# hello, I want to convert this code into legendre differential equation, can someone help?

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Rajveer Kalsi el 22 de Abr. de 2021
Respondida: Vaibhav el 30 de Mayo de 2024
clc
clear all
syms x z rhs rhs2 t x1 z2 y k4 k5
%%%euler legender starts
f=input ('enter the function f(x), which is coefficient of D terms:');
z2=log(f);
a=diff(f,x)
a
k1=input('enter the constant k1');
k2=input('enter the constant k2');
k3=input('enter constant k3');
rhs=input ('enter the rhs function, which is coefficient of D terms:');
eqn=f==exp(z);
x=solve(eqn,x);
x
rhs2=subs(rhs,x)
rhs2
a0=k1*a^2;
b0=(k2*a)-(k1*a*a)
c0=k3
a1=a0/a0;b1=b0/a0;c1=c0/a0;
eq=a1*x1^2+b1*x1+c1
r=solve(eq,'x1')
if imag(r)~=0
y1=exp(real(r(1))*z)*cos(imag(r(1))*z);
y2=exp(real(r(1))*z)*sin(abs(imag(r(1)))*z)
else if r(1)==r(2)
y1=exp(r(1)*z)
y2=z*exp(r(1)*z)
else
y1=exp(r(1)*z);
y2=exp(r(2)*z)
end
end
y_h=k4*y1+k5*y2;
W=simplify(y1*diff(y2)-y2*diff(y1));
g=rhs2;
y_p=-y1*int(y2*g/W)+y2*int(y1*g/W);
y=simplify(y_h+y_p);
y
simplify(subs(y,z,log(f)))
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### Respuestas (1)

Vaibhav el 30 de Mayo de 2024
Hi Rajveer
To adapt this code for solving the Legendre differential equation, we can remove the user input for the general equation and hard-code the specifics of the Legendre differential equation. Here is the code for your reference:
clc
clear all
syms x y(x) n
% Legendre Differential Equation
LDE = (1 - x^2)*diff(y, x, 2) - 2*x*diff(y, x) + n*(n+1)*y == 0;
% Solve the Legendre Differential Equation
ySol = dsolve(LDE);
% Display the solution
disp(ySol)
Above code defines the Legendre differential equation and then uses dsolve function to find the solution. The variable n is kept as a symbolic variable, which means the solution will be in terms of n. You may specify a particular value of n if you are interested in a specific solution corresponding to that value.
Hope it helps!
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