Answer to Question #140579 in Differential Equations for Nikhil

Question

Find a homogeneous linear differential equation with constant coefficients that's satisfied by x[e^-x(sin2x) + 3e^-x(cos2x)].

Answer 1

"\\displaystyle\n\n\\textsf{A differential equation whose solution is}\\\\ e^{-x}(\\sin(2x) + 3e^{-x}\\cos(2x)) \\, \\textsf{has the auxiliary equations}\\\\\n\nm = -1 \\pm 2j \\, \\textsf{since if the solution to the auxiliary equation}\\\\\n\\textsf{is}\\, m = \\alpha \\pm j \\beta,\\, \\textsf{the general solution to such differential}\\\\\n\\textsf{equation is}\\, e^{\\alpha x}(A\\cos(\\beta x) + B\\sin(\\beta x)) \\\\\n\n\n\n\\textsf{If}\\, m = -1 \\pm j2,\\, \\textsf{the quadratic is}\\\\\n\nm^2 - (-1 + j2 - 1 - j2)m + (-1 + j2)(-1 - j2) = 0\\\\\n\nm^2 + 2m + 5= 0\\\\\n\n\\textsf{To obtain a homogeneous differential equation whose solution is}\\\\ \\textsf{form}\\, xe^{-x}(\\sin(2x) + 3e^{-x}\\cos(2x)), \\,\\textsf{The auxiliary equation must have}\\\\ \\textsf{repeated roots as its solution, so}\\\\\ne^{-x}(\\sin(2x) + 3e^{-x}\\cos(2x))\\, \\textsf{can be}\\\\ \\textsf{multiplied by}\\, x\\, \\textsf{to serve as another}\\\\\\textsf{solution to the differential equation.}\\\\\n\n\\textsf{This implies that}\\, m = -1 \\pm j2 \\, \\textsf{twice}\\\\\n\n\n\\therefore (m\u00b2 + 2m + 5)\u00b2 = 0\\\\\n\n\n\\implies m^4 + 4m^3 + 14m^2 + 20m + 25 = 0\\\\\n\\textsf{A differential equation with the}\\\\\\textsf{above auxiliary equation is}\\\\\ny^{(iv)} + 4y^{(iii)}+ 14y" + 20y' + 25y = 0\\\\\n\n\\therefore y^{(iv)} + 4y^{(iii)}+ 14y" + 20y' + 25y = 0\n\\\\\\textsf{is the homogeneous linear}\\\\\n\\textsf{differential equation with constant}\\\\\n\\textsf{coefficients that is satisfied by}\\, xe^{-x}(\\sin(2x) + 3e^{-x}\\cos(2x))."