y\" + 4y = 5e4x + 3 - sec2 2x. Solution First let us find the homogenous equatio

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First let us find the homogenous equation solution r^2 + 4 = 0 => Yc = A sin2x + B cos2x First let us find the particular solution for 5e^(4x) + 3 => Yp1 = A e^(4x) + B 16A e^4x + 4A e^4x + 4B = 5e^(4x) + 3 => A =1/ 4 and B = 3/4 Yp1 = e^(4x)/4 + 3/4 Now let us find the particular solution for (sec(2x))^2 => Yp2 = A(x) sin2x + B(x) cos2x Yp2' = A'(x) sin2x + B'(x) cos2x + 2(A(x) cos2x - B(x) sin2x) A'(x) sin2x + B'(x) cos2x = 0 Yp2'' = 2[A'(x) cos2x - B'(x) sin2x - 2(A(x) sin2x + B(x) cos2x)] Yp2'' - 4Yp2 = (sec(2x))^2 => 2(A'(x) cos2x - B'(x) sin2x ) = (sec(2x))^2 A'(x) ( (cos(2x))^2 + (sin(2x))^2 ) = (sec(2x))^2 * cos2x /2 => A'(x) = sec(2x) /2 B'(x) = - tan(2x) * sec(2x) / 2 A(x) = (ln|sec2x + tan2x|)/ 4 and B(x) = sec(2x)/4 Yp2 = (ln|sec2x + tan2x|)/ 4 * sin(2x) + (1/4) General Solution = A sin2x + B cos2x + Yp1 + Yp2

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