What is the General Solution of the Differential Equation #y''-6y'+10y = 0#?

We have;

# y''-6y'+10y = 0 #

This is a Second Order Homogeneous Differential Equation which we solve as follows:

We look at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, i.e.

# m^2-6m+10 = 0#

This quadratic does not factorise to I will solve by completing the square (you could equally use the quadratic formula)

# (m-3)^2-3^2+10 = 0#
# :. (m-3)^2 = -1#
# :. m-3 = +-i#
# :. m-3 = 3+-i#

Because this has two distinct complex solutions #p+-qi#, the solution to the DE is;

# y = e^(pt)(Acosqt+Bsinqt)#

Where #A,B# are arbitrary constants. and so the General Solution is;

# y = e^(3t)(Acost+Bsint)#

The general solution for this kind of differential equation (homogeneous linear with constant coefficients) is

#y = C e^(lambda t)#

substituting into the differential equation we have

#C(lambda^2-6lambda +10)e^(lambda t)=0#

but #C e^(lambda t) ne 0# so the feasible #lambda's# obey the condition

#lambda^2-6lambda +10=0# so #lambda_1=3-i# and #lambda_2=3+i# then the solution is a linear combination of both

#y=C_1e^((3-i)t)+C_2e^((3+i)t)# or

#y=(C_1e^(-it)+C_2e^(it))e^(3t)#
with #C_2=\tilde C_1# where #tilde( (cdot))# means that #C_1# and #C_2# are conjugate.

Using de Moivre's identity

#e^(it)= cost+isint# and substituting we have

#y=((C_2+C_1)cost+i(C_2-C_1)sint)e^(3t)# or

#y=(K_1cost+K_2sint)e^(3t)#

Here #K_1=C_1+\tilde C_1# and #K_2=i(C_1-tilde C_1)#