Math Exam 3 Solutions

Transcription

1 Math 6 - Exam 3 Solutions Thursday, July 3rd, 0 Recast the following higher-order differential equations into first order systems If the equation is linear, be sure to give the coefficient matrix At and the forcing ft a + t y + e t y sin ty cos t b v + cos v v v v + v 3 0 a We define x y x x y x 3 y taking its derivative and using the ODE, we find x et x x 3 sin t x +t sin t +t 0 et x +t + +t cos t +t x cos t +t b Define Therefore x x v x x x 3 v v x v x x 3 x cos x 3 x + x x x 3 Correction in red Answers with or without the correction are accepted Consider the following initial value problem y + 3y 6y ft, y0, y 0 where ft { t 0 t < te t+ t

2 Using the Laplace transform method, find the Laplace transformed solution Y s L[y]s You do not need to find the inverse Laplace transform We first Laplace transform the differential equation using L[y ]s s Y s sy0 y 0 L[y ]s sy s y0 We will denote F s L[f]s the Laplace transform of the forcing The differential equation becomes s + 3s 6Y s s F s, which can be solved for Y s, Y s s F s s + 3s 6 To derermine F s, we write ft using step functions ut, ft tut ut + te t+ ut t + te t+ + t ut The first term is easy to compute using the table L[ t]s s s, the second term is of the form jt ut, where Using the tables we see that Js L[j]s jt t + e t + t s + + s + + s and therefore using the shift property from the table, L[ut jt ]s e s s + + s + + s It follows that F s is s s + e s s + + s + + s and so Y s is Y s s s s + e s s + 3s 6 + s+ s+ + s

3 3 Find a general solution to each of the following systems d x 3 3 x a dt y y d x 7 x b dt y 3 y a Solution The characteristic polynomial of A p A z z z 5 z + 3z 5 Therefore the eigenvalues of A are 3, 5 Since 6 3 A + 3I, A 5I 3 3 is given by 3, 6 we can read off the eigenpairs as 3,, 3 5, A general solution is therefore xt c e 3t + c e 5t 3 Solution The characteristic polynomial of A p A z z z 5 z 3 3 is given by Therefore µ, δ 6 > 0, ν 6 Therefore the matrix exponential is e ta e t cosh ti + A I cosh t 0 3 e t + e t 0 cosh t 3 e t cosh t + cosh t 3

4 A general solution is then given by xt e ta c e t cosh t + c e t cosh t + 3 cosh t + c e t b Solution The characteristic polynomial of A p A z z 0z + 5 z 5 Therefore there is one eigenvalue λ 5 We see that A 5I, and so the eigenpair is 5, c c 3 cosh t 7 is given by 3 One solutions is then given by x t e 5t To get another solution, we choose a vector w orthogonal to, which can be gotten as a multiple of the top row of the matrix A 5I shown above Another solution is then given by A general solution is then given by x t e 5t w + te 5t A 5Iw e 5t + te 5t e 5t 5te 5t e 5t + 0te 5t + 5t e 5t + 0t xt c e 5t + 5t + c e 5t + 0t

5 Solution The characteristic polynomial of A p A z z 0z + 5 z 5 7 is given by 3 The only eigenvalue is therefore 5 Using the general formula, e ta e 5t I + ta 5I 0 e 5t + t 0 + t t e 5t t t Therefore a general solution is + t xt e ta c e 5t t t + c e 5t t Consider the matrices A 3i + i, B 3 i 7 Compute the following matricies a A b AB c B d A B a The hermition transpose of A is A 3i 3 + i i b The product is 3i + i AB 3 i 7 6i i i + + i 6 i + i i + i 0 i 6 i 5

