def test_heuristic3():
a, b = symbols("a b")
df = f(x).diff(x)
eq = x**2 * df + x * f(x) + f(x)**2 + x**2
i = infinitesimals(eq, hint='bivariate')
assert i == [{eta(x, f(x)): f(x), xi(x, f(x)): x}]
assert checkinfsol(eq, i)[0]
eq = x**2 * (-f(x)**2 + df) - a * x**2 * f(x) + 2 - a * x
i = infinitesimals(eq, hint='bivariate')
assert checkinfsol(eq, i)[0]
def test_heuristic_abaco2_similar():
a, b = symbols("a b")
F = Function('F')
eq = f(x).diff(x) - F(a * x + b * f(x))
i = infinitesimals(eq, hint='abaco2_similar')
assert i == [{eta(x, f(x)): -a / b, xi(x, f(x)): 1}]
assert checkinfsol(eq, i)[0]
eq = f(x).diff(x) - (f(x)**2 / (sin(f(x) - x) - x**2 + 2 * x * f(x)))
i = infinitesimals(eq, hint='abaco2_similar')
assert i == [{eta(x, f(x)): f(x)**2, xi(x, f(x)): f(x)**2}]
assert checkinfsol(eq, i)[0]
def test_heuristic_linear():
a, b, m, n = symbols("a b m n")
eq = x**(n * (m + 1) - m) * (f(x).diff(x)) - a * f(x)**n - b * x**(n *
(m + 1))
i = infinitesimals(eq, hint='linear')
assert checkinfsol(eq, i)[0]
def test_heuristic_abaco2_unique_unknown():
a, b = symbols("a b")
F = Function('F')
eq = f(x).diff(x) - x**(a - 1)*(f(x)**(1 - b))*F(x**a/a + f(x)**b/b)
i = infinitesimals(eq, hint='abaco2_unique_unknown')
assert i == [{eta(x, f(x)): -f(x)*f(x)**(-b), xi(x, f(x)): x*x**(-a)}]
assert checkinfsol(eq, i)[0]
eq = f(x).diff(x) + tan(F(x**2 + f(x)**2) + atan(x/f(x)))
i = infinitesimals(eq, hint='abaco2_unique_unknown')
assert i == [{eta(x, f(x)): x, xi(x, f(x)): -f(x)}]
assert checkinfsol(eq, i)[0]
eq = (x*f(x).diff(x) + f(x) + 2*x)**2 -4*x*f(x) -4*x**2 -4*a
i = infinitesimals(eq, hint='abaco2_unique_unknown')
assert checkinfsol(eq, i)[0]
def test_heuristic1():
a, b, c, a4, a3, a2, a1, a0 = symbols("a b c a4 a3 a2 a1 a0")
df = f(x).diff(x)
eq = Eq(df, x**2*f(x))
eq1 = f(x).diff(x) + a*f(x) - c*exp(b*x)
eq2 = f(x).diff(x) + 2*x*f(x) - x*exp(-x**2)
eq3 = (1 + 2*x)*df + 2 - 4*exp(-f(x))
eq4 = f(x).diff(x) - (a4*x**4 + a3*x**3 + a2*x**2 + a1*x + a0)**Rational(-1, 2)
eq5 = x**2*df - f(x) + x**2*exp(x - (1/x))
eqlist = [eq, eq1, eq2, eq3, eq4, eq5]
i = infinitesimals(eq, hint='abaco1_simple')
assert i == [{eta(x, f(x)): exp(x**3/3), xi(x, f(x)): 0},
{eta(x, f(x)): f(x), xi(x, f(x)): 0},
{eta(x, f(x)): 0, xi(x, f(x)): x**(-2)}]
i1 = infinitesimals(eq1, hint='abaco1_simple')
assert i1 == [{eta(x, f(x)): exp(-a*x), xi(x, f(x)): 0}]
i2 = infinitesimals(eq2, hint='abaco1_simple')
assert i2 == [{eta(x, f(x)): exp(-x**2), xi(x, f(x)): 0}]
i3 = infinitesimals(eq3, hint='abaco1_simple')
assert i3 == [{eta(x, f(x)): 0, xi(x, f(x)): 2*x + 1},
{eta(x, f(x)): 0, xi(x, f(x)): 1/(exp(f(x)) - 2)}]
i4 = infinitesimals(eq4, hint='abaco1_simple')
assert i4 == [{eta(x, f(x)): 1, xi(x, f(x)): 0},
{eta(x, f(x)): 0,
xi(x, f(x)): sqrt(a0 + a1*x + a2*x**2 + a3*x**3 + a4*x**4)}]
i5 = infinitesimals(eq5, hint='abaco1_simple')
assert i5 == [{xi(x, f(x)): 0, eta(x, f(x)): exp(-1/x)}]
ilist = [i, i1, i2, i3, i4, i5]
for eq, i in (zip(eqlist, ilist)):
check = checkinfsol(eq, i)
assert check[0]
# This ODE can be solved by the Lie Group method, when there are
# better assumptions
eq6 = df - (f(x)/x)*(x*log(x**2/f(x)) + 2)
i = infinitesimals(eq6, hint='abaco1_product')
assert i == [{eta(x, f(x)): f(x)*exp(-x), xi(x, f(x)): 0}]
assert checkinfsol(eq6, i)[0]
eq7 = x*(f(x).diff(x)) + 1 - f(x)**2
i = infinitesimals(eq7, hint='chi')
assert checkinfsol(eq7, i)[0]
def test_heuristic_function_sum():
eq = f(x).diff(x) - (3 * (1 + x**2 / f(x)**2) * atan(f(x) / x) +
(1 - 2 * f(x)) / x + (1 - 3 * f(x)) * (x / f(x)**2))
i = infinitesimals(eq, hint='function_sum')
assert i == [{eta(x, f(x)): f(x)**(-2) + x**(-2), xi(x, f(x)): 0}]
assert checkinfsol(eq, i)[0]
def test_kamke():
a, b, alpha, c = symbols("a b alpha c")
eq = x**2 * (a * f(x)**2 + (f(x).diff(x))) + b * x**alpha + c
i = infinitesimals(eq, hint='sum_function') # XFAIL
assert checkinfsol(eq, i)[0]
