Solutions to Final Exam Sample Problems, Math 246, Spring 2011

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1 Solutions to Final Exam Sample Problems, Math 246, Spring 2 () Consider the differential equation dy dt = (9 y2 )y 2 (a) Identify its equilibrium (stationary) points and classify their stability (b) Sketch how solutions move in the interval 5 y 5 (its phase-line portrait) (c) If y() =, how does the solution y(t) behave as t? Solution (a,b) The right-hand side factors as ( + y)( y)y 2 The stationary solutions are y =, y =, and y = A sign analysis of ( + y)( y)y 2 shows that the phase-line portrait for this equation is therefore + + y unstable semistable stable Solution (c) The phase-line shows that if y() = then y(t) as t (2) Solve (possibly implicitly) each of the following initial-value problems Identify their intervals of definition (a) dy dt + 2ty + t 2 = t2, y() = (b) dy dx + ex y + 2x 2y + e x =, y() = Solution (a) This equation is linear and is already in normal form An integrating factor is ( t ) 2s exp + s ds = exp ( log( + t 2 ) ) = + t 2, 2 so that the integrating factor form is d ( ( + t 2 )y ) = ( + t 2 )t 2 = t 2 + t 4 dt Integrate this to obtain ( + t 2 )y = t + 5 t5 + c The initial condition y() = implies that c = (+ 2 ) 5 5 = Therefore y = + t + 5 t5 + t 2 This solution exists for every t, so its interval of definition is (, ) Remark Because this equation is linear, you can see that the interval of definition of its solution is (, ) without solving it because both its coefficient and forcing are continuous over (, ) Solution (b) The initial-value problem is dy dx + ex y + 2x 2y + e x =, y() =

2 2 Express this equation in the differential form This differential form is exact because (e x y + 2x) dx + (2y + e x ) dy = y (e x y + 2x) = e x = x (2y + e x ) = e x We can therefore find H(x, y) such that x H(x, y) = e x y + 2x, y H(x, y) = 2y + e x The first equation implies H(x, y) = e x y + x 2 + h(y) Plugging this into the second equation gives e x + h (y) = 2y + e x, which yields h (y) = 2y Taking h(y) = y 2, the general solution is e x y + x 2 + y 2 = c The initial condition y() = implies that c = e = Therefore y 2 + e x y + x 2 = The quadratic formula then yields the explicit solution y = ex + e 2x 4x 2 2 Here the positive square root is taken because that solution satisfies the initial condition Its interval of definition is the largest interval (x L, x R ) containing the initial time over which e 2x > 4x 2 You cannot find the endpoints of this interval explicitly () Consider the following MATLAB function M-file function [t,y] = solveit(ti, yi, tf, n) t = zeros(n +, ); y = zeros(n +, ); t() = ti; y() = yi; h = (tf - ti)/n; for i = :n t(i + ) = t(i) + h; y(i + ) = y(i) + h*((t(i))ˆ4 + (y(i))ˆ2); end Suppose that the input values are ti =, yi =, tf = 5, and n = 4 (a) What is the initial-value problem being approximated numerically? (b) What is the numerical method being used? (c) What is the timestep (step size)? (d) What are the output values for the first two approximations, y(2) and y()? Solution (a) The initial-value problem being approximated numerically is dy dt = t4 + y 2, y() = Solution (b) The forward Euler (explicit Euler) method is being used Solution (c) The time step is tf ti h = n = 5 4 = 4 4 = =

3 Solution (d) By carrying out the for loop in the MATLAB code for i = and i = 2 you obtain the output values t(2) = t() + h = + =, y(2) = y() + h*((t())ˆ4 + (y())ˆ2) = + ( ) = + 2 = 2 t() = t(2) + h = + = 2, y() = y(2) + h*((t(2))ˆ4 + (y(2))ˆ2) = 2 + ( () 4 + (2) 2) You DO NOT have to work out the arithmetic to compute y()! If you did then you would obtain y() = 494 (4) Give an explicit real-valued general solution of the following equations (a) y 2y + 5y = te t + cos(2t) (b) u u u = te 2t (c) v + 9v = cos(t) Solution (a) This is a constant coefficient, nonhomogeneous, linear equation Its characteristic polynomial is p(z) = z 2 2z + 5 = (z ) = (z ) This has the conjugate pair of roots ± i2, which yields a general solution of the associated homogeneous problem y H (t) = c e t cos(2t) + c 2 e t sin(2t) A particular solution y P (t) can be found by either the method of Key Identity Evaluations or the method of Undetermined Coefficients The characteristics of the forcing terms te t and cos(2t) are r + is = and r + is = i2 respectively Because these characteristics are different, they should be treated separately Key Indentity Evaluations The forcing term t e t has degree d = and characteristic r + is =, which is a root of p(z) of multiplicity m = Because m = and m + d =, you need the Key Identity and its first derivative L(e zt ) = (z 2 2z + 5)e zt, L(t e zt ) = (z 2 2z + 5)t e zt + (2z 2) e zt Evaluate these at z = to find L(e t ) = 4e t and L(t e t ) = 4t e t Dividing the second of these equations by 4 yields L( 4 t et ) = t e t, which implies y P (t) = 4 t et The forcing term cos(2t) has degree d = and characteristic r + is = i2, which is a root of p(z) of multiplicity m = Because m = m + d =, you only need the Key Identity, L(e zt ) = (z 2 2z + 5)e zt Evaluating this at z = i2 to find L(e i2t ) = ( i4)e i2t and dividing by i4 yeilds e i2t L = e i2t i4

