### 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

a y′′+b y′+c y=0, (1)

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

  y=C1 y1+C2 y2

  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):

  y2=y1 | ∫ | y1−2. (2)

  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−3/16x2)y.

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

⎡
⎢
⎣

x

1
4

e

2

√

x

,x

1
4

e

−2

√

x

⎤
⎥
⎦

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

√

t2−t

+2 t−1

⎞
⎟
⎠

⎞
⎟
⎠

1
4

,

⎛
⎜
⎝

t

⎛
⎝

t−1

⎞
⎠

⎛
⎜
⎝

2

√

t2−t

−2 t+1

⎞
⎟
⎠

⎞
⎟
⎠

1
4

⎤
⎥
⎥
⎥
⎥
⎦

so the general solution is, for C1,C2∈ℝ,

x(t)=C1

∜

t(t−1) (1−2t−2

√

t2−t

)

+

C2

∜

t(t−1) (1−2t+2

√

t2−t

)

.

Find a particular solution to y′′=4x6−8x5+12x4+4x3+7x2−20x+4/4x4 y.

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

⎡
⎢
⎢
⎢
⎢
⎣

⎛
⎝

−1+x2

⎞
⎠

e

1
2

⎛
⎝

x3−2 x2−2

⎞
⎠

x

x

√

x

⎤
⎥
⎥
⎥
⎥
⎦

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

Solve y′′+y′=6y/x2.

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

⎡
⎢
⎢
⎣

⎛ ⎝ | 12+6 x+x2 | ⎞ ⎠ | e−x

x2

⎤
⎥
⎥
⎦

Solve the Titchmarsh equation y′′+(19−x2)y=0.

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

⎡
⎢
⎢
⎣

⎛
⎜
⎜
⎝

945
16

x−

315
2

x3+

189
2

x5−18 x7+x9

⎞
⎟
⎟
⎠

e

−

x2
2

⎤
⎥
⎥
⎦

This is only a single, particular solution.

Find the general solution of Halm’s equation (1+x2)2 y′′(x)+3y(x)=0.

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

⎡
⎢
⎢
⎢
⎢
⎣

−1+x2

√

x2+1

⎤
⎥
⎥
⎥
⎥
⎦

The other basic solution is obtained by using (2).

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

−1+x2

√

x2+1

,−

x

√

x2+1

x2+1

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′′−27y/36 (x−1)2=x+4. First you need to find the general solution to the corresponding homogeneous equation yh′′−27yh/36 (x−1)2=0.

sols:=kovacicsols(y''-y*27/(36*(x-1)^2),x,y)

⎡
⎢
⎢
⎢
⎢
⎣

−2 x+x2

√

x−1

⎤
⎥
⎥
⎥
⎥
⎦

Call this solution y1 and find the other basic independent solution by using (2).

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

−

√

x−1

2 x−2

So the general solution of the homogeneous equation is

yh=C1 y1+C2 y2=

C1(x2−2x)+C2

√

x−1

,C1,C2∈ℝ.

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

yp=−y1

∫

y2 f(x)
W

d x+y2

∫

y1 f(x)
W

d x,

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

W=y1 y2′−y2 y1′≠ 0.

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

4 x3+72 x2−156 x+80
21

Hence yp=1/21 (4x3+72x2−156x+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

true |  |  |  |  |  |  |  |  |  |

Solve the equation y′′=(3/16x(x−1)−2/9 (x−1)2−3/16x2)y from the original Kovacic’s paper.

r:=-3/(16 x^2)-2/(9*(x-1)^2)+3/(16 x*(x-1)):; kovacicsols(y''=r*y)

−ω_4 x4 | ⎛ ⎝ | x−1 | ⎞ ⎠ | 4

+

ω_3 x3 | ⎛ ⎝ | x−1 | ⎞ ⎠ | 3 | ⎛ ⎝ | 7x−3 | ⎞ ⎠

3

−

ω_2 x2 | ⎛ ⎝ | x−1 | ⎞ ⎠ | 2 | ⎛ ⎝ | 48x2−41x+9 | ⎞ ⎠

24

+

ω_x | ⎛ ⎝ | x−1 | ⎞ ⎠ | ⎛ ⎝ | 320x3−409x2+180x−27 | ⎞ ⎠

432

+

−2048x4+3484x3−2313x2+702x−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 48t(t+1) (5t−4)y′′+8 (25t+16) (t−2)y′−(5t+68)y=0.

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

1
20736

ω_4

⎛
⎝

135t4−616t3−144t2+3072t−4096

⎞
⎠

−

1
54

ω_3 t

⎛
⎝

t+1

⎞
⎠

⎛
⎝

23t2−92t+128

⎞
⎠

−

1
24

ω_2 t2

⎛
⎝

t+1

⎞
⎠

⎛
⎝

15t3−80t2+80t+256

⎞
⎠

+

2
3

ω_t3

⎛
⎝

t−4

⎞
⎠

⎛
⎝

t+1

⎞
⎠

2

⎛
⎝

5t+8

⎞
⎠

−t4

⎛
⎝

t+1

⎞
⎠

2

⎛
⎝

t+4

⎞
⎠

⎛
⎝

5t+4

⎞
⎠