t 2 + c 2 + t 6.

3 cosh2t) 2 sinh2t) ) [] Given that e ta sinh2t) cosh2t) 2 d x x A, dt y y) ), solve the initial-value problem x0) y0) ) 2. ) Solution. The solution of the initial-value problem is given by xt) 2 cosh2t) 2 sinh2t) 2 yt) ta sinh2t) cosh2t) 2 2 cosh2t) + 6 sinh2t). sinh2t) + cosh2t) 6) [9] Consider two interconnected tanks filled with brine salt water). The first tank contains 60 liters and the second contains 0 liters. Brine with a concentration of grams of salt per liter flows into the first tank at a rate of liters per hour. Well stirred brine flows from the first tank into the second at a rate of liters per hour, from the second into the first at a rate of 2 liters per hour, from the first into a drain at a rate of liter per hour, and from the second into a drain at a rate of liters per hour. At t 0 there are grams of salt in the first tank and 7 grams in the second. Give an initial-value problem that governs the amount of salt in each tank as a function of time. Solution: The rates work out so there will always be 60 liters of brine in the first tank and 0 liters in the second. Let S t) be the grams of salt in the first tank and S 2 t) be the grams of salt in the second tank. These are governed by the initial-value problem ds + S 2 dt 0 2 S 60 S 60, S 0), ds 2 S dt 60 S S 2 0, S 20) 7. You could leave the answer in the above form. It can however be simplified to ds dt ds 2 dt 2 + S 2 20 S 0, S 0), S 2 S 2 8, S 20) 7.

4 7) [2] Find a general solution for each of the following systems. a) d x 2 x dt y) ) y 2 Solution. The characteristic polynomial of A is given by pz) z 2 tra)z + deta) z 2 + z + z + 2) The eigenvalues of A are the roots of this polynomial, which are the conjugate pair 2 ± i. One therefore has [ e ta 2t I cost) + A + 2I ) ] sint) [ ] 0 2 sint) 2t cost) + 0 cost) + 2t sint) 2 sint) ) sint) cost) sint). A general solution is therefore given by xt) yt) ta c cost) + c 2t sint) 2 sint) ) c 2 sint) cost) sint) c 2 cost) + c e 2t sint) ) sint) + c 2 e 2t 2 sint) ) cost) sint). b) d x dt y) 2 x 2 ) y Solution. The characteristic polynomial of A 2 is given by 2 pz) z 2 tra)z + deta) z 2 6z + 9 z ) 2. The eigenvalues of A are the roots of this polynomial, which is the double real root. One therefore has e ta t[ I + A I ) t ] [ ] t + t t 2t t. 2t + 2t A general solution is therefore given by xt) yt) ta c 2t 2t c t c 2 2t + 2t c 2 2t 2t c e t + c 2t 2 e t. + 2t

5 8) [8] Sketch a phase-plane portrait for each of the two systems in the previous problem. For each portrait identify its type and give a reason why the origin is either attracting, stable, unstable, or repelling. a) Solution. Because the characteristic polynomial of A is pz) z +2) 2 +9, one sees that µ 2 and δ 9. There are no real eigenpairs. Because µ 2 < 0, δ 9 < 0, and a 2 > 0 the phase portrait is a counterclockwise spiral sink. The origin is thereby attracting and also stable). The phase portrait should indicate a family of counterclockwise spiral trajectories that approach the origin. b) Solution. Because the characteristic polynomial of A is pz) z ) 2, one sees that µ and δ 0. Because 2 2 A I, 2 2 we see that the eigenvectors associated with have the form α for some α 0. ) The associated solutions will therefore lie on the line y x. Because µ > 0, δ 0, and a 2 2 < 0 the phase portrait is a clockwise twist source. The origin is thereby repelling and also unstable). The phase portrait should show that on the line y x there is one trajectory that emerges from each side of the origin. Every other trajectory emerges from the origin with a clockwise twist. 9) [9] Consider the system d x y dt y x + x. a) Find all of its stationary points. Solution. Stationary points satisfy 0 y, 0 x + x xx 2 ) xx + 2)x 2). The top equation shows that y 0 while the bottom equation shows that either x 0 or x 2 or x 2. The stationary points of the system are therefore 0, 0), 2, 0), 2, 0). b) Find a nonconstant function hx, y) such that every trajectory of the system satisfies hx, y) c for some constant c. Solution. The associated first-order equation is dy dx x + x y This equation is separable, so can be integrated as y dy x + x dx,.

