Differential equations
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1 Differential equations Math 7 Spring Practice problems for April Exam Problem Use the method of elimination to find the x-component of the general solution of x y = 6x 9x + y = x 6y 9y Soln: The system can be written (D 3) x y = = L x + L y x + (D + 3) y = = L 3 x + L 4 y whose operational determinant L L 4 L L 3 is (D + 3) (D 3) = (D )(D 8) so the solutions for x(t) are x = c e t + c e t + c 3 e 8 t + c 4 e 8 t
2 Problem A system of 4 tanks is set up in such a way that solution passes from tank to tank at a rate of gallon per minute, from tank to each of tanks 3 and 4 at a rate of gallon per minute each, from tank 3 to tanks and 4 at a rate of gallon per minute each and from tank 4 to tanks and 3 at a rate of gallon per minute. Each tank contains gallon of solution. Find a matrix A that describes the system Soln: The flow equations are dx dt = Ax x = x + x 3 + x 4 x = x x x 3 = x x 3 + x 4 x 4 = x + x 3 x 4 The system therefore can be expressed in matrix form as dx dt = A x; A = x x x 3 x 4
3 Problem 3 Provide a solution of the system x (t) x (t) = x 3(t) 3 6 x () having initial displacement vector x () x 3 () x (t) x (t) x 3 (t) = + cos t Soln: Multiplying by the inverse of the mass matrix gives the sytem d x dt = + cos t / 3/ x (t) x (t) x 3 (t) A particular solution x p (t) = v cos t will satisfy v 4v cos t = cos t / 3/ v v cos t 3 or, upon dividing through by cos t, v v + v 4 v v 3 = v / v + 3v 3 / + v v 3 or = v + v + v + 4v + 3v 3 + v + v 3 + So v = ( + v )/ and then ( + v ) + 8v + 6v = or 6v 3 + 7v = 3 and v = v 3 so 6v 3 + 7( v 3 ) = 3 or v 3 = 4/5 then v = 3/5 and v = 9/5. NB: The eigenvalues of A are messy so best not to consider the frequencies. 3
4 Problem 4 Two matrices A and B are said to commute if AB = BA. Which pairs among the following of matrices commute? Is it true that if AB = BA and CB = BC then AC = CA? Is it true that if AB = BA and CB BC then AC CA? A = 3 A = 3 6 A 3 = A 4 = A 5 = A 6 = Problem 5 Find the general solution of the system dx dt = x 4
5 Problem 6 A system of 4 masses and 5 springs is as follows: m =, m = 4, m 3 = 8, m 4 = 6 and k =, k =, k 3 = 4 and k 4 = 8 and k 5 = 6. What are the natural frequencies of the system? The determinant of a 4 4 matrix A whose (i, j)-entry is a ij is given by det A = a det A(, ) a det A(, ) + a 3 det A(, 3) a 4 det A(, 4) where A(i, j) is the 3 3 matrix obtained by deleting the ith row and jth column of A. Soln: The system has the form d x dt = M where M = diag (/, /4, /8, /6) so that A = M K = so that A has characteristic polynomial x = ( 3 λ)[( 3 λ)[( 3 λ) ] ( 3 λ)] (( 3 λ) ) = (3 + λ) [(λ + 3) 4] + 4 (3 + λ) = (3 + λ) 4 6(3 + λ) + 4 so if µ = (3 + λ) then µ = 3 ± 5 and the eigenvalues of A are λ = 3± µ so the natural frequencies are 3 ± µ 3 ± 3 ± 5 ω = = Note that there are four frequencies, one for each combination of ± 5
6 Problem 7 Find the general solution of the following system 7 dx dt = 9 5 x 4 4 Soln: The eigenvalue 8 has an eigenvector [, 3, ] T. The eigenvalue λ = has an ordinary eigenvector [, 3, 4] and a defect one generalized eigenvector [5/, /4, /3] T. Then one has a general solution e 8t c 3 + e t c c 3 t t t 8 6
