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

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

The kovacicsols command finds Louivillian solutions of ordinary linear homogeneous second-order differential equations of the form

a y″+b y′+c y=0, (12)

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 (12), 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 (12). 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. (13)

  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.

  Input:
  kovacicsols(y’’=y*(1/x-3/16x^2))
  Output:

  ⎡
  ⎢
  ⎣

  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.

  Input:
  kovacicsols(x’’+3*(t^2-t+1)/(16*(t-1)^2*t^2)*x,t,x)
  Output:

  ⎡
  ⎢
  ⎢
  ⎢
  ⎢
  ⎣

  ⎛
  ⎜
  ⎝

  −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) (1−2 t−2

  √

  t2−t

  )

  +

  C2

  ∜

  t (t−1) (1−2 t+2

  √

  t2−t

  )

  .

- 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:

  ⎡
  ⎢
  ⎢
  ⎢
  ⎢
  ⎣

  ⎛
  ⎝

  −1+x2

  ⎞
  ⎠

  e

  1
  2

  ⎛
  ⎝

  x3−2 x2−2

  ⎞
  ⎠

  x

  x

  √

  x

  ⎤
  ⎥
  ⎥
  ⎥
  ⎥
  ⎦

  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

  .

  Input:
  kovacicsols(y’’+y’=6y/x^2)
  Output:

  ⎡
  ⎢
  ⎢
  ⎣

  ⎛ ⎝ | 12+6 x+x2 | ⎞ ⎠ | e−x

  x2

  ⎤
  ⎥
  ⎥
  ⎦

- Solve the Titchmarsh equation

  y″+(19−x2) y=0

  Input:
  kovacicsols(y’’+(19-x^2)*y=0,x,y)
  Output:

  ⎡
  ⎢
  ⎢
  ⎣

  ⎛
  ⎜
  ⎜
  ⎝

  945
  16

  x−

  315
  2

  x3+

  189
  2

  x5−18 x7+x9

  ⎞
  ⎟
  ⎟
  ⎠

  e

  −

  x2
  2

  ⎤
  ⎥
  ⎥
  ⎦

  This is only one, particular solution.
- 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:

  ⎡
  ⎢
  ⎢
  ⎢
  ⎢
  ⎣

  −1+x2

  √

  x2+1

  ⎤
  ⎥
  ⎥
  ⎥
  ⎥
  ⎦

  The other basic solution is obtained by using (13).
  Input:
  y1:=sol[0]; y2:=normal(y1*int(y1^-2,x))
  Output:

  −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″−

  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.
  Input:
  sols:=kovacicsols(y’’-y*27/(36*(x-1)^2),x,y)
  Output:

  ⎡
  ⎢
  ⎢
  ⎢
  ⎢
  ⎣

  −2 x+x2

  √

  x−1

  ⎤
  ⎥
  ⎥
  ⎥
  ⎥
  ⎦

  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:

  −

  √

  x−1

  2 x−2

  So 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 x3+72 x2−156 x+80
  21

  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:

  true

- Solve the equation:

  y″=

  ⎛
  ⎜
  ⎜
  ⎝

  3
  16 x (x−1)

  −

  2
  9 (x−1)2

  −

  3
  16 x2

  ⎞
  ⎟
  ⎟
  ⎠

  y.

  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:

  1
  20736

  ω_4 | ⎛ ⎝ | 135 t4−616 t3−144 t2+3072 t−4096 | ⎞ ⎠

  −

  1
  54

  ω_3 t

  ⎛
  ⎝

  t+1

  ⎞
  ⎠

  ⎛
  ⎝

  23 t2−92 t+128

  ⎞
  ⎠

  −

  1
  24

  ω_2 t2

  ⎛
  ⎝

  t+1

  ⎞
  ⎠

  ⎛
  ⎝

  15 t3−80 t2+80 t+256

  ⎞
  ⎠

  +

  2
  3

  ω_ t3

  ⎛
  ⎝

  t−4

  ⎞
  ⎠

  ⎛
  ⎝

  t+1

  ⎞
  ⎠

  2

  ⎛
  ⎝

  5 t+8

  ⎞
  ⎠

  −t4

  ⎛
  ⎝

  t+1

  ⎞
  ⎠

  2

  ⎛
  ⎝

  t+4

  ⎞
  ⎠

  ⎛
  ⎝

  5 t+4

  ⎞
  ⎠