y" - y = e2x[3 tan e+ 3 (sec e*)2]
y′′−y=e2x(3tane+3(sece)2)=Ce2xm2−1=0(m+1)(m−1)=0m=−1, 1The complementary factor is y=Acoshx+BsinhxThe particular integral is y=De2x/y′′=4De2x, y=De2x4De2x−De2x=Ce2x3De2x=Ce2xD=C3y=e2x(3tane+3(sece)2)3∴y=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}.\\y′′−y=e2x(3tane+23(sece))=Ce2xm2−1=0(m+1)(m−1)=0m=−1,1The complementary factor isy=Acoshx+BsinhxThe particular integral isy=De2x/y′′=4De2x,y=De2x4De2x−De2x=Ce2x3De2x=Ce2xD=3Cy=3e2x(3tane+23(sece))∴y=C.F+P.Iy=Acoshx+Bsinhx+3e2x(3tane+23(sece)).
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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.
y" - y = e2x[3 tan e+ 3 (sec e)^2]
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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.
y" - y = e2x[3 tan e+ 3 (sec e)^2]
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