Question

d^2x/dt^2 - 4dx/dt -5x =te^(2t)cos 3t

Answer 1

Question #8148 $\frac{d^2x}{dt^2} - 4\frac{dx}{dt} - 5x = te^{2t}\cos 3t$

Solution. The general solution of linear non homogeneous equation is sum of general solution of the respective homogeneous equation and any solution of non-homogeneous. The general solution of homogeneous is $x(t) = C_1e^{5t} + C_2e^{-t}$ . The solution of non-homogeneous should be found in the form $e^{2t}(ax + b)\cos 3t + e^{2t}(cx + d)\sin 3t$ . It can be verified by substituting, that $x_0(t) = -1 / 54e^{2t}(3t\cos (3t) - \sin 3t)$ . Thus, the general solution is $x(t) = C_1e^{5t} + C_2e^{-t} - 1 / 54e^{2t}(3t\cos (3t) - \sin 3t)$ .

Answer. $x(t) = C_1e^{5t} + C_2e^{-t} - 1 / 54e^{2t}(3t\cos (3t) - \sin 3t)$