From part a we have that y t cos t so y t sin t then

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Unformatted text preview: From Part (a) we have that y ( t ) = cos( t ), so y ′ ( t ) = − sin( t ), then u = bracketleftbigg cos( t ) − sin( t ) bracketrightbigg and du dt = bracketleftbigg − sin( t ) − cos( t ) bracketrightbigg , which equals bracketleftbigg 1 − 1 0 bracketrightbiggbracketleftbigg cos( t ) − sin( t ) bracketrightbigg = bracketleftbigg − sin( t ) − cos( t ) bracketrightbigg and u (0) = bracketleftbigg 1 bracketrightbigg , showing that this vector solution u solves the differential equation and has the correct initial conditions. Problem 12 If A is invertible then a particular solution to du dt = Au − b, will be u a constant if and only if du dt = 0 or 0 = Au − b or u = A − 1 b . Part (a): For du dt = 2 u − 8. The particular solution is given by 2 u = 8 (or u = 4), and the homogeneous solution is given by du dt = 2 u ⇒ u = Ce 2 t . Thus the complete solution is given by u ( t ) = 4 + Ce 2 t . Part (b): For du dt = bracketleftbigg 2 0 0 3 bracketrightbigg u − bracketleftbigg 8 6 bracketrightbigg . Then a particular solution is given by (again assuming u is a constant) bracketleftbigg 2 0 0 3 bracketrightbigg u = bracketleftbigg 8 6 bracketrightbigg ⇒ bracketleftbigg u 1 u 2 bracketrightbigg = bracketleftbigg 4 2 bracketrightbigg a particular solution is given by the solution to du dt = bracketleftbigg 2 0 0 3 bracketrightbigg u . The coefficient matrix A is then given by bracketleftbigg 2 0 0 3 bracketrightbigg , which has eigenvalues 2 and 3, with eigenvectors bracketleftbigg 1 bracketrightbigg and bracketleftbigg 1 bracketrightbigg , then the total solution is then c 1 bracketleftbigg 1 bracketrightbigg e 2 t + c 2 bracketleftbigg 1 bracketrightbigg e 3 t , so that the total solution (particular plus the homogeneous) is given by u = c 1 bracketleftbigg 1 bracketrightbigg e 2 t + c 2 bracketleftbigg 1 bracketrightbigg e 3 t + bracketleftbigg 4 2 bracketrightbigg . Problem 13 Assume that c is not an eigenvalue of A . Let u = e ct v , where v is a constant vector. Then du dt = ce ct v and Au = Ae ct v = e ct Av , so that the equation du dt = Au − e ct b becomes ce ct e ct Av − e ct b cv = Av − b ( A − cI ) v = b v = ( A − cI ) − 1 b. Since c is not an eigenvector of A A − cI is invertible, showing that u = e ct v = e ct ( A − cI ) − 1 b is a particular solution to the differential equation du dt = Au − e ct b. If c is an eigenvector of A , then A − cI is not invertible and there exists a nonzero v such that Av = cv , so that when e ct v is substituted into our differential equation we have cv = Av − b or 0 = − b a contradiction. Problem 14 For a differential equation to be stable we require that u → 0 as t → ∞ . For the differential equation du dt = Au , when A is a matrix, this will happen when all the eigenvalues of A have negative real parts. For a two by two systems, this eigenvalue condition breaks down into conditions on the trace ( T ) and determinant ( D ) of A . The conditions are that T...
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