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

  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 r_k, 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.

Therefore, the general solution is y=C₁ x^1/4 e^2 √x+C₂ x^1/4 e^−2 √x.

Solve x′(t)+3 (t²−t+1)/16 (t−1)² t² x(t)=0.

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.

Hence y=(x²−1) x^−3/2 e^{x3−2 x2−2/2 x} is a solution to the given equation.

Solve y′′+y′=6 y/x².

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

This is only a single, particular solution.

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

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

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 y_h′′−27 y_h/36 (x−1)²=0.

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

So the general solution of the homogeneous equation is

A particular solution y_p 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.

Hence y_p=1/21 (4 x³+72 x²−156 x+80). Now y=y_p+y_h. You can checking that it is indeed the general solution of the given equation.

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

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.