Answer to Question #176282 in Differential Equations for Isaax Numoah Boateng

Question

y" - y = e2x[3 tan e+ 3 (sec e*)2]

Answer 1

$\displaystyle y'' - y = e^{2x}\left(3\tan{e}+ \frac{3(\sec{e})}{2}\right) = Ce^{2x} \\ m^2 - 1 = 0 \\ (m + 1)(m - 1) = 0\\ m = -1,\,\, 1\\ \textsf{The complementary factor is}\,\,\, y = A\cosh{x} + B\sinh{x} \\ \textsf{The particular integral is}\,\,\, y = De^{2x} \\/ y'' = 4De^{2x},\,\, y = De^{2x} \\ 4De^{2x} - De^{2x} = Ce^{2x} \\ 3De^{2x} = Ce^{2x} \\ D = \frac{C}{3}\\ y = \frac{e^{2x}\left(3\tan{e}+ \frac{3(\sec{e})}{2}\right)}{3} \\ \therefore y = C.F + P.I \\ y = A\cosh{x} + B\sinh{x} + \frac{e^{2x}\left(3\tan{e}+ \frac{3(\sec{e})}{2}\right)}{3}.\\$