-6 Even-Numbered Homework Solutions Suppose that the matric B has λ = + 5i as an eigenvalue with eigenvector Y 0 = solution to dy = BY Using Euler s formula, we can write the complex-valued solution Y c t as Y c t = e +5it i = e t e 5it i = e t cos5t + i sin5t i = e t cos5t + ie t sin5t cos5t + sin5t sin5t cos5t Compute the general i So, we have Y re t = e t cos5t and Y cos5t + sin5t im t = e t sin5t Hence, the general sin5t cos5t solution is: Yt = k e t cos5t + k cos5t + sin5t e t sin5t sin5t cos5t The following linear system has complex eigenvalues: dy = Y, with initial condition Y 6 0 =, a The characteristic polynomial is λ 8λ + 0 = 0 Use the quadratic formula: λ = 8 ± 8 0 So, the eigenvalues are λ = + i and λ = i = 8 ± 6 80 = 8 ± 6 = 8 ± i = ± i b The real parts of the eigenvalues are positive, so the origin is a spiral source See page 05 of your textbook c The natural period of the system is given by π β, and in this system β = Thus, the natural period for this system is π The natural frequency is given by π d The direction field for the system which you can find using the DE Tools software indicates that the solutions spiral out from the origin going clockwise
6 The following linear system has complex eigenvalues: dy 0 = Y, with initial condition Y 0 =, a The characteristic polynomial is λ + λ + = 0 Aftering using the quadratic formula, we get λ = + i and λ = i b The real parts of the eigenvalues are negative, so the origin is a spiral sink See page 05 of your textbook c The natural period of the system is given by π β, and in this system β = Thus, the natural period for this π system is The natural frequency is given by π 8 The following linear system has complex eigenvalues: dy = Y, with initial condition Y 0 =, a The characteristic polynomial is λ λ + = 0 Use the quadratic formula to get λ = i 7 λ = + i 7 b The real parts of the eigenvalues are positive, so the origin is a spiral source See page 05 of your textbook c The natural period of the system is given by π 7 β, and in this system β = Thus, the natural period for this π 7 system is The natural frequency is given by 7 π and 0 Using the same system as in exercise : a Find the general solution For λ = + i: i x i y = ix + y = 0 x + iy = 0 = ix + y = 0 x + iy = 0 = 0 0 Letting x = in the first equation gives that is an associated eigenvector Letting x = in the + i second equation gives that is an eigenvector Your answers should be the same regardless of your i choice of eigenvector For the first possible eigenvector, the corresponding complex solution is: Y t = e +it = e + i t cos t + isin t + icos t + isin t = e t cos t + isin t cos t + isin t + icos t + isin t = e t cos t + ie cos t sin t t sin t sin t + cos t It is easily checked that these solutions are linearly independent; the initial condition t = 0 gives for the
0 real part and for the imaginary part Thus, the general solution is: Yt = k e t cos t + k cos t sin t e t sin t sin t + cos t b Find the particular solution with the given initial value Using the initial condition Y 0 =,, we get: 0 = k + k This implies that k = and k = 0 So, the particular solution is: Yt = e t cos t cos t sin t Using the same system as in exercise 6: a The associated eigenvector for λ = + i i λ = i is Using the eigenvector Yt = e Using Euler s formula, write Yt = e cos Yt = e t/ cos Yt = k e t/ t cos t + i sin t cos t + i is, and the associated eigenvector for + i we obtain the complex-valued solution: t i t + i sin t Taking real and imaginary parts, form the general solution: t sin t cos t b Using your initial condition, find that k = and k = 0 + i, which we rewrite as: cos + t sin t ie t/ sin t cos + k t sin t e t/ sin t
5 The following linear system has one eigenvalue and one line of eigenvectors: dy 0 = Y a The characteristic polynomial is λ + λ + = λ + λ +, so the eigenvalue is λ = x 0 b We have =, which gives us that x + y = 0 and x y = 0 Thus x = y and y 0 is an eigenvector c cpng d dpng e Sketch the xt- and yt-graphs of the solution with initial condition Y 0 =, 0 epng
