Weber-Hermite differential equation
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I was solving a quantum mechanics problem (harmonic oscillateur) and i need to solve this Weber-Hermite differential equation in an analytic method:
$$y"-x^2(y)=0$$
I know the solution of this equation is the parabolic cylinder function but i need to solve it analyticaly (to understand how they get the solution)
And thank you!!
differential-equations parabolic-pde hermite-polynomials
edited Nov 16 at 4:22
Isham
12.7k3929
asked Nov 16 at 3:23
Yassine Sifeddine
1
add a comment |
up vote
-1
down vote
favorite
I was solving a quantum mechanics problem (harmonic oscillateur) and i need to solve this Weber-Hermite differential equation in an analytic method:
$$y"-x^2(y)=0$$
I know the solution of this equation is the parabolic cylinder function but i need to solve it analyticaly (to understand how they get the solution)
And thank you!!
differential-equations parabolic-pde hermite-polynomials
edited Nov 16 at 4:22
Isham
12.7k3929
asked Nov 16 at 3:23
Yassine Sifeddine
1
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I was solving a quantum mechanics problem (harmonic oscillateur) and i need to solve this Weber-Hermite differential equation in an analytic method:
$$y"-x^2(y)=0$$
I know the solution of this equation is the parabolic cylinder function but i need to solve it analyticaly (to understand how they get the solution)
And thank you!!
differential-equations parabolic-pde hermite-polynomials
edited Nov 16 at 4:22
Isham
12.7k3929
asked Nov 16 at 3:23
Yassine Sifeddine
1
I was solving a quantum mechanics problem (harmonic oscillateur) and i need to solve this Weber-Hermite differential equation in an analytic method:
$$y"-x^2(y)=0$$
I know the solution of this equation is the parabolic cylinder function but i need to solve it analyticaly (to understand how they get the solution)
And thank you!!
differential-equations parabolic-pde hermite-polynomials
differential-equations parabolic-pde hermite-polynomials
edited Nov 16 at 4:22
Isham
12.7k3929
asked Nov 16 at 3:23
Yassine Sifeddine
1
edited Nov 16 at 4:22
Isham
12.7k3929
asked Nov 16 at 3:23
Yassine Sifeddine
1
edited Nov 16 at 4:22
Isham
12.7k3929
edited Nov 16 at 4:22
Isham
12.7k3929
edited Nov 16 at 4:22
Isham
12.7k3929
12.7k3929
asked Nov 16 at 3:23
Yassine Sifeddine
1
asked Nov 16 at 3:23
Yassine Sifeddine
1
asked Nov 16 at 3:23
Yassine Sifeddine
1
1
add a comment |
add a comment |
1 Answer
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I do not know if this answers the question.
The general Weber differential equation (the solution of which being $D_nu (x)$) is
$$y''+y left(nu +frac{1}{2}-frac{t^2}{4}right)y=0$$ So, for your case, $nu=-frac{1}{2}$ and you need to redefine $x=t sqrt 2$ to get your equation.
answered Nov 16 at 5:22
Claude Leibovici
116k1156131
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I do not know if this answers the question.
The general Weber differential equation (the solution of which being $D_nu (x)$) is
$$y''+y left(nu +frac{1}{2}-frac{t^2}{4}right)y=0$$ So, for your case, $nu=-frac{1}{2}$ and you need to redefine $x=t sqrt 2$ to get your equation.
answered Nov 16 at 5:22
Claude Leibovici
116k1156131
add a comment |
up vote
0
down vote
I do not know if this answers the question.
The general Weber differential equation (the solution of which being $D_nu (x)$) is
$$y''+y left(nu +frac{1}{2}-frac{t^2}{4}right)y=0$$ So, for your case, $nu=-frac{1}{2}$ and you need to redefine $x=t sqrt 2$ to get your equation.
answered Nov 16 at 5:22
Claude Leibovici
116k1156131
add a comment |
up vote
0
down vote
up vote
0
down vote
I do not know if this answers the question.
The general Weber differential equation (the solution of which being $D_nu (x)$) is
$$y''+y left(nu +frac{1}{2}-frac{t^2}{4}right)y=0$$ So, for your case, $nu=-frac{1}{2}$ and you need to redefine $x=t sqrt 2$ to get your equation.
answered Nov 16 at 5:22
Claude Leibovici
116k1156131
I do not know if this answers the question.
The general Weber differential equation (the solution of which being $D_nu (x)$) is
$$y''+y left(nu +frac{1}{2}-frac{t^2}{4}right)y=0$$ So, for your case, $nu=-frac{1}{2}$ and you need to redefine $x=t sqrt 2$ to get your equation.
answered Nov 16 at 5:22
Claude Leibovici
116k1156131
answered Nov 16 at 5:22
Claude Leibovici
116k1156131
answered Nov 16 at 5:22
Claude Leibovici
116k1156131
answered Nov 16 at 5:22
Claude Leibovici
116k1156131
116k1156131
add a comment |
add a comment |
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