6 6 whereby you find that You can thereby set 2 y2 2x 2 + x + c. hx, y) 2 y2 + 2x 2 x. Alternative Solution. An alternative approach is to notice that x fx, y) + y gx, y) x y + y x + x ) 0. The system is therefore Hamiltonian with hx, y) such that y hx, y) y, x hx, y) x + x. Integrating the first equation above yields hx, y) 2 y2 + cx). Substituting this into the second equation gives c x) x + x. Integrating this equation yields cx) 2x 2 x, whereby hx, y) 2 y2 + 2x 2 x. 0) [8] Compute the Laplace transform of ft) ut ) e t from its definition. Here u is the unit step function.) Solution. The definition of Laplace transform gives L[f]s) lim T 0 e st ut ) e t dt lim T e s+)t dt. When s this limit diverges to + because in that case one has for every T > that e s+)t dt dt T, which clearly diverges to + as T. When s > one has for every T > that e s+)t dt e s+)t T s + e s+)t s + + e s+) s +, whereby [ ] L[f]s) lim e s+)t T s + + e s+) s+) s + s +.

7 ) [9] Find the Laplace transform Y s) of the solution yt) of the initial-value problem where d 2 y dt 2 + 6dy dt + 8y ft), y0) 2, y 0), ft) { for 0 t <, e t for t. You may refer to the table on the last page. DO NOT take the inverse Laplace transform to find yt), just solve for Y s)! Solution. The Laplace transform of the initial-value problem is where L[y ]s) + 6L[y ]s) + 8L[y]s) L[f]s), L[y]s) Y s), L[y ]s) sy s) y0) sy s) 2, L[y ]s) s 2 Y s) sy0) y 0) s 2 Y s) 2s +. To compute L[f]s), first write f as ft) ut) ut ) ) t + ut )e t ut) t + ut ) e t t) t + ut ) e t t ) ). Referring to the table on the last page, item 6 with c and ht) t t shows that L [ ut ) e t t ) )] s) L[ut )ht )]s) s L[h]s), while item with a and n 0, with a 0 and n, and with a 0 and n 0 shows that L[e t ]s) s +, L[t]s) s, L[]s) 2 s, whereby L[ t]s) s, L[h]s) s + s 2 s. Therefore L[f]s) s + 2 e s s + s ). 2 s The Laplace transform of the initial-value problem then becomes s 2 Y s) 2s + ) + 6 sy s) 2 ) + 8Y s) e s + e s s 2 s + e s s, which becomes s 2 + 6s + 8)Y s) 2s + 2 e s s 2 Hence, Y s) is given by Y s) 2s e s s 2 + 6s + 8 s 2 + e s s + e s s. ) + e s s + e s. s 7

8 8 2) [8] Find the inverse Laplace transforms of the function Fs) s s + s 2 s + 6. You may refer to the table on the last page. Solution. The denominator factors as s )s 2), so the partial fraction decomposition is s + s 2 s + 6 s + s )s 2) 8 s + 7 s 2. Referring to the table on the last page, item with a and n 0, and with a 2 and n 0 shows that L[e t ]s) s, L[e2t ]s) s 2. You therefore obtain [ ] [ ] [ ] L s + t) 8L 7L 8e t 7e 2t. s 2 s + 6 s s 2 Then item 6 with c and ht) 8e t 7e 2t shows that [ [ L e s s + ]t) ut ) L s 2 s + 6 s + s 2 s + 6 ut ) 8e t ) 7e 2t )). ] t ) A Short Table of Laplace Transforms L[e at t n n! ]s) s a) n+ for s > a, L[e at s a cosbt)]s) s a) 2 + b 2 for s > a, L[e at sinbt)]s) b s a) 2 + b 2 for s > a, L[e at ht)]s) Hs a) where Hs) L[ht)]s), L[t n ht)]s) ) n H n) s) where Hs) L[ht)]s), L[ut c)ht c)]s) cs Hs) where Hs) L[ht)]s) and u is the step function.