Basics of Fluid Mechanics, 2014a

A.2. ORDINARY DIFFERENTIAL EQUATIONS (ODE) 585
Where the real part is
and the imaginary number is
α = −b
2 a
√
4 ac− b
2
β =
2 a
(A.73)
(A.74)
Example A.6:
Solve the following ODE
d 2 u
dt 2 +7du +10u =0
(1.VI.a)
dt
Solution
The characteristic equation is
s 2 +7s +10=0
The solution of equation (1.VI.b) are −2, and −5. Thus, the solution is
u = k 1 e −2 t + k 2 e −5 t
(1.VI.b)
(1.VI.c)
End Solution
A.2.4.1
Non–Homogeneous Second ODE
Homogeneous equation are equations that equal to zero. This fact can be used to solve
non-homogeneous equation. Equations that not equal to zero in this form
a d2 u
dt 2 + b du + cu= l(x)
dt (A.75)
The solution of the homogeneous equation is zero that is the operation L(u h )=0,
where L is Linear operator. The additional solution of L(u p ) is the total solution as
=0
{ }} {
L (u total )= L (u h )+L (u p )=⇒ u total = u h + u p
(A.76)
Where the solution u h is the solution of the homogeneous solution and u p is the solution
of the particular function l(x). If the function on the right hand side is polynomial than
the solution is will
n∑
u total = u h +
(A.77)
The linearity of the operation creates the possibility of adding the solutions.
i=1
u pi