Question

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

Answer 1

$\displaystyle \textsf{A differential equation whose solution is}\\ e^{-x}(\sin(2x) + 3e^{-x}\cos(2x)) \, \textsf{has the auxiliary equations}\\ m = -1 \pm 2j \, \textsf{since if the solution to the auxiliary equation}\\ \textsf{is}\, m = \alpha \pm j \beta,\, \textsf{the general solution to such differential}\\ \textsf{equation is}\, e^{\alpha x}(A\cos(\beta x) + B\sin(\beta x)) \\ \textsf{If}\, m = -1 \pm j2,\, \textsf{the quadratic is}\\ m^2 - (-1 + j2 - 1 - j2)m + (-1 + j2)(-1 - j2) = 0\\ m^2 + 2m + 5= 0\\ \textsf{To obtain a homogeneous differential equation whose solution is}\\ \textsf{form}\, xe^{-x}(\sin(2x) + 3e^{-x}\cos(2x)), \,\textsf{The auxiliary equation must have}\\ \textsf{repeated roots as its solution, so}\\ e^{-x}(\sin(2x) + 3e^{-x}\cos(2x))\, \textsf{can be}\\ \textsf{multiplied by}\, x\, \textsf{to serve as another}\\\textsf{solution to the differential equation.}\\ \textsf{This implies that}\, m = -1 \pm j2 \, \textsf{twice}\\ \therefore (m² + 2m + 5)² = 0\\ \implies m^4 + 4m^3 + 14m^2 + 20m + 25 = 0\\ \textsf{A differential equation with the}\\\textsf{above auxiliary equation is}\\ y^{(iv)} + 4y^{(iii)}+ 14y" + 20y' + 25y = 0\\ \therefore y^{(iv)} + 4y^{(iii)}+ 14y" + 20y' + 25y = 0 \\\textsf{is the homogeneous linear}\\ \textsf{differential equation with constant}\\ \textsf{coefficients that is satisfied by}\, xe^{-x}(\sin(2x) + 3e^{-x}\cos(2x)).$