Page 306 - DMTH504_DIFFERENTIAL_AND_INTEGRAL_EQUATION

Unit 20: Higher Order Equations with Constant Coefficients and Monge’s Method

Self Assessment Notes

1. Solve
2
(D 3 6D D 11D D 2 6D 3 )z 0
2. Solve
2r 5s 2t 0

2 z 2 z 2 z
where r , s , t
x 2 x y y 2

20.2 Case when the Auxiliary Equation has Equal Roots

Consider the equation

2
(D mD ) z = 0 ...(9)
Put (D mD )z = u.

Equation (9) becomes

(D mD )u = 0
The solution is

u = F (y mx )

Therefore
(D mD )z = F (y mx )

z z
or m = F (y mx )
x y
The subsidiary equations are

dx dy dz
=
1 m ( F y mx )
From the first two terms we get

y mx = a
and from first and last term we have

dz ( F y mx )dx = 0

a
or dz F ( )dx = 0
So the solution is

a
z = xF ( ) b

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