4 4 Because cos(2t) = Re(e i2t ), the above equation implies e i2t ( + i4)e i2t y P2 (t) = Re = Re i = 7 Re( ( + i4)e i2t) = 7( cos(2t) 4 sin(2t) ) Combining these two particular solutions with the general solution of the associated homogeneous problem found earlier yields the general solution y = y H (t) + y P (t) + y P2 (t) = c e t cos(2t) + c 2 e t sin(2t) + 4 t et + 7 cos(2t) 4 7 sin(2t) Undetermined Coefficients The forcing term t e t has degree d = and characteristic r + is =, which is a root of p(z) of multiplicity m = Because m = and m + d =, you seek a particular solution of the form Because one sees that y P (t) = A t e t + A e t y P (t) = A t e t + (A + A )e t, y P (t) = A t e t + (2A + A )e t, Ly P (t) = y P (t) 2y P (t) + 5y P(t) = ( A t e t + (2A + A )e t) 2 ( A t e t + (A + A )e t) + 5 ( A t e t + A e t) = 4A t e t + 4A e t Setting 4A t e t + 4A e t = t e t, we see that 4A = and 4A =, whereby A = 4 and A = Hence, a particular solution is y P (t) = 4 t et The forcing term cos(2t) has degree d = and characteristic r + is = i2, which is a root of p(z) of multiplicity m = Because m = and m + d =, you seek a particular solution of the form Because one sees that y P2 (t) = A cos(2t) + B sin(2t) y P2(t) = 2A sin(2t) + 2B cos(2t), y P2 (t) = 4A cos(2t) 4B sin(2t), Ly P2 (t) = y P2 (t) 2y P2 (t) + 5y P2(t) = ( 4A cos(2t) 4B sin(2t) ) 2 ( 2A sin(2t) + 2B cos(2t) ) + 5 ( A cos(2t) + B sin(2t) ) = (A 4B) cos(2t) + (B + 4A) sin(2t) Setting (A 4B) cos(2t) + (B + 4A) sin(2t) = cos(2t), we see that A 4B =, B + 4A =

5 This system can be solved by any method you choose to find A = and B = 4 7 whereby a particular solution is y P2 (t) = 7 cos(2t) 4 7 sin(2t) Combining these two particular solutions with the general solution of the associated homogeneous problem found earlier yields the general solution y = y H (t) + y P (t) + y P2 (t) = c e t cos(2t) + c 2 e t sin(2t) + 4 t et + 7 cos(2t) 4 7 sin(2t) Solution (b) The equation is u u u = te 2t This is a constant coefficient, nonhomogeneous, linear equation Its characteristic polynomial is p(z) = z 2 z = (z 5)(z + 2) This has the two real roots 5 and 2, which yields a general solution of the associated homogeneous problem u H (t) = c e 5t + c 2 e 2t A particular solution u P (t) can be found by either the method of Key Identity Evaluations or the method of Undetermined Coefficients Key Indentity Evaluations The forcing term t e 2t has degree d = and characteristic r+is = 2, which is a root of p(z) of multiplicity m = Because m = and m + d = 2, you will need the first and second derivative of the Key Identity, which are computed from the Key Identity as L(e zt ) = (z 2 z ) e zt, L(t e zt ) = (z 2 z ) t e zt + (2z ) e zt, L(t 2 e zt ) = (z 2 z ) t 2 e zt + 2(2z ) t e zt + 2 e zt Evaluate the last two of these at z = 2 to find By adding 2 7 L(t e 2t ) = 7e 2t, L(t 2 e 2t ) = 4t e 2t + 2e 2t of the first to the second you get By dividing this by 4 you obtain L ( t 2 e 2t t e 2t) = 4t e 2t L ( 4 t2 e 2t 2 98 t e 2t) = t e 2t, whereby a particular solution is u P (t) = 4 t2 e 2t 2 98 t e 2t A general solution is therefore u(t) = u H (t) + u P (t) = c e 5t + c 2 e 2t 4 t2 e 2t 2 98 t e 2t 5 7,

6 6 Undetermined Coefficients The forcing term t e 2t has degree d = and characteristic r + is = 2, which is a root of p(z) of multiplicity m = Because m = and m + d = 2, you seek a particular solution of the form Because u P (t) = A t 2 e 2t + A t e 2t u P (t) = 2A t 2 e 2t + (2A 2A )t e 2t + A e 2t, u P(t) = 4A t 2 e 2t + ( 8A + 4A )t e 2t + (2A 4A )e 2t, one sees that Lu P (t) = u P (t) u P (t) u P (t) = ( 4A t 2 e 2t + ( 8A + 4A )t e 2t + (2A 4A )e 2t) ( 2A t 2 e 2t + (2A 2A )t e 2t + A e 2t) ( A t 2 e 2t + A t e 2t) = 4A t e 2t + (2A 7A )e 2t By setting 4A t e 2t + (2A 7A )e 2t = t e 2t we see that 4A =, 2A 7A = This system can be solved by any method you choose to find A = and A 4 = 2 whereby a particular solution is u P (t) = 4 t2 e 2t 2 98 t e 2t A general solution is therefore u(t) = u H (t) + u P (t) = c e 5t + c 2 e 2t 4 t2 e 2t 2 98 t e 2t Solution (c) The equation is v + 9v = cos(t) This is a constant coefficient, nonhomogeneous, linear equation Its characteristic polynomial is p(z) = z = z This has the conjugate pair of roots ±i, which yields a general solution of the associated homogeneous problem v H (t) = c cos(t) + c 2 sin(t) A particular solution v P (t) can be found by either the method of Key Identity Evaluations or the method of Undetermined Coefficients Key Indentity Evaluations The forcing term cos(t) has degree d = and characteristic r + is = i, which is a root of p(z) of multiplicity m = Because m = and m + d =, you need the first derivative of the Key Identity, which is found as L(e zt ) = (z 2 + 9) e zt, L(t e zt ) = (z 2 + 9) t e zt + 2z e zt 98,

