For second order des, the roots of the characteristic equation may be real or complex. if the roots are real, the complementary solution is the weighted sum of real exponentials. use c1 and c2 for the weights, where c1 is associated with the root with smaller magnitude. if the roots are complex, the complementary solution is the weighted sum of complex conjugate exponentials, which can be written as a constant times a decaying exponential times a cosine with phase. use c1 for the constant and phi for the phase. (note: some equations in the text give the constant multiplying the decaying exponential as 2c1. this was done for the derivation. the constant for this problem should be c1 alone.)

all numerical angles(phases) should be given in radian angles (not degrees). given the differential equation y′′+10y′+61y=6u(t)

write the functional form of the complementary solution, yc(t)
yc(t)=(c1e^(-5t)cos(6t+phi))u(t) < - solved correct

find the particular solution, yp(t).
yp(t)=(6/61)u(t) < -solved, correct

find the total solution, y(t)y(t) for the initial conditions y(0)=9 and y′(0)=4
y(t)=

all numerical angles(phases) should be given in radian angles (not degrees).