(x, y, t, r)
return [Eq(x(t), sol1), Eq(y(t), sol2)]
r"""
     The equations of this type of ode are '\\frac{d}{d t} x{\\left (t \\right )} = f{\\left (t \\right )} ' \
    'x{\\left (t \\right )} + g{\\left (t \\right )} y{\\left (t \\right )}' and
    '\\frac{d}{d t} y{\\left (t \\right )} = f{\\left (t \\right )} y{\\left (t \\right )} - g{\\left (t \\right )} x{\\left (t \\right )}'.
    The solution is given by
    'F \\left(C_{1} \\cos{\\left (G \\right )} + C_{2} \\sin{\\left (G \\right )}\\right)' and
    'F \\left(- C_{1} \\sin{\\left (G \\right )} + C_{2} \\cos{\\left (G \\right )}\\right)'
    where F and G are integrals of f(t) and g(t).
r"""
     The equations of this type of ode are '\\frac{d}{d t} x{\\left (t \\right )} = f{\\left (t \\right )} ' \
    'x{\\left (t \\right )} + g{\\left (t \\right )} y{\\left (t \\right )}' and
    '\\frac{d}{d t} y{\\left (t \\right )} = f{\\left (t \\right )} y{\\left (t \\right )} - g{\\left (t \\right )} x{\\left (t \\right )}'.
    The solution is given by
    'F \\left(C_{1} \\cos{\\left (G \\right )} + C_{2} \\sin{\\left (G \\right )}\\right)' and
    'F \\left(- C_{1} \\sin{\\left (G \\right )} + C_{2} \\cos{\\left (G \\right )}\\right)'
    where F and G are integrals of f(t) and g(t).
r The equations of this type of ode are \\ frac { d } { d t } x { \\ left ( t \\ right ) } = f { \\ left ( t \\ right ) } \ x { \\ left ( t \\ right ) } + g { \\ left ( t \\ right ) } y { \\ left ( t \\ right ) } and \\ frac { d } { d t } y { \\ left ( t \\ right ) } = f { \\ left ( t \\ right ) } y { \\ left ( t \\ right ) } - g { \\ left ( t \\ right ) } x { \\ left ( t \\ right ) } . The solution is given by F \\ left ( C_ { 1 } \\ cos { \\ left ( G \\ right ) } + C_ { 2 } \\ sin { \\ left ( G \\ right ) } \\ right ) and F \\ left ( - C_ { 1 } \\ sin { \\ left ( G \\ right ) } + C_ { 2 } \\ cos { \\ left ( G \\ right ) } \\ right ) where F and G are integrals of f ( t ) and g ( t ) .
def _linear_2eq_order1_type4(x, y, t, r):
    r"""
     The equations of this type of ode are '\\frac{d}{d t} x{\\left (t \\right )} = f{\\left (t \\right )} ' \
    'x{\\left (t \\right )} + g{\\left (t \\right )} y{\\left (t \\right )}' and
    '\\frac{d}{d t} y{\\left (t \\right )} = f{\\left (t \\right )} y{\\left (t \\right )} - g{\\left (t \\right )} x{\\left (t \\right )}'.
    The solution is given by
    'F \\left(C_{1} \\cos{\\left (G \\right )} + C_{2} \\sin{\\left (G \\right )}\\right)' and
    'F \\left(- C_{1} \\sin{\\left (G \\right )} + C_{2} \\cos{\\left (G \\right )}\\right)'
    where F and G are integrals of f(t) and g(t).
    """
    C1, C2, C3, C4 = symbols('C1:5')
    if r['b'] == -r['c']:
        F = exp(C.Integral(r['a'], t))
        G = C.Integral(r['b'], t)
        sol1 = F*(C1*cos(G) + C2*sin(G))
        sol2 = F*(-C1*sin(G) + C2*cos(G))
    elif r['d'] == -r['a']:
        F = exp(C.Integral(r['c'], t))
        G = C.Integral(r['d'], t)
        sol1 = F*(-C1*sin(G) + C2*cos(G))
        sol2 = F*(C1*cos(G) + C2*sin(G))
    return [Eq(x(t), sol1), Eq(y(t), sol2)]
def _linear_2eq_order1_type4 ( x , y , t , r ) : C1 , C2 , C3 , C4 = symbols ( 'C1:5' ) if r [ 'b' ] == - r [ 'c' ] : F = exp ( C . Integral ( r [ 'a' ] , t ) ) G = C . Integral ( r [ 'b' ] , t ) sol1 = F * ( C1 * cos ( G ) + C2 * sin ( G ) ) sol2 = F * ( - C1 * sin ( G ) + C2 * cos ( G ) ) elif r [ 'd' ] == - r [ 'a' ] : F = exp ( C . Integral ( r [ 'c' ] , t ) ) G = C . Integral ( r [ 'd' ] , t ) sol1 = F * ( - C1 * sin ( G ) + C2 * cos ( G ) ) sol2 = F * ( C1 * cos ( G ) + C2 * sin ( G ) ) return [ Eq ( x ( t ) , sol1 ) , Eq ( y ( t ) , sol2 ) ]