7 7 Evaluate the first derivative of the Key Identity at z = i to find that L(t e it ) = i6 e it Because cos(t) = Re(e it ), upon dividing by i6 and taking the real part you see that a particular solution is v P (t) = Re ( i6 t eit) = Re ( i6 t (cos(t) + i sin(t))) = 6 t sin(t) A general solution is therefore v(t) = v H (t) + v P (t) = c cos(t) + c 2 sin(t) + 6 t sin(t) Undetermined Coefficients The forcing term cos(t) has degree d = and characteristic r + is = i, which is a root of p(z) of multiplicity m = Because m = and m + d =, you seek a particular solution of the form Because v P (t) = At cos(t) + Bt sin(t) v P(t) = At sin(t) + Bt cos(t)a cos(t) + B sin(t), v P (t) = 9At cos(t) 9Bt sin(t) 6A sin(t) + 6B cos(t), one sees that Lv P (t) = v P(t) + 9v P (t) = ( 9At cos(t) 9Bt sin(t) 6A sin(t) + 6B cos(t) ) + 9 ( At cos(t) + Bt sin(t) ) = 6A sin(t) + 6B cos(t) By setting 6A sin(t)+6b cos(t) = cos(t) we see that A = and B = 6, whereby a particular solution is A general solution is therefore v P (t) = 6 t sin(t) v(t) = v H (t) + v P (t) = c cos(t) + c 2 sin(t) + 6 t sin(t) Remark Because of the simple form of this equation, if you had tried to solve it by either the Green Function or Variation of Parameters method then integrals that arise are not too difficult However, it is not generally a good idea to use these methods for such problems because evaluating the integrals that arise often involve much more work that the methods shown above (5) Solve the following initial-value problems (a) w + 4w + 2w = 5e 2t, w() =, w () = 7 (b) y 4y + 4y = e2t + t, y() =, y () = 5 You should try to evaluate any definite integrals that arise Solution (a) This is a constant coefficient, nonhomogeneous, linear equation Its characteristic polynomial is p(z) = z 2 + 4z + 2 = (z + 2) = (z + 2)

8 8 This has the conjugate pair of roots 2 ± i4, which yields a general solution of the associated homogeneous problem w H (t) = c e 2t cos(4t) + c 2 e 2t sin(4t) A particular solution w P (t) can be found by either the method of Key Identity Evaluations or the method of Undetermined Coefficients Key Indentity Evaluations The forcing term 5e 2t has degree d = and characteristic r + is = 2, which is a root of p(z) of multiplicity m = Because m = and m + d =, you only need the Key Identity, Evaluate this at z = 2 to find that Upon multiplying this by 5 2 L(e zt ) = (z 2 + 4z + 2) e zt L(e 2t ) = ( )e 2t = 2e 2t you see that a particular solution is w P (t) = 5 2 e2t Undetermined Coefficients The forcing term 5e 2t has degree d = and characteristic r + is = 2, which is a root of p(z) of multiplicity m = You therefore seek a particular solution of the form Because you see that w P (t) = Ae 2t w P (t) = 2Ae2t, w P (t) = 4Ae2t, Lw P (t) = w P (t) + 4w P (t) + 2w P (t) = 4Ae 2t + 4(2Ae 2t ) + 2Ae 2t = 2Ae 2t By setting 2Ae 2t = 5e 2t, you see that A = 5, whereby a particular solution is 2 w P (t) = 5 2 e2t Solving the Initial-Value Problem By either method you find that a general solution is Because w(t) = w H (t) + w P (t) = c e 2t cos(4t) + c 2 e 2t sin(4t) e2t w (t) = 2c e 2t cos(4t) 4c e 2t sin(4t) 2c 2 e 2t sin(4t) + 4c 2 e 2t cos(4t) e2t, the initial conditions then yield = w() = c + 5 2, 7 = w () = 2c + 4c Upon solving this system we find that c = 9 and c 2 2 =, whereby the solution 2 of the initial-value problem is w(t) = 9 2 e 2t cos(4t) 2 e 2t sin(4t) e2t

9 9 Solution (b) The initial-value problem is y 4y + 4y = e2t + t, y() =, y () = 5 This is a constant coefficient, nonhomogeneous, linear equation in normal form Its characteristic polynomial is p(z) = z 2 4z + 4 = (z 2) 2 This has the double real root 2, which yields a general solution of the associated homogeneous problem y H (t) = c e 2t + c 2 t e 2t A particular solution y P (t) cannot be found by either the method of Key Identity Evaluations or the method of Undetermined Coefficients Rather, you must use either the Green Function or the Variation of Parameters method Green Function The associated Green function g(t) satisfies g 4g + 4g =, g() =, g () = A general solution of this equation is g(t) = c e 2t + c 2 t e 2t Because = g() = c, we see that g(t) = c 2 t e 2t Then g (t) = c 2 e 2t + 2c 2 t e 2t Because = g () = c 2, the Green function is g(t) = t e 2t The particular solution y P (t) that satisfies y P () = y P () = is given by Because t t y P (t) = you find that t = e 2t t g(t s) e2s + s ds = t s + s ds = e2t t + s ds = log( + s) t s + s ds = t t t ds = t log + s 2t 2s e2s (t s)e + s ds t ds e2t + s s + s ds ( + t = log( + t) log() = log + t, y P (t) = e 2t t log( + t) e2t( t log( + t)) A general solution of the equation is therefore y(t) = c e 2t + c 2 t e 2t + y P (t) Because y (t) = 2c e 2t + 2c 2 t e 2t + c 2 e 2t + y P(t), and because y P () = y P () =, the initial conditions imply = y() = c, 5 = y () = 2c + c 2 ),

