- Find the general solution to:
Input:
kovacicsols(y’’=y*(1/x-3/16x^2))
Output:
Therefore, the general solution is
y=C1 x1/4 e2 √x+C2 x1/4 e−2 √x.
- Solve:
Input:
kovacicsols(x’’+3*(t^2-t+1)/(16*(t-1)^2*t^2)*x,t,x)
Output:
⎡
⎢
⎢
⎢
⎢
⎣ | ⎛
⎜
⎝ | −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.
|
Input:
r:=(4x^6-8x^5+12x^4+4x^3+7x^2-20x+4)/(4x^4) |
kovacicsols(y’’=r*y)
|
Output:
Hence y=(x2−1) x−3/2 ex3−2 x2−2/2 x is
a solution to the given equation.
- Solve
Input:
kovacicsols(y’’+y’=6y/x^2)
Output:
- Solve the Titchmarsh equation
Input:
kovacicsols(y’’+(19-x^2)*y=0,x,y)
Output:
⎡
⎢
⎢
⎣ | ⎛
⎜
⎜
⎝ | | x− | | x3+ | | x5−18 x7+x9 | ⎞
⎟
⎟
⎠ | e | | ⎤
⎥
⎥
⎦ |
This is only one, particular solution.
- Find the general solution of Halm’s equation
Input:
sol:=kovacicsols((1+x^2)^2*y’’+3y=0,x,y)
Output:
The other basic solution is obtained by
using (13).
Input:
y1:=sol[0]; y2:=normal(y1*int(y1^-2,x))
Output:
Therefore,
y=C1 x2−1/√x2+1+C2 x/√x2+1,
where C1,C2∈ℝ.
- Find the general solution of the
non-homogeneous equation
First you need to find the general solution to the corresponding
homogeneous equation
yh″−27 yh/36 (x−1)2=0.
Input:
sols:=kovacicsols(y’’-y*27/(36*(x-1)^2),x,y)
Output:
Call this solution y1 and find the other basic independent solution
by using (13).
Input:
y1:=sols[0]:; y2:=y1*int(1/y1^2,x)
Output:
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 as
where f(x)=x+4 and W is the Wronskian of y1 and y2, i.e.
Input:
W:=y1*y2’-y2*y1’:; f:=x+4:; |
yp:=normal(-y1*int(y2*f/W,x)+y2*int(y1*f/W,x))
|
Output:
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.
Input:
purge(C1,C2):; ysol:=yp+C1*y1+C2*y2:; |
normal(diff(ysol,x,2)-27/(36*(x-1)^2)*ysol)==f
|
Output:
- Solve the equation:
from the original Kovacic’s paper:
Input:
r:=-3/(16x^2)-2/(9*(x-1)^2)+3/(16x*(x-1)):; |
kovacicsols(y’’=r*y)
|
Output:
-omega_^4*x^4*(x-1)^4+
omega_^3*x^3*(x-1)^3*(7*x-3)/3- omega_^2*x^2*(x-1)^2*(48*x^2-41*x+9)/24+ omega_*x*(x-1)*(320*x^3-409*x^2+180*x-27)/432+ (-2048*x^4+3484*x^3-2313*x^2+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.
|
Input:
kovacicsols([48t*(t+1)*(5t-4),8*(25t+16)*(t-2),-(5t+68)],t)
Output:
| | 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 | ⎞
⎠ |
| | | | | | | | | |
|