Advanced Calculus fi..

Chapter 9 Ordinary Differential EquationsPROBLEMS1. Find the general solutions:Find the general solution of the system:by solving the equivalent system:dx dy dz dw .&- = z, - = W, - =x-4y, -=y-x.dt d t d t d tb) Find the general solution of the equationby solving the equivalent systeni:--.3. One may be able to obtain the solution of (9.38) for t 2 0 with initial value xo for t = 0by applying Laplace transforms as in Problem 16 following Section 9.4. For Example 1in the text, let (x(t), y(t)) be the solution with initial value (1,2) and let 49[x(t)] = X(s),Y[y(t)] = Y(s). Then we obtain sX(s) - 1 = X(s) - 2Y(s), sY(s) - 2 = -2X(s) + Y(s).We solve these simultaneous equations for X(s), Y(s), obtaining X(s) = (3)(s + I)-' -(f)(s - 3)~'. ~(s)=(f)(s - 3)-' +(i)(s + I)-', so that x =($e-' -(f)e3'.y =(&)e3' + (%)e-'. Apply the method to the following equations with initial conditions.[The rule 2'[tkea'] = k!(s - a)-k-' will be found helpful. In parts (e) and (f) thereare multiple eigenvalues.]a) Equations of Problem l(a), with x(0) = 2, y(0) = -1.b) Equation of Problem I(b), with x[(O) = 3, x2(0) = 2.C) Equation of Problem l(c), with x(0) = 0.d) Equation of Problem l(d), with x(0) = col(1, 1, 1).. -4. In the text a general solution for Example 2 is obtained. Verify that this is indeed the generalsolution by showing that the vector functions xl(t), x2(t), x3(t) are linearly independentfor -00 c t < oo and that each function is a solution of the differential equation for-00 < t < 00.5. For a matrix function A(t) = (aij(?)), a 5 t 5 b, one defines the derivative dA/dt to be thematrix function (daij/dt), provided that all aij(t) have derivatives. One defines 1 A(?) dtto be the matrix (jaij(t) dt); here we assume all aij(t)aij(t) dt) and S,b A(?) dt to be (hb