10 You find that c = and c 2 = 5, whereby the solution of the initial-value problem is y(t) = 5t e 2t + e 2t t log( + t) e2t( t log( + t)) Variation of Parameters The equation is already in normal form We therefore seek a particular solution of the form such that y P (t) = e 2t u (t) + t e 2t u 2 (t), e 2t u (t) + t e2t u 2 (t) =, 2e 2t u (t) + (2t e 2t + e 2t )u 2(t) = e2t + t This system can be solved to find that u (t) = t + t, u 2(t) = + t These can be integrted to obtain t u (t) = + t dt = + t dt = t + log( + t) + c, u 2 (t) = + t dt = log( + t) + c 2, whereby a general solution is y(t) = c e 2t e 2t( t log( + t) ) + c 2 t e 2t + t e 2t log( + t) Because y (t) = 2c e 2t 2e 2t( t log( + t) ) ( e 2t ) + t + 2c 2 t e 2t + c 2 e 2t + 2t e 2t log( + t) + e 2t log( + t) + t e 2t + t, the initial conditions imply that = y() = c + log(), 5 = y () = 2c + 6 log() + c 2 + log() You can solve this system to find that c = log() and c 2 = 5 log() The solution of the initial-value problem is therefore y(t) = log()e 2t e 2t( t log( + t) ) + ( 5 log() ) t e 2t + t e 2t log( + t) (6) When a mass of 2 kilograms is hung vertically from a spring, it stretches the spring 5 meters (Gravitational acceleration is 98 m/sec 2 ) At t = the mass is set in motion from meters below its equilibrium (rest) position with a upward velocity of 2 m/sec It is acted upon by an external force of 2 cos(5t) Neglect drag and assume that the spring force is proportional to its displacement Formulate an initial-value problem that governs the motion of the mass for t > (DO NOT solve this initialvalue problem; just write it down!)

11 Solution Let h(t) be the displacement (in meters) of the mass from its equilibrium (rest) position at time t (in seconds), with upward displacements being positive The governing initial-value problem then has the form m d2 h dt 2 + kh = 2 cos(5t), h() =, h () = 2, where m is the mass and k is the spring constant The problem says that m = 2 kilograms The spring constant is obtained by balancing the weight of the mass (mg = 2 98 Newtons) with the force applied by the spring when it is stetched 5 m This gives k 5 = 2 98, or k = = 4 98 Newtons/m The governing initial-value problem is therefore 2 d2 h dt h = 2 cos(5t), h() =, h () = 2 Had you chosen downward displacements to be positive then the sign of the initial data would change! You should make your convention clear! (7) Give an explicit general solution of the equation h + 2h + 5h = Sketch a typical solution for t If this equation governs a damped spring-mass system, is the system over, under, or critically damped? Solution This is a constant coefficient, homogeneous, linear equation Its characteristic polynomial is p(z) = z 2 + 2z + 5 = (z + ) This has the conjugate pair of roots ± i2, which yields a general solution When c 2 + c 2 2 h(t) = c e t cos(2t) + c 2 e t sin(2t) > this can be put into the amplitute-phase form h(t) = Ae t cos(2t δ), where A > and δ < 2π are determined from c and c 2 by A = c 2 + c2 2, cos(δ) = c A, sin(δ) = c 2 A In other words, (A, δ) are the polar coordinates for the point in the plane whose Cartesian coordinates are (c, c 2 ) The sketch should show a decaying oscillation with amplitude Ae t and quasiperiod 2π = π A sketch might be given during the 2 review session The equation governs an under damped spring-mass system because its characteristic polynomial has a conjugate pair of roots

12 2 (8) Find the Laplace transform Y (s) of the solution y(t) to the initial-value problem where y + 4y + 8y = f(t), y() = 2, y () = 4 f(t) = { 4 for t < 2, t 2 for 2 t You may refer to the Laplace table in the book (DO NOT take the inverse Laplace transform to find y(t); just solve for Y (s)!) Solution The Laplace transform of the initial-value problem is where L[y ](s) + 4L[y ](s) + 8L[y](s) = L[f](s), L[y](s) = Y (s), L[y ](s) = sy (s) y() = sy (s) 2, L[y ](s) = s 2 Y (s) sy() y () = s 2 Y (s) 2s 4 To compute L[f](s), first write f as where f(t) = ( u(t 2) ) 4 + u(t 2)t 2 = 4 u(t 2)4 + u(t 2)t 2 = 4 + u(t 2)(t 2 4) = 4 + u(t 2)j(t 2), j(t) = (t + 2) 2 4 = t 2 + 4t Referring to the table of Laplace transforms in the book, item with c = 2 and f(t) = j(t), item, and item with n = and n = 2 then show that L[f](s) = 4L[](s) + L [ u(t 2)j(t 2) ] (s) = 4L[](s) + e 2s L [ j(t) ] (s) = 4L[](s) + e 2s L[4t + t 2 ](s) = 4L[](s) + 4e 2s L[t](s) + e 2s L[t 2 ](s) = 4 s + 4e 2s s 2 + e 2s 2 s The Laplace transform of the initial-value problem then becomes ( s 2 Y (s) 2s 4 ) + 4 ( sy (s) 2 ) + 8Y (s) = 4 s + e 2s 4 s 2 + e 2s 2 s, which becomes (s 2 + 4s + 8)Y (s) 2s 2 = 4 s + e 2s 4 s 2 + e 2s 2 s Hence, Y (s) is given by Y (s) = (2s s s 2 + 4s s e 2s + 2s ) 2 e 2s

13 (9) Find the function y(t) whose Laplace transform Y (s) is given by e s 4 (a) Y (s) = s 2 6s + 5, (b) Y (s) = e 2s s s 2 + 4s + 8 You may refer to the table in Section 62 of the book Solution (a) The denominator factors as (s 5)(s ), so the partial fraction decomposition is 4 s 2 6s + 5 = 4 (s 5)(s ) = s 5 s Referring to the table of Laplace transforms in the book, item with n = and a = 5, and with n = and a = gives [ ] [ ] L (t) = e 5t, L (t) = e t, s 5 s whereby [ ] [ L 4 (t) = L s 2 6s + 5 s 5 ] = e 5t e t s It follows from item with c = and f(t) = e 5t e t that [ ] y(t) = L [Y (s)](t) = L e s 4 (t) s 2 6s + 5 [ ] = u(t )L 4 (t ) = u(t ) ( e 5(t ) e t ) s 2 6s + 5 Solution (b) The denominator does not have real factors The partial fraction decomposition is s s 2 + 4s + 8 = s (s + 2) = s + 2 (s + 2) (s + 2) Referring to the table of Laplace transforms in the book, items and 9 with a = 2 and b = 2 give [ [ ] L s + 2 ](t) = e 2t cos(2t), L 2 (t) = e 2t sin(2t), (s + 2) (s + 2) whereby [ [ [ L s ](t) = L s + 2 ](t) L s 2 + 4s + 8 (s + 2) = e 2t( cos(2t) sin(2t) ) 2 (s + 2) ] (t) It follows from item with c = 2 and f(t) = e 2t( cos(2t) sin(2t) ) that [ [ ] y(t) = L [Y (s)](t) = L e 2s s ](t) = u(t 2)L s (t 2) s 2 + 4s + 8 s 2 + 4s + 8 = u(t 2)e 2(t 2)( cos(2(t 2)) sin(2(t 2)) )