8 Using the same system as in 5: a Find the general solution The general solution has the form Yt = e λt V 0 + te λt V Let V 0 = Y 0 be an arbitrary initial condition x0 x0 + y x 0, y 0 Then V = A λiv 0 = A + IV 0 = = 0 Thus the general solution y 0 x 0 y 0 is Yt = e t x0 + te y t x0 + y 0 0 x 0 y 0 b Find the particular solution for the initial condition Y 0 =, 0 We have V 0 =, 0, so by above V =, 0 Evaluate the limit of te λt as t if a λ > 0 Since λ > 0 and t, te λt will approach infinity b λ < 0 Replace λ = µ, where µ is positive Then as t, te λt = t will approach 0 e µt 8 The following system has zero as an eigenvalue: dy = Y 6 a Find the eigenvalues The characteristic equation for this system is λ6 λ = λ 8λ + = λ 8λ Setting this equal to zero gives λ = 0 and λ = 8 as eigenvalues b Find the eigenvectors x 0 For λ = 0, = gives that x = y Thus, is an eigenvector associated with λ = 0 6 y 0 6 x 0 For λ = 8, = gives that x = y 0 y Thus, is an eigenvector associated with λ = 8 c Sketch the phase portrait cpng 5
d Sketch the xt- and yt-graphs of the solution with initial condition Y 0 =, 0 dpng e Find the general solution The general solution will take the form Yt = k Y t + k Y t, where Y t = e 0t and Y t = e 8t f Find the particular solution for the initial condition Y 0 =, 0 We need to find k and k such that k + k = Solving this gives k 0 = 8 and k = 8 Our final answer is thus Yt = + 8 8 e8t This can, of course, be simplified a b 0 Let A = c d a Suppose that at least one of the eigenvalues is zero both could be zero The characteristic polynomial of A is λ a + dλ + ad bc = 0 Since λ = 0 is one eigenvalue, the λ terms vanish and leave ad bc = 0 But note that deta = ad bc = 0, and we re done b Now suppose that deta = 0 Then the characteristic polynomial is λ a + dλ = 0 We can factor λ out, so at least one eigenvalue λ = 0 Find the general solution for the linear systems: a dy 0 = Y 0 0 The eigenvalue λ = 0 is repeated for this sytem, with eigenvector You ll find that y = 0 must be true, but 0 x can be any number; I chose x = There is only one line of eigenvectors, so use the theorem on page 9 to obtain the general solution: x0 y0 Yt = e + te y 0 0 b dy = 0 Y 0 0 Like in part a, the eigenvalue λ = 0 is repeated for this sytem, with eigenvector eigenvectors, so use the theorem on page 9 again to obtain the general solution: x0 y0 Yt = e + te 0 y 0 0 There is only one line of 6
Use the system: dx = x y dy = x + y a The eigenvalue for the system is λ = repeated Since λ < 0, the equilibrium point at the origin is a sink See page 0 of your textbook b For the eigenvalue λ =, we have eigenvector v = So, the straight line solution of this system is given by Yt = e t 7
6 d y dy y = 0 The characteristic polynomial is s s, so the eigenvalues are λ = and λ = Hence, the general solution is yt = k e t + k e t d y dy + y = 0 The characteristic polynomial is s s +, so the eigenvalue is λ = repeated Hence, the general solution is yt = k e t + k te t 6 d y dy + 9y = 0 The characteristic polynomial is s s + 9, so the eigenvalues are λ = + 5i and λ = 5i Hence, the general solution is yt = k e t cos5t + k e t sin5t 8 Find the solution of the initial value problem d y + dy 5y = 0 y0 =, y 0 = 7 Guess yt = e st as a solution, where s is a constant Substituting the guess into the left-hand side of the equation gives: d e st + dest 5e st = s e st + se st 5e st = e st s + s 5 Since e st is never zero, we must set s + s 5 = 0 and solve for s The roots are s = and s = 5 So, we have the general solution yt = k e t + k e 5t Using your initial conditions, you should find that your solution is yt = 8e t + e 5t 0 Find the solution of the initial value problem d y + dy + 0y = 0 y0 =, y 0 = 8 Guess yt = e st as a solution, where s is a constant Substituting the guess into the left-hand side of the equation gives: d e st + dest + 0e st = s e st + se st + 0e st = e st s + s + 0 Since e st is never zero, we must set s + s + 0 = 0 and find s using the quadratic formula Your roots will be s = ± i So, we have the general solution yt = k e t cos t + k e t sin t Using your initial conditions, you should find that your solution is yt = e t cos t + e t sin t 8
