13.4.3 Solving linear homogeneous second-order ODE with rational coefficients
The kovacicsols command
finds Louivillian solutions of ordinary linear homogeneous second-order
differential equations of the form
where a, b and c are rational functions of the independent
variable. kovacicsols uses Kovacic’s algorithm.
-
kovacicsols takes one mandatory argument and two
optional arguments:
-
kode, an equality of the form of
equation (1), an expression for the left-hand side,
or a list of the coefficients [a,b,c].
- Optionally, x the independent variable (by default x).
- Optionally, y, the dependent variable (by default y).
This option should not be used if the first argument is a list
of coefficients.
- kovacicsols(kode ⟨,x,y ⟩)
returns a Liouvillian solution of equation (1).
This can be a list or an expression. An empty list means
that there are no Liouvillian solutions to the equation. A
non-empty list will contain one or two independent
solutions to the differential equation. If
the list contains two solutions y1 and y2, the
general solution will be
where C1,C2∈ℝ are arbitrary constants. However, for
some equations only one solution y1 is returned, in which case the
other solution can be obtained as (using reduction of order):
If kovacicsols returns an expression, it will give the
solution of the differential equation implicitly. In that case
the return value is a polynomial P of order n∈{4,6,12} in
the variable omega_ (denoted here by ω) with
rational coefficients rk, k=0,1,2,…,n. If P(ω0)=0
for some ω0, then y=exp(∫ω0) is a
solution to the differential equation.
Examples
Find the general solution to y′′=(1/x−3/16 x2) y.
kovacicsols(y''=y*(1/x-3/16x^2)) |
Therefore, the general solution is
y=C1 x1/4 e2 √x+C2 x1/4 e−2 √x.
Solve x′(t)+3 (t2−t+1)/16 (t−1)2 t2 x(t)=0.
kovacicsols(x''+3*(t^2-t+1)/(16*(t-1)^2*t^2)*x,t,x) |
|
| ⎡
⎢
⎢
⎢
⎢
⎣ | ⎛
⎜
⎝ | −t | ⎛
⎝ | t−1 | ⎞
⎠ | ⎛
⎜
⎝ | 2 | √ | | +2 t−1 | ⎞
⎟
⎠ | ⎞
⎟
⎠ | | , | ⎛
⎜
⎝ | t | ⎛
⎝ | t−1 | ⎞
⎠ | ⎛
⎜
⎝ | 2 | √ | | −2 t+1 | ⎞
⎟
⎠ | ⎞
⎟
⎠ | | ⎤
⎥
⎥
⎥
⎥
⎦ |
| | | | | | | | | | |
|
so the general solution is, for C1,C2∈ℝ,
Find a particular solution to
y′′=4 x6−8 x5+12 x4+4 x3+7 x2−20 x+4/4 x4 y.
r:=(4x^6-8x^5+12x^4+4x^3+7x^2-20x+4)/(4x^4);
kovacicsols(y''=r*y) |
Hence y=(x2−1) x−3/2 ex3−2 x2−2/2 x is
a solution to the given equation.
Solve y′′+y′=6 y/x2.
kovacicsols(y''+y'=6y/x^2) |
Solve the Titchmarsh equation y′′+(19−x2) y=0.
kovacicsols(y''+(19-x^2)*y=0,x,y) |
|
| ⎡
⎢
⎢
⎣ | ⎛
⎜
⎜
⎝ | | x− | | x3+ | | x5−18 x7+x9 | ⎞
⎟
⎟
⎠ | e | | ⎤
⎥
⎥
⎦ |
| | | | | | | | | | |
|
This is only a single, particular solution.
Find the general solution of Halm’s equation (1+x2)2 y′′(x)+3 y(x)=0.
sol:=kovacicsols((1+x^2)^2*y''+3y=0,x,y) |
The other basic solution is obtained by
using (2).
y1:=sol[0]; y2:=normal(y1*int(y1^-2,x)) |
Therefore,
y=C1 x2−1/√x2+1+C2 x/√x2+1,
where C1,C2∈ℝ.