14 4 () Consider two interconnected tanks filled with brine (salt water) The first tank contains 8 liters and the second contains liters Brine flows with a concentration of grams of salt per liter flows into the first tank at a rate of 2 liters per hour Well stirred brine flows from the first tank to the second at a rate of 6 liters per hour, from the second to the first at a rate of 4 liters per hour, and from the second into a drain at a rate of 2 liters per hour At t = there are 7 grams of salt in the first tank and 25 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 8 liters of brine in the first tank and 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 dt = 2 + S 2 4 S 8 6, S () = 7, ds 2 dt = S 8 6 S 2 4 S 2 2, S 2() = 25 You could leave the answer in the above form It can however be simplified to ds dt = , S () = 7, ds 2 dt = S 4 S 5 2, S 2 () = 25 ( t () Consider the real vector-valued functions x (t) =, x t) 2 (t) = + t 4 (a) Compute the Wronskian W[x,x 2 ](t) (b) Find A(t) such that x, x 2 is a fundamental set of solutions to the linear system x = A(t)x (c) Give a general solution to the system you found in part (b) Solution (a) The Wronskian is given by t W[x,x 2 ](t) = det t + t 4 = ( + t 4 ) t t = + t 4 t 4 = Solution (b) Set Ψ(t) = ( x (t) x 2 (t) ) t = t + t 4 If x (t) and x 2 (t) satisfy the linear system then the matrix-valued function Ψ(t) must satisfy dψ (t) = A(t)Ψ(t) dt

15 Because det(ψ(t)) = W[x,x 2 ](t) =, we see that Ψ(t) is a fundamental matrix of the linear system with A(t) given by A(t) = dψ t 2 + t 4 t dt (t)ψ(t) = 4t t = t t 2 t t t 4 t = 2 t 4 t Therefore x (t), x 2 (t) is a fundamental set of solutions for the linear system whose coefficient matrix is this A(t) Solution (c) Because x (t), x 2 (t) is a fundamental set of solutions for the linear system whose coefficient matrix is A(t), a general solution is given by t x(t) = c x (t) + c 2 x 2 (t) = c + c t 2 + t 4 (2) Give a general real vector-valued solution of the linear planar system x = Ax for (a) A =, (b) A = 4 2 Solution (a) The characteristic polynomial of A is p(z) = z 2 tr(a)z + det(a) = z 2 6z 6 = (z ) 2 25 = (z ) The eigenvalues of A are the roots of this polynomial, which are ± 5, or simply 2 and 8 One therefore has [ e ta = e t cosh(5t)i + sinh(5t) ] (A I) 5 = e [cosh(5t) t + sinh(5t) ] ( cosh(5t) + = e t sinh(5t) 4 sinh(5t) ) sinh(5t) cosh(5t) sinh(5t) 5 5 A general solution is therefore given by ( x(t) = e ta c cosh(5t) + = c c e t sinh(5t) ) ( sinh(5t) + c 2 e t 4 sinh(5t) ) 5 cosh(5t) sinh(5t) 5 5 Alternative Solution (a) The characteristic polynomial of A is p(z) = z 2 tr(a)z + det(a) = z 2 6z 6 = (z ) 2 25 = (z ) The eigenvalues of A are the roots of this polynomial, which are ± 5, or simply 2 and 8 Because A + 2I =, A 8I =,

16 6 you see that A has the eigenpairs ( ) 2, 2, ( ( 2 8, )) Form these eigenpairs you construct the solutions 2 x (t) = e 2t, x 2 2 (t) = e 8t, A general solution is therefore x(t) = c x (t) + c 2 x 2 (t) = c e 2t 2 + c 2 e 8t ( 2 ) Solution (b) The characteristic polynomial of A is p(z) = z 2 tr(a)z + det(a) = z 2 2z + 5 = (z ) = (z ) The eigenvalues of A are the roots of this polynomial, which are ±i2 One therefore has [ e ta = e t cos(2t)i + sin(2t) ] (A I) 2 = e [cos(2t) t + sin(2t) ] cos(2t) sin(2t) = e t sin(2t) cos(2t) A general solution is therefore given by x(t) = e ta c cos(2t) sin(2t) = c c e t + c 2 sin(2t) 2 e t cos(2t) Alternative Solution (b) The characteristic polynomial of A is p(z) = z 2 tr(a)z + det(a) = z 2 2z + 5 = (z ) = (z ) The eigenvalues of A are the roots of this polynomial, which are ± i2 Because i2 2 i2 2 A ( + i2)i =, A ( i2)i =, 2 i2 2 i2 we see that A has the eigenpairs ( ( + i2,, i)) ( i2, ) i Because cos(2t) + i sin(2t) e (+i2)t = e i t, sin(2t) + i cos(2t) two real solutions of the system are cos(2t) sin(2t) x (t) = e t, x sin(2t) 2 (t) = e t cos(2t)