Find the solution of the initial value problem d y dy + y = 0 y0 =, y 0 = Guess yt = e st as a solution, where s is a constant Substituting the guess into the left-hand side of the equation gives: d e st dest + e st = s e st se st + e st = e st s s + Since e st is never zero, we must set s s + = 0 and solve for s The roots are s = repeated Then we have the general solution yt = k e t + k te t Using your initial conditions, you should find that your solution is yt = e t te t Consider harmonic oscillators with mass m, spring constant k, and damping coefficient b such that: m =, k = 8, b = 6, with inital conditions y0 =, v0 = 0 a The second-order differential equation is given by d y + 6dy + 8y = 0 The corresponding first-order system is found by setting dy = v Thus our first-order system is: dy = v dv = 6v 8y dy b The linear system is matrix form is given by: 0 yt dv = 8 6 vt Use the traditional method to find the eigenvalues: λ = and λ = The eigenvector associated with λ = is, and the eigenvector associated with λ = is, c Note that b km = 6 8 =, so the oscillator is overdamped There are no oscillations 6 Consider harmonic oscillators with mass m, spring constant k, and damping coefficient b such that: m =, k = 8, b = 0, with inital conditions y0 =, v0 = a The second-order differential equation is given by d y by setting dy = v Thus our first-order system is: dy = v dv = 8y + 8y = 0 The corresponding first-order system is found dy b The linear system is matrix form is given by: 0 yt dv = 8 0 vt Use the traditional method to find the eigenvalues: you should get λ = i and λ = i The i eigenvectors are i and, respectively c Note that b km = 0 8 =, so the oscillator is underdamped The period is π 9
8 Consider harmonic oscillators with mass m, spring constant k, and damping coefficient b such that: m = 9, k =, b = 6, with inital conditions y0 =, v0 = a The second-order differential equation is given by 9 d y +6dy +y = 0, which is the same as d y + dy + 9 y = 0 The corresponding first-order system is found by setting dy = v Thus our first-order system is: dy = v dv = v y 9 b The linear system is matrix form is given by: dy 0 yt dv = vt 9 Use the traditional method to find the eigenvalues: you should get λ = for your single eigenvalue The eigenvector corresponding to λ = is c Note that b km = 6 9 = 0, so the oscillator is critically damped In the case of a critically damped oscillator, the mass tends to its rest position but does not oscillate 0 Consider harmonic oscillators with mass m, spring constant k, and damping coefficient b such that: m =, k =, b =, with inital conditions y0 = 0, v0 = a The second-order differential equation is given by d y + dy +y = 0, which is the same as d y + dy + y = 0 The corresponding first-order system is found by setting dy = v Thus our first-order system is: dy = v dv = v b The linear system is matrix form is given by: dy 0 yt dv = vt Use the traditional method to find the eigenvalues: you should get λ = ± i ± i with eigenvectors 6 c Note that b 8π km = =, so the oscillator is underdamped The period is Use the same system as in exercise a The second-order equation is d y + 6dy + 8y = 0, so the characteristic equation is s + 6s + 8 = 0, which has roots s = and s = So the general solution is yt = k e t + k e t b To find the particular solution, use the initial conditions y0 =, v0 = 0 to find k = and k = Use the same system as in exercise 6 a The second-order equation is d y + 8y = 0, so the characteristic equation is s + 8 = 0, which has roots s = ±i A complex-valued solution is y c t = e it = cos t + i sin t, so the general solution is yt = k cos t + k sin t 0
b To find the particular solution, use the initial conditions y0 =, v0 = to find k = and k = 6 Use the same system as in exercise 8 a The second-order differential equation is given by 9 d y +6dy +y = 0, which is the same as d y + dy + 9 y = 0 The characteristic equation is s + s + 9 = 0, which has repeated root s = So the general solution is yt = k e t/ + k te t/ b To find the particular solution, use the initial conditions y0 =, v0 = to find k = and k = 8 Use the same system as in exercise 0 a The second-order differential equation is given by d y + dy +y = 0, which is the same as d y + dy + y = 0 The characteristic equation is s + s + = 0, which has roots λ = ± i The general solution is t t yt = k e t/ sin + k e t/ cos b To find the particular solution, use the initial conditions y0 =, v0 = to find k = and k = 0