6 c Note that B has an inverse since detb det Therefore the inverse of B is B detb d The difference is given by 3i + i 3 i i 3 + i i 0 5 Correction shown in red 0 points given for this mistake cos t Given that e ta sin t sin t + cos t a Find A b Find the solution to the initial value problem d x x x0 A, dt y y y0 a We may find A using the fact that e ta is a fundamental solution, and satisfies e ta Ae ta Multiplying both sides of this equation on the right by e ta and using the fact that for the matrix exponential we have e ta e ta, gives A e ta e ta sin t cos t cos t sin t cos t sin t sin t cos t 0 sin t + cos t cos t + sin t b The solution to the initial value problem is just xt e ta x I cos t sin t sin t cos t cos t sin t sin t cos t 6

7 6 Consider the following MATLAB code [y; y3; -y + y^ - y^]; [T,Y] ode5f,[0 ], [, -, 0]; which solves an initial value problem for a system of differential equations a What the is first order differential system that is being solved? Be sure to state the initial conditions b This problem is also solving an initial value problem for a third order differential equation Give a third order differential equation along with initial conditions that is being sovled a The code is solving the following first order linear system y d y y 0 y y 3, y 0 dt y 3 y + y y y b Choosing y y, we see that the first two equations of the system say that y y, y 3 y and the last equation is the following third order equation y y + y y with initial conditions y0, y 0, y Find the inverse Laplace transform of the following functions a F s s + 5 3s b F s s s 6 c F s s e 3s s s + 5 a From the table we see that L[te 5t ]s s + 5 Multiplying through by we conclude that [ ] L t te 5t s + 5 7

8 b The denominator factors as s 3s +, therefore the partial fraction decomposition is 3s s s 6 3s s 3s + 9/5 s 3 + 6/5 s + From the table, we see that therefore L[e 3t ]s s 3, L[e t ]s s +, [ ] L 3s t 9 s s 6 5 e3t e t c Completing the square in the denominator gives s + We know from the table that L[e t s cos t]s s + Next, using the shift property, with jt e t cos t, the table shows us that L[ut 3e t 3 cos t 3]s e 3s s s + Therefore we conclude that [ ] L e 3s s ut 3e t 3 cos t 3 s s Find eigenpairs for the following matrices 3 a A 5 b A a The characteristic polynomial is p A z z z + z Therefore there is one eigenvalue λ An eigenvector can be gotten from the matrix A I, whence the eigenpair is, 8

9 b The characterisitc polynomial is p A z z + 6 Which has two roots ±i Therefore the eigenvalues are λ ±i We compute the following matricies i 5 + i 5 A ii, A + ii, i + i whereby we can read off the eigenpairs + i i,, i i, 9 Consider two interconnected tanks filled with brine salt water The first tank contains 60 liters and the second tank contans 90 liters Brine with a concentration of 9 grams of salt per liter flows into the first tank at a rate of 6 liters per hour The well-stirred brine flows from the first tank into the second tank at a rate of 5 liters per hour, from the second tank into the first tank at a rate 3 liters per hour, from the first into a drain at a rate of liters per hour, and from the second into a drain at a rate of liters per hour At t 0 there are 90 grams of salt in the first tank and 30 grams of salt in the second tank Give an initial value problem that governs the amount of salt in each tank as a function of time You do not need to solve this problem Let S t, S t denote the mass of salf the first and second tank respectively The volume flow rate into and out of each tank are such that the volume of each tank remains, in particular the volume of the first tank is V 60 L and the volume of the second tank is V 90 L The rate of change of salt in the first tank is then ds dt S 90 5S 60 S 60 while for the second tank ds dt 5S 60 3S 90 S 90 with initial conditions S 0 90, S

10 The following table of Laplace Transforms has been included for your convenience L[e at t n n! ]s s a n+ for s > a, L[e at s a cos bt]s s a + b for s > a, L[e at b sin bt]s s a + b for s > a, L[e at jt]s Js a where Js L[jt]s, L[t n jt]s n J n s L[ut cjt c]s e cs Js where Js L[jt]s, where Js L[jt]s 0