我想評估ODE的解決方案phi(+oo) 在哪里的價值phi(xi)Eq(Derivative(phi(xi), (xi, 2)), (-K + xi**2)*phi(xi))并且K是一個已知的實變量。通過dsolve，我得到了解決方案：Eq(phi(xi), -K*xi**5*r(3)/20 + C2*(K**2*xi**4/24 - K*xi**2/2 + xi**4/12 + 1) + C1*xi*(xi**4/20 + 1) + O(xi**6))r()在右側的第一項中具有未知函數。這是我的代碼：import numpy as npimport matplotlib.pyplot as pltimport sympyfrom sympy import I, pi, oosympy.init_printing()def apply_ics(sol, ics, x, known_params):    """    Apply the initial conditions (ics), given as a dictionary on    the form ics = {y(0): y0, y(x).diff(x).subs(x, 0): yp0, ...},    to the solution of the ODE with independent variable x.    The undetermined integration constants C1, C2, ... are extracted    from the free symbols of the ODE solution, excluding symbols in    the known_params list.    """    free_params = sol.free_symbols - set(known_params)    eqs = [(sol.lhs.diff(x, n) - sol.rhs.diff(x, n)).subs(x, 0).subs(ics)            for n in range(len(ics))]    sol_params = sympy.solve(eqs, free_params)    return sol.subs(sol_params)K = sympy.Symbol('K', positive = True)xi = sympy.Symbol('xi',real = True)phi = sympy.Function('phi')ode = sympy.Eq( phi(xi).diff(xi, 2), (xi**2-K)*phi(xi))ode_sol = sympy.dsolve(ode)ics = { phi(0):1, phi(xi).diff(xi).subs(xi,0): 0}phi_xi_sol = apply_ics(ode_sol, ics, xi, [K])ode_sol解決方案在哪里，phi_xi_sol是應用初始條件后的解決方案。由于r()在 NumPy 中未定義，我無法通過以下方式評估結果for g in [0.9, 0.95, 1, 1.05, 1.2]:    phi_xi = sympy.lambdify(xi, phi_xi_sol.rhs.subs({K:g}), 'numpy')有誰知道這個功能是什么r()意思，我應該如何處理它？