Differential Equations: Second Order (Auxiliary Equation with 2 Different Real Roots)

Worked Solution:

The differential equation is second order, homogeneous and has constant coefficients.
We can therefore use the auxiliary equation: 3 lambda

² – 14

– 5 = 0
It factorises: (3 lambda

+ 1) (

– 5) = 0

The two roots of the auxiliary equation are

–	1

	3

and

The general solution is

y = A e^{– 1/3 z} + B e^{5 z} where A and B are arbitrary constants.

We can find A and B using the initial conditions:

The general solution for y at z = 0:
A + B =
7
4
Differentiating the general solution to find
dy
dz
at z = 0 :
– 1
3
A + 5 B =
41
12
Solving these simultaneous equations:
A = 1 , B =
3
4

The particular solution is

y =

e^{– 1/3 z}

e^{5 z}