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

Question #176282

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


1
Expert's answer
2021-03-30T08:49:08-0400

yy=e2x(3tane+3(sece)2)=Ce2xm21=0(m+1)(m1)=0m=1,  1The complementary factor is   y=Acoshx+BsinhxThe particular integral is   y=De2x/y=4De2x,  y=De2x4De2xDe2x=Ce2x3De2x=Ce2xD=C3y=e2x(3tane+3(sece)2)3y=C.F+P.Iy=Acoshx+Bsinhx+e2x(3tane+3(sece)2)3.\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}.\\


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Comments

Assignment Expert
09.04.21, 14:50

Dear Isaax Numoah, the initial math formulae in the question were not clear enough. Please type correctly and submit a new question with a help of the panel.

Isaax Numoah Boateng
02.04.21, 03:12

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

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