7 Problem 8 Find the exponential of the matrix ta = t Soln: Problem 9 For each of the following matrices, find a matrix A such that the given matrix has the form e ta or explain why there cannot be such a matrix. e πt e t I. e e t t e t te t II e t one million t e te t e t [ ] [ ] te III t cos t + e t sin t e t sin t e e t sin t te t cos t e t IV t cos 3t e t sin 3t sin t e t sin 3t e t cos t Soln: Any matrix of the form e ta has to have the property that e A = e = I. π This excludes III. In case I we can take e. For the one million other matrices, [ recall that the ] columns [ are solutions ] of dx/dt = A x. For e example, x = t cos 3t e e t and x sin 3t = t sin 3t e t satisfy cos 3t [ ] [ ] [ ] dx cos 3t 3 sin 3t 3 e = e t = t cos 3t dt sin 3t 3 cos 3t 3 e t sin 3t [ ] [ ] [ ] dx sin 3t + 3 cos 3t 3 e = e t = t sin 3t dt cos 3t 3 sin 3t 3 e t cos 3t [ ] 3 so we see that A = works for IV. Similar approaches apply to 3 the other matrices. 7
8 Problem Find the exponential e ta of the matrix 4 A = 4 = U TU where U =, T =, U = Soln: The structure of A allows us to write A k = k U T k U and the structure of T allows us to write k ( ) k T k = (I + N) k = N m k(k ) = I + kn + N = k k(k ) k m m= when k since N = ; N = so e ta = I + ta + t A = e t U (I+tN+t N )U = e t I + t ; N 3 = N 4 = = Ak + + tk k! + = (t) k U T k k! U k= + t ; Problem Solve the initial value problem [ dx 3 9 dt = 5 3 ] x; x() = [ ] 8
9 Soln: For the given matrix, det (A λi) = 3 λ λ = (3 λ)(3 + λ) + 45 = λ + 36 so the eigenvalues are ±6i. The eigenvector for λ = 6i satisfies [ ] [ ] [ ] 3 6i 9 c = 5 3 6i so ( i)c = 3c or c = ( 3 3 i) c so c = [3, i] works and then the eigenvector for λ = 6i is the conjugate c = [3, + i]. Then [ ] e 6it 3 c = (cos 6t + i sin 6t) i [ ] [ ] 3 cos 6t 3 sin 6t = + i cos 6t + sin 6t sin 6t cos 6t [ ] 3 The real and imaginary parts are solutions having initial values and [ ] so, to satisfy the initial conditions one takes minus one-half imaginary solution, which gives Problem [ 3 sin 6t sin 6t cos 6t Use the method of undetermined coefficients to solve ]. x = 5x 6y; y = 4x + 3y + sin t, x() = 4/8, y() = 5/8 c Problem 3 Use the Laplace transform to solve the initial value problem x + 4x + 8x = e t ; x() =, x () =. 9
10 Soln: Taking the Laplace transform of both sides gives so that s X(s) + 4sX(s) + 8X(s) = s + X(s)(s + 4s + 8) = s + + or X(s) = s + 3 (s + )((s + ) + 4) One has the partial fraction decomposition s + 3 (s + )((s + ) + 4) = A s + + Bs + C (s + ) + 4 Taking s = yields A = /4 and taking s = + i yields or s+3 = A((s+) +4)+(Bs+C)(s+). + i = (C B + ib)(i) or C B =, B=-/4, C=/. This gives So that X(s) = /4 s + / + 4(s + ) (s + ) + 4 x(t) = 4 e t 4 e t cos t + e t sin t which verifies x() = and x () =.
11 Problem 4 Find the Laplace transform of the function f(t) = t [[t]], (t ) where [[t]] denotes the integer part of t, that is, [[t]] = n if n t < n +. Soln: This is a homework exercise in the book.
12 Problem 5 Find the solution of the system 4 5 dx dt = x, x() = Soln: The eigenvalues are { 4, 4, 6} with eigenvectors 5 v = 4, v = 6, and w = 5 3 gen 4, where the last is a generalized eigenvector of defect one. The corresponding eigensolutions are x (t) = e 4t v, x (t) = e 6t v and x 3 (t) = e 6t (tv + w ). The general solution of the homogeneous system is c x + c x + c 3 x 3. The initial condition requires c 5c =, c + 5/3c 3 = and /3c 3 = so c 3 = 3/ then c = (3) /5 and c = 5c = (3).
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