Download Solutions of Ordinary Differential Equations

Review - Solutions of Ordinary Differential Equations
Author: John M. Cimbala, Penn State University
Latest revision: 07 November 2016
First-Order, Separable, Ordinary Differential Equations
Example:
y = y(t)
y
dy
=C
dt
Example:
y = y(t)
ty
dy
=C
dt
•
Separate variables: ydy = Cdt
•
2 Ct + C1
Solve: y 2 / =
•
Separate variables: ydy =
•
y 2 / 2 C ln ( t ) + C1
Solve: =
C
dt
t
First-Order, Linear, Non-Homogeneous, Ordinary Differential Equations
Example:
y = y(x)
dy
Q( x)
+ P( x) y =
dx
( ∫ Pdx )
•
Multiply each term by an integrating factor, v = exp
•
Solve: This leads to an exact differential on the LHS, which can be solved
easily.
Higher-Order, Linear, Homogeneous, Ordinary Differential Equations with Constant Coefficients
Example:
y = y(x)
d3y
dx3
Example:
y = y(x)
d2y
Example:
y = y(x)
2
dx 2
−4
d2y
dx 2
+
dy
+ 6y =
0
dx
− a2 y =
0
Rewrite in terms of a linear operator, D n y ≡
•
•
(D3 - 4D2 + D + 6) y = 0
Factor the LHS: Here, (D + 1) (D - 2) (D - 3) y = 0
Find the roots of the LHS: Here the roots are -1, 2, and 3
•
Solve: y = C1e − x + C2 e 2 x + C3e3 x
•
•
•
Rewrite in terms of the linear operator: Here, (D2 - a) y = 0
Factor the LHS: Here, (D - a) (D + a) y = 0
Find the roots of the LHS: Here the roots are a and -a
=
y C1e ax + C2=
e − ax or y C3 cosh(ax) + C4 sinh(ax)
Solve:
•
d y
+ a2 y =
0
dx 2
dny
•
•
•
•
•
dx n
: Here,
Rewrite in terms of the linear operator: Here, (D2 + a) y = 0
Factor the LHS: Here, (D - ia) (D + ia) y = 0
Find the roots of the LHS: Here the roots are ia and -ia
− iax
=
y C1eiax + C2 e=
Solve:
or y C3 cos(ax) + C4 sin(ax)
Higher-Order, Nonlinear, Non-Homogeneous, Ordinary Differential Equations
Example:
y = y(x)
d3y
dx
3
+ P( x)
2
•
 dy 
+
Q( x)
  =
2
dx
 dx 
d2y
Define N unknowns, where N is the highest order derivative of the
original equation: Here, N = 3, and define
•
d2y
dy
, F2 ≡
, F3 ≡ y
dx
dx 2
Split equation into N first-order equations, one for each F:
F1 ≡
2
•
•
dF3 dy
dF2 d 2 y
dF1 d 3 y
d 2 y  dy 
= 2 ,
=
=
Q
(
x
)
−
P
(
x
)
−
=
,
 
dx
dx dx
dx
dx
dx3
dx 2  dx 
Define derivative array in standard Runge-Kutta form, in terms of the
N unknowns, F1, F2, ..., FN:
dF
dF
dF
D1 ≡ 1 = Q( x) − P( x) F1 − F2 2 , D2 ≡ 2 =
F1 , D3 ≡ 3 =
F2
dx
dx
dx
Solve simultaneously for F1, F2, ..., FN using the Runge-Kutta (R-K)
numerical technique and the boundary conditions. [See a separate online learning module from Professor Cimbala that explains the R-K
technique.]