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5.57.3  Solving linear homogeneous second-order ODE with rational coefficients : kovacicsols

kovacicsols uses Kovacic’s algorithm to find a Liouvillian solution of an ordinary linear homogeneous second-order differential equation

a y″+b y′+c y=0,    (6)

where a, b and c are rational functions of the independent variable. The command takes from one to three arguments :

The dependent variable should not be specified if the equation (6) is entered as a list of coefficients.

The return value can be a list or an expression. An empty list means that there are no Liouvillian solutions to the input equation. If a non-empty list is returned, it contains one or two independent solution(s) y1 (and y2) to the equation (6). The general solution to (6) is then

y=C1 y1+C2 y2

where C1,C2∈ℝ are arbitrary constants. However, for some equations only y1 is returned, in which case y2 can be obtained as (using reduction of order) :

y2=y1 y1−2.    (7)

If kovacicsols returns an expression, it means that the solution to (6) is given 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 P0)=0 for some ω0, then y=exp(∫ω0) is a solution to the equation (6).

Examples.

In the first example we find the general solution to the equation

y″=


1
x
3
16 x2



 y

Input :

kovacicsols(y''=y*(1/x-3/16x^2))

Output :

[x^(1/4)*exp(2*sqrt(x)),x^(1/4)*exp(-2*sqrt(x))]

Therefore, y=C1 x1/4 e2 √x+C2 x1/4 e−2 √x is the general solution.

In the following example we solve the equation

x″(t)+
3 (t2t+1)
16 (t−1)2 t2
 x(t)=0. 

Input :

kovacicsols(x''+3*(t^2-t+1)/(16*(t-1)^2*t^2)*x,t,x)

Output :

[(-t*(t-1)*(2*t+2*sqrt(t^2-t)-1))^(1/4), (t*(t-1)*(-2*t+2*sqrt(t^2-t)+1))^(1/4)]

Now for arbitrary C1,C2∈ℝ we have

x(t)=C1 
t (t−1) (1−2 t−2 
t2t
)
+
C2 
t (t−1) (1−2 t+2 
t2t
)
.

In the next example we find a particular solution to the equation

y″=
x6−8 x5+12 x4+4 x3+7 x2−20 x+4
x4
 y

Input :

r:=(4x^6-8x^5+12x^4+4x^3+7x^2-20x+4)/(4x^4):; kovacicsols(y''=r*y)

Output :

[(x^2-1)/(x*sqrt(x))*exp((x^3-2*x^2-2)/(2*x))]

Hence y=(x2−1) x−3/2 ex3−2 x2−2/2 x is a solution to the given equation.

A similar output is obtained when solving the equation

y″+y′=
y
x2

Input :

kovacicsols(y''+y'=6y/x^2)

Output :

[(x^2+6*x+12)*exp(-x)/x^2]

To solve Titchmarsh equation

y″+(19−x2y=0, 

input :

kovacicsols(y''+(19-x^2)*y=0,x,y)

We obtain a particular solution

y=


x9−18 x7+
189 x5
2
315 x3
2
+
945 x
16



 exp


x2
2



To find the general solution of Halm’s equation

(1+x2)2 y″(x)+3 y(x)=0, 

input :

sol:=kovacicsols((1+x^2)^2*y''+3y=0,x,y)

Output :

[(x^2-1)/(sqrt(x^2+1))]

The other basic solution is obtained by using (7). Input :

y1:=sol[0]; y2:=normal(y1*int(y1^-2,x))

Output :

(x^2-1)/(sqrt(x^2+1)),-x/(sqrt(x^2+1))

Therefore, y=C1 x2−1/√x2+1+C2 x/√x2+1, where C1,C2∈ℝ.

In the following example we find the general solution of the non-homogeneous equation

y″−
27 y
36 (x−1)2
=x+4. 

First we 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 :

[(x^2-2*x)/(sqrt(x-1))]

We call the obtained solution y1 and find the other basic independent solution by using (7). Input :

y1:=sols[0]:; y2:=y1*int(1/y1^2,x)

Output :

-1/(sqrt(x-1)*2)

Now the general solution of the homogeneous equation is

yh=C1 y1+C2 y2=
C1 (x2−2 x)+C2
x−1
,   C1,C2∈ℝ. 

A particular solution yp of the non-homogeneous equation can be obtained by variation of parameters as

yp=−y1 
y2 f(x)
W
 dx+y2 
y1 f(x)
W
 dx

where f(x)=x+4 and W is the Wronskian of y1 and y2, i.e. 

W=y1 y2′−y2 y1′≠ 0.

Input :

W:=y1*y2’-y2*y1’:; f:=x+4:; yp:=normal(-y1*int(y2*f/W,x)+y2*int(y1*f/W,x))

Output :

(4*x^3+72*x^2-156*x+80)/21

Hence yp=1/21 (4 x3+72 x2−156 x+80). Now y=yp+yh. We proceed by 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 :

true

In the next example we attempt to solve the equation from the original Kovacic’s paper :

y″=


3
16 x (x−1)
2
9 (x−1)2
3
16 x2



 y

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 one can try the Ferrari method to obtain ω0.

We get similar output while trying to solve the equation

48 t (t+1) (5 t−4) y″+8 (25 t+16) (t−2) y′−(5 t+68) y=0. 

Input :

de:=[48t*(t+1)*(5t-4),8*(25t+16)*(t-2),-(5t+68)]:; kovacicsols(de,t)

Output :

omega_^4*(135*t^4-616*t^3-144*t^2+3072*t-4096)/20736- omega_^2*t^2*(t+1)*(15*t^3-80*t^2+80*t+256)/24- t^4*(t+1)^2*(t+4)*(5*t+4)+ 2*omega_*t^3*(t+1)^2*(t-4)*(5*t+8)/3- omega_^3*t*(t+1)*(23*t^2-92*t+128)/54

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