Answer to Question #177693 in Differential Equations for Isaax Numoah Boateng

Question

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

Answer 1

y"-y = e^2x[3tan e+ $3sec e\over 2$ ]

=Ce^2x

$\implies$ m²-1=0

(m+1)(m-1)=0

m= 1 , -1

complementary factor y

y= Acosh x + Bsinh x

The PI is y = De^2x , y" = 4De^2x

$\implies$ 4De^2x- De^2x = Ce^2x

$\implies$ D=$C\over3$

$\implies$

y= $e^{2x}[3tan e+ {3sece \over2} ]\over3$

We know

y = C.F + P.I

y= Acosh x + Bsinh x +

$e^{2x}[3tan e+ {3sece \over2} ]\over3$