17 7 A general solution is therefore cos(2t) x(t) = c x (t) + c 2 x 2 (t) = c e t sin(2t) sin(2t) + c 2 e t cos(2t) () What answer will be produced by the following MATLAB command? >> A = [ 4; 2]; [vect, val] = eig(sym(a)) You do not have to give the answer in MATLAB format Solution The MATLAB command will produce the eigenpairs of A = The characteristic polynomial of A is p(z) = z 2 tr(a)z + det(a) = z 2 z = (z 5)(z + 2), so its eigenvalues are 5 and 2 Because 4 4 A 5I =, A + 2I = you can read off that the eigenpairs are ( ( 5,, )) ( 2, ) 4 4, (4) A real 2 2 matrix A has eigenvalues 2 and with associated eigenvectors ( and ) 2 (a) Give a general solution to the linear planar system x = Ax (b) Compute e ta (c) Sketch a phase-plane portrait for this system and identify its type Classify the stability of the origin (Carefully mark all sketched trajectories with arrows!) Solution (a) Use the given eigenpairs to construct the solutions x (t) = e 2t, x 2 (t) = e t 2 A general solution is therefore x(t) = c x (t) + c 2 x 2 (t) = c e 2t ( ) + c 2 e t 2 Solution (b) The matrix A can be diagonalized as A = VDV where 2 V =, D =, V 2 = 2 7 Then e ta = Ve td V = 7 = 7 ( e 2t 2 ) ( e 2t e t 2 e 2t 2e t ) 2 e t = 7 ( 6e 2t + e t e 2t e t 2e 2t 2e t e 2t + 6e t )

18 8 Alternative Solution (b) By part (a) a fundamental matrix is Ψ(t) = ( x (t) x 2 (t) ) e 2t e = t e 2t 2e t Then ) ( e e ta = Ψ(t)Ψ() 2t e = t e 2t 2e t 2 = e 2t e t 2 7 e 2t 2e t = 6e 2t + e t e 2t e t 7 2e 2t 2e t e 2t + 6e t Solution (c) The matrix A has two real eigenvalues of opposite sign The origin is therefore a saddle and is thereby unstable There is one trajectory moves away from (, ) along each half of the line x = y, and one trajectory moves towards(, ) along each half of the line y = 2x (These are the lines of eigenvectors) Every other trajectory sweeps away from the line y = 2x and towards the line x = y A phase-plane portrait might be sketched during the review session (5) Solve the initial-value problem x = Ax, x() = x I and describe how its solution behaves as t for the following A and x I (a) A =, x 5 7 I = (b) A =, x 5 2 I = Solution (a) The characteristic polynomial of A = is 5 7 p(z) = z 2 tr(a)z + det(a) = z 2 + 4z + 29 = (z + 2) The eigenvlues of A are therefore 2 ± i5 Then [ e ta = e 2t cos(5t)i + sin(5t) ] (A + 2I) 5 = e [cos(5t) 2t + sin(5t) ] cos(5t) + sin(5t) 2 sin(5t) = e 2t sin(5t) cos(5t) sin(5t) The solution of the initial-value problem is therefore cos(5t) + sin(5t) 2 sin(5t) x(t) = e ta x I = e 2t sin(5t) cos(5t) sin(5t) 2 cos(5t) + sin(5t) = e 2t 2 cos(5t) + sin(5t) This solution decays to zero as t

19 Alternative Solution (a) The characteristic polynomial of A = p(z) = z 2 tr(a)z + det(a) = z 2 + 4z + 29 = (z + 2) The eigenvlues of A are therefore 2 ± i5 Because 5 i5 A ( 2 + i5)i =, 5 5 i5 you can read off that A has the conjugate eigenpairs ( ) ( ) + i i 2 + i5,, 2 i5, Because + i e 2t+i5t = e 2t( cos(5t) + i sin(5t) ) + i a fundamental set of real solutions is cos(5t) sin(5t) x (t) = e 2t cos(5t) A fundamental matrix Ψ(t) is then given by Because = e 2t ( cos(5t) sin(5t) + i cos(5t) + i sin(5t) cos(5t) i sin(5t) is 5 7 ), cos(5t) + sin(5t), x 2 (t) = e 2t sin(5t) Ψ(t) = ( x (t) x 2 (t) ) ( cos(5t) sin(5t) cos(5t) + sin(5t) = e 2t cos(5t) sin(5t) Ψ() = = =, you see that cos(5t) sin(5t) cos(5t) + sin(5t) e ta = Ψ(t)Ψ() = e 2t cos(5t) sin(5t) cos(5t) + sin(5t) 2 sin(5t) = e 2t sin(5t) cos(5t) sin(5t) The solution of the initial-value problem is therefore cos(5t) + sin(5t) 2 sin(5t) x(t) = e ta x I = e 2t sin(5t) cos(5t) sin(5t) 2 cos(5t) + sin(5t) = e 2t 2 cos(5t) + sin(5t) This solution decays to zero as t Remark After you have constructed the fundmental set of solutions x (t) and x 2 (t), you could also have solved the initial-value problem by finding constants c and c 2 such that x(t) = c x (t) + c 2 x 2 (t) satisfies the initial condition Had you done this using the x (t) and x 2 (t) constructed above, you would have found that c = 2 and c 2 = ) 9

20 2 Solution (b) The characteristic polynomial of A = 8 5 is 5 2 p(z) = z 2 tr(a)z + det(a) = z 2 6z + 9 = (z ) 2 The ony eigenvalue of A is Then e ta = e t[ I + t (A I) ] [ = e t ( + 5t 5t = e t 5t 5t ] t 5 5 ) The solution of the initial-value problem is therefore ( + 5t 5t + 2t x(t) = e ta x I = e t = e 5t 5t) t + 2t This solution grows like 2t e t as t Alternative Solution (b) The characteristic polynomial of A = p(z) = z 2 tr(a)z + det(a) = z 2 6z + 9 = (z ) 2 The only eigenvalue of A is Because A I = 5 5, 5 5 you can read off that A has the eigenpair ( (, )) 8 5 is 5 2 You can use this eigenpair to construct the solution x (t) = e t A second solution can be constructed by x 2 (t) = e t w + t e t (A I)w, where w is any nonzero vector that is not an eigenvector associated with For example, taking w = ( )T yields 5 + 5t x 2 (t) = e t + t e t = e 5 5)( t 5t A fundamental matrix Ψ(t) is then given by Ψ(t) = ( x (t) x 2 (t) ) + 5t = e t 5t Because Ψ() = = =,