Find the general solution of the
non-homogeneous equation y′′−27 y/36 (x−1)2=x+4.
First you need to find the general solution to the corresponding
homogeneous equation
yh′′−27 yh/36 (x−1)2=0.
sols:=kovacicsols(y''-y*27/(36*(x-1)^2),x,y) |
Call this solution y1 and find the other basic independent solution
by using (2).
y1:=sols[0]:;
y2:=y1*int(1/y1^2,x) |
So the general solution of the homogeneous equation is
yh=C1 y1+C2 y2= | | , C1,C2∈ℝ.
|
A particular solution yp of the non-homogeneous equation can be
obtained by variation of parameters:
where f(x)=x+4 and W is the Wronskian of y1 and y2, i.e.
W:=y1*y2'-y2*y1':; f:=x+4:;
yp:=normal(-y1*int(y2*f/W,x)+y2*int(y1*f/W,x)) |
Hence yp=1/21 (4 x3+72 x2−156 x+80). Now y=yp+yh.
You can checking that it is indeed the general solution of the
given equation.
purge(C1,C2):; ysol:=yp+C1*y1+C2*y2:;
normal(diff(ysol,x,2)-27/(36*(x-1)^2)*ysol)==f |
Solve the equation
y′′=(3/16 x (x−1)−2/9 (x−1)2−3/16 x2)y
from the original Kovacic’s paper.
r:=-3/(16x^2)-2/(9*(x-1)^2)+3/(16x*(x-1)):;
kovacicsols(y''=r*y) |
| | + | ω_3 x3 | ⎛
⎝ | x−1 | ⎞
⎠ | 3 | ⎛
⎝ | 7 x−3 | ⎞
⎠ |
|
|
3 |
| − | ω_2 x2 | ⎛
⎝ | x−1 | ⎞
⎠ | 2 | ⎛
⎝ | 48 x2−41 x+9 | ⎞
⎠ |
|
|
24 |
|
| | | | | | | | | |
| + | ω_ x | ⎛
⎝ | x−1 | ⎞
⎠ | ⎛
⎝ | 320 x3−409 x2+180 x−27 | ⎞
⎠ |
|
|
432 |
|
| | | | | | | | | |
| + | −2048 x4+3484 x3−2313 x2+702 x−81 |
|
20736 |
|
| | | | | | | | | |
|
The solution is y=exp(∫ω0), where ω0 is
a zero of the above expression, thus being a root of a fourth-order
polynomial in ω. In similar cases you can try the Ferrari method to obtain ω0.
Solve the equation 48 t (t+1) (5 t−4) y′′+8 (25 t+16) (t−2) y′−(5 t+68) y=0.
kovacicsols([48t*(t+1)*(5t-4),8*(25t+16)*(t-2),-(5t+68)],t) |
| | | ω_4 | ⎛
⎝ | 135 t4−616 t3−144 t2+3072 t−4096 | ⎞
⎠ | − | | ω_3 t | ⎛
⎝ | t+1 | ⎞
⎠ | ⎛
⎝ | 23 t2−92 t+128 | ⎞
⎠ |
| | | | | | | | | |
| − | | ω_2 t2 | ⎛
⎝ | t+1 | ⎞
⎠ | ⎛
⎝ | 15 t3−80 t2+80 t+256 | ⎞
⎠ | + | | ω_ t3 | ⎛
⎝ | t−4 | ⎞
⎠ | ⎛
⎝ | t+1 | ⎞
⎠ | 2 | ⎛
⎝ | 5 t+8 | ⎞
⎠ | −t4 | ⎛
⎝ | t+1 | ⎞
⎠ | 2 | ⎛
⎝ | t+4 | ⎞
⎠ | ⎛
⎝ | 5 t+4 | ⎞
⎠ |
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