Solutions and Zeros of Differential Equations: Lagrangian, Hermite Equation

1. The differential equation is
y’’ + (3-sinx)y = 0
The possible Lagrangian of the equation is given by,
L(y’,y,x) = ½ (y’^2 – y^2(sinx – 3))
Now, the graph of individual solution of the equation is given in wolframalpha as,
A sample of solutions for different initial conditions of the above equation is given
below.

Hence, it is seen that the sample solutions have zeros for different x’s as given above.
Hence, all the solutions will have zeros for infinite numbers of x values in the range
[0,2π].
3. The Hermite equation is given by the form,
y’’ – 2xy’ +2αy = 0
i) Now, putting α = 3 in the equation we get
y’’ – 2xy’ + 6y = 0
This equation can be converted in the form of Sturm-Lioville equation which is of
the form
d
dx ( y' ( x )
ex2
) – 6 y ( x )
e
( x2 ) =0
Now, the solution of the equation is given by,
y(x) = c 2 ex2
¿
where, erf(x) is the gauss error function given by,
erfi(x) = imaginary error function given by,
erfi(x) = -ierf(ix), where i = √−1.
c1 and c2 are the constants determined by the initial conditions of the differential
equation.
Now, as the first expression in the solution vanishes only when
¿ ¿ ¿
And the second expression vanishes when (8 x3−12 x) = 0. Now, the above two
expressions vanishes for finite number of zeros, so the solution has finite number
of zeros.
ii) Now, putting y(0) = 0 in the solution of the differential equation we get,