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

2 Answers
Feb 7, 2017

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)

Feb 7, 2017

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

Explanation:

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)