21 you see that e ta = Ψ(t)Ψ() = + 5t + 5t 5t = 5t 5t 5t The solution of the initial-value problem is therefore + 5t 5t + 2t x(t) = e ta x I = e t = e 5t 5t t + 2t This solution grows like 2t e t as t Remark After you have constructed the fundmental set of solutions x (t) and x 2 (t), you could also have solved the initial-value problem by finding constants c and c 2 such that x(t) = c x (t) + c 2 x 2 (t) satisfies the initial condition Had you done this using the x (t) and x 2 (t) constructed above, you would have found that c = and c 2 = 4 (6) Consider the nonlinear planar system x = y +, y = 4x x 2 (a) Find all of its equilibrium (critical, stationary) points (b) Find a nonconstant function H(x, y) such that every trajectory of the system satisfies H(x, y) = c for some constant c (c) Sketch a phase-plane portrait of the system by sketching contour lines of H(x, y) (Carefully mark all sketched trajectories with arrows!) (d) Classify the type and stability of each equilibrium (critical, stationary) point Solution (a) Stationary points satisfy = y +, = 4x x 2 = x(4 x) The top equation shows that y = while the bottom equation shows that either x = or x = 4 The stationary points of the system are therefore (, ), (4, ) Solution (b) The associated first-order equation is dy dx = 4x x2 y + This equation is separable, so can be integrated as y + dy = 4x x 2 dx, 2 whereby you find that You can thereby set (y + 2 )2 = 2x 2 x + c H(x, y) = 2 (y + )2 2x 2 + x

22 22 Alternative Solution (b) An alternative approach is to notice that x f(x, y) + y g(x, y) = x (y + ) + y (4x x 2 ) = The system is therefore Hamiltonian with H(x, y) such that y H(x, y) = y +, x H(x, y) = 4x x 2 Integrating the first equation above yields H(x, y) = 2 (y + )2 + h(x) Substituting this into the second equation gives h (x) = 4x x 2 Integrating this equation yields h(x) = 2x 2 + x, whereby H(x, y) = 2 (y + )2 2x 2 + x Solution (c) Solving H(x, y) = c for y, you see that trajectories lie on the curves y = ± 2 ( c + 2x 2 x), wherever c + 2x 2 x Each cubic in the family p c (x) = c + 2x 2 x has a local minimum at x = with value p c () = c and a local maximum at x = 4 with value p c (4) = c = c + (2 4 )42 = c = c + 2 On the side, sketch five of these cubics for c < 2, c = 2, 2 < c <, c =, and c > You can see those points x for which each of these cubics p c (x) is nonnegative A phase portrait is obtained by first sketching y = ± 2p c (x) over those points x for which each p c (x) is nonnegative, and then adding arrows to indicate the direction of the trajectories The arrows go to the right for y > and to the left for y < This might be illustrated during the review The curves y = ± 2p c (x) will hit the stationary point (, ) when c = This point will be a saddle, and therefore unstable The stationary point (4, ) is a isolated point on y = ± 2p c (x) with c = 2 This point will be a center point, and therefore stable Remark You can sketch a phase portrait with MATLAB as follows The values of H(x, y) at the stationary points (, ) and (4, ) are H(, ) = 2 ( + ) =, H(4, ) = 2 ( + ) = ( )42 = 2 6 = 2, You should then pick three values c, c, and c 5 such that c < 2 < c < < c 5 and use contour to plot the five level sets H(x, y) = 2, H(x, y) =, H(x, y) = c, H(x, y) = c, H(x, y) = c 5 Solution (d) As indicated above, a correct phase portrait will give you the answer to this part However, you can also get the answer without the phase portait as follows The Hessian matrix H(x, y) of second partial derivatives is H(x, y) = ( xx H(x, y) xy H(x, y) yx H(x, y) yy H(x, y) ) = ( 4 + 2x )

23 Evaluating this at the stationary points yields 4 H(, ) =, H(4, ) = 4 Because the matrix H(, ) is diagonal, you can easily see that its eigenvalues are 4 and Because these have different signs, the stationary point (, ) is a saddle and is therefore unstable Similarly, because the matrix H(4, ) is diagonal, you can easily see that its eigenvalues are 4 and Because these have the same sign, the stationary point (4, ) is a center an is therefore stable (7) Consider the nonlinear planar system x = 5y, y = x 4y x 2 (a) Find all of its equilibrium (critical, stationary) points (b) Compute the Jacobian matrix (derivative matrix, linearization coefficient matrix) at each equilibrium (critical, stationary) point (c) Classify the type and stability of each equilibrium (critical, stationary) point (d) Sketch a plausible global phase-plane portrait (Carefully mark all sketched trajectories with arrows!) Solution (a) Stationary points satisfy = 5y, = x 4y x 2 The first equation implies y =, whereby the second equation becomes = x x 2 = x( x), which implies either x = or x = All the stationary points of the system are therefore (, ), (, ) Solution (b) Because f(x, y) 5y = g(x, y) x 4y x 2, the matrix of partial derivatives is x f(x, y) y f(x, y) = x g(x, y) y g(x, y) 5 2x 4 Evaluating this matrix at each stationary point yields the coefficient matrices 5 5 A = at (, ), A = at (, ) 4 4 Solution (c) The coefficient matrix A at (, ) has eigenvalues that satisfy = det(zi A) = z 2 tr(a)z + det(a) = z 2 + 4z + 5 = (z + 2) The eigenvalues are thereby 2 ± i Because a 2 = >, the stationary point (, ) is therefore a counterclockwise spiral sink, which is asymptotically stable or attracting This is one of the generic types, so it describes the phase-plane portrait of the nonlinear system near (, ) 2

24 24 The coefficient matrix A at (, ) has eigenvalues that satisfy = det(zi A) = z 2 tr(a)z + det(a) = z 2 + 4z 5 = (z + 2) 2 2 The eigenvalues are thereby 2±, or simply 5 and The stationary point (, ) is therefore a saddle, which is unstable This is one of the generic types, so it describes the phase-plane portrait of the nonlinear system near (, ) Solution (d) The nullcline for x is the line y = This line partitions the plane into regions where x is increasing or decreasing as t increases The nullcline for y is the parabola y = (x 4 x2 ) This curve partitions the plane into regions where y is increasing or decreasing as t increases Neither of these nullclines is invariant The stationary point (, ) is a counterclockwise spiral sink The stationary point (, ) is a saddle The coefficient matrix A has eigenvalues 5 and Because it has the eigenpairs A + 5I = 5 5, A I = ( ( 5,, )) (, 5, 5 ) 5 Near (, ) there is one trajectory that emerges from (, ) tangent to each side of the line x = 5y There is also one trajectory that approaches (, ) tangent to each side of the line y = x These trajectories are separatrices A global phase-plane portrait might be sketched during the review session Remark The global phase-plane portrait becomes clearer if you are able to observe that H(x, y) = 2 x y2 x satisfies d dt H(x, y) = xh(x, y) x + y H(x, y) y = (x x 2 )( 5y) + 5y(x 4y x 2 ) = 2y 2 The trajectories of the system are thereby seen to cross the contour lines of H(x, y) so as to decrease H(x, y) You would not be expected to see this on the Final Exam (8) Consider the nonlinear planar system x = x( x + 2y), y = y(6 x y) Do parts (a-d) as for the previous problem (e) Why do solutions that start in the first quadrant stay in the first quadrant? Solution (a) Stationary points satisfy = x( x + 2y), = y(6 x y) The first equation implies either x = or x +2y =, while the second equation implies either y = or 6 x y = If x = and y = then (, ) is a stationary point If x = and 6 x y = then (, 6) is a stationary point If x + 2y = and y = then (, ) is a stationary point If x + 2y = and 6 x y =

25 then upon solving these equations one finds that (, ) is a stationary point All the stationary points of the system are therefore (, ), (, 6), (, ), (, ) 25 Solution (b) Because (f(x, ) y) x x = 2 + 2xy g(x, y) 6y xy y 2, the matrix of partial derivatives is x f(x, y) y f(x, y) 6x + 2y 2x = x g(x, y) y g(x, y) y 6 x 2y Evaluating this matrix at each stationary point yields the coefficient matrices 5 A = at (, ), A = at (, 6), A = at (, ), A = at (, ) 5 Solution (c) The coefficient matrix A at (, ) is diagonal, so you can read-off its eigenvalues as and 6 The stationary point (, ) is thereby a nodal source, which is unstable (or even better is repelling) This is one of the generic types, so it describes the phase-plane portrait of the nonlinear system near (, ) The coefficient matrix A at (, 6) is triangular, so you can read-off its eigenvalues as 6 and 5 The stationary point (, 6) is thereby a saddle, which is unstable This is one of the generic types, so it describes the phase-plane portrait of the nonlinear system near (, 6) The coefficient matrix A at (, ) is triangular, so you can read-off its eigenvalues as and 5 The stationary point (, ) is thereby a saddle, which is unstable This is one of the generic types, so it describes the phase-plane portrait of the nonlinear system near (, ) The coefficient matrix A at (, ) has eigenvalues that satisfy = det(zi A) = z 2 tr(a)z + det(a) = z 2 + 2z + 45 = (z + 6) Its eigenvalues are thereby 6±i Because a 2 = <, the stationary point (, ) is therefore a clockwise spiral sink, which is asymptotically stable or attracting This is one of the generic types, so it describes the phase-plane portrait of the nonlinear system near (, ) Solution (d) The nullclines for x are the lines x = and x + 2y = These lines partition the plane into regions where x is increasing or decreasing as t increases The nullclines for y are the lines y = and 6 x y = These lines partition the plane into regions where y is increasing or decreasing as t increases Next, observe that the lines x = and y = are invariant A trajectory that starts on one of these lines must stay on that line Along the line x = the system reduces to y = y(6 y)

26 26 Along the line y = the system reduces to x = x( x) The arrows along these invariant lines can be determined from a phase-line portrait of these reduced systems The stationary point (, ) is a nodal source The coefficient matrix A has eigenvalues and 6 Because it has the eigenpairs A I =, A 6I = ( (,, )) ( ( 6, )), Near there is one trajectory that emerges from (, ) along each side of the invariant lines y = and x = Every other trajectory emerges from (, ) tangent to the line y =, which is the line corresponding to the eigenvalue with the smaller absolute value The stationary point (, 6) is a saddle The coefficient matrix A has eigenvalues 6 and 5 Because 2 A + 6I =, A 5I =, it has the eigenpairs ( ( 6,, )) ( ) 7 5, 2 Near (, 6) there is one trajectory that approaches (, 6) along each side of the invariant line x = There is also one trajectory that emerges from (, 6) tangent to each side of the line y = 6 2 x These trajectories are separatrices 7 The stationary point (, ) is a saddle The coefficient matrix A has eigenvalues and 5 Because A + I =, A 5I =, 8 it has the eigenpairs ( (,, )) ( ( 5, 4)) Near (, ) there is one trajectory that emerges from (, ) along each side of the invariant line y = There is also one trajectory that approaches (, ) tangent to each side of the line y = 4(x ) These trajectories are also separatrices Finally, the stationary point (, ) is a clockwise spiral sink All trajectories in the positive quadrant will spiral into it A phase-plane global portrait might be sketched during the review session Be sure your portrait be correct near each stationary point! Solution (e) Because the lines x = and y = are invariant, the uniqueness theorem implies that no other trajectories can cross them

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