(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} x{\\left (t \\right )} = f{\\left (t \\right )} y{\\left (t \\right )} + ' \
    'g{\\left (t \\right )} x{\\left (t \\right )}'.
    The solution of such equations is given by
    '\\left(C_{1} e^{G} + C_{2} e^{- G}\\right) e^{F}' and
    '\\left(C_{1} e^{G} - C_{2} e^{- G}\\right) e^{F}' where C1 and C2 are arbitary constants, and
    'F = \\int f{\\left (t \\right )}\\, dt' and 'G = \\int g{\\left (t \\right )}\\, dt'.
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} x{\\left (t \\right )} = f{\\left (t \\right )} y{\\left (t \\right )} + ' \
    'g{\\left (t \\right )} x{\\left (t \\right )}'.
    The solution of such equations is given by
    '\\left(C_{1} e^{G} + C_{2} e^{- G}\\right) e^{F}' and
    '\\left(C_{1} e^{G} - C_{2} e^{- G}\\right) e^{F}' where C1 and C2 are arbitary constants, and
    'F = \\int f{\\left (t \\right )}\\, dt' and 'G = \\int g{\\left (t \\right )}\\, dt'.
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 } x { \\ left ( t \\ right ) } = f { \\ left ( t \\ right ) } y { \\ left ( t \\ right ) } + \ g { \\ left ( t \\ right ) } x { \\ left ( t \\ right ) } . The solution of such equations is given by \\ left ( C_ { 1 } e^ { G } + C_ { 2 } e^ { - G } \\ right ) e^ { F } and \\ left ( C_ { 1 } e^ { G } - C_ { 2 } e^ { - G } \\ right ) e^ { F } where C1 and C2 are arbitary constants and F = \\ int f { \\ left ( t \\ right ) } \\ dt and G = \\ int g { \\ left ( t \\ right ) } \\ dt .
def _linear_2eq_order1_type3(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} x{\\left (t \\right )} = f{\\left (t \\right )} y{\\left (t \\right )} + ' \
    'g{\\left (t \\right )} x{\\left (t \\right )}'.
    The solution of such equations is given by
    '\\left(C_{1} e^{G} + C_{2} e^{- G}\\right) e^{F}' and
    '\\left(C_{1} e^{G} - C_{2} e^{- G}\\right) e^{F}' where C1 and C2 are arbitary constants, and
    'F = \\int f{\\left (t \\right )}\\, dt' and 'G = \\int g{\\left (t \\right )}\\, dt'.
    """
    C1, C2, C3, C4 = symbols('C1:5')
    F = C.Integral(r['a'], t)
    G = C.Integral(r['b'], t)
    sol1 = exp(F)*(C1*exp(G) + C2*exp(-G))
    sol2 = exp(F)*(C1*exp(G) - C2*exp(-G))
    return [Eq(x(t), sol1), Eq(y(t), sol2)]
def _linear_2eq_order1_type3 ( x , y , t , r ) : C1 , C2 , C3 , C4 = symbols ( 'C1:5' ) F = C . Integral ( r [ 'a' ] , t ) G = C . Integral ( r [ 'b' ] , t ) sol1 = exp ( F ) * ( C1 * exp ( G ) + C2 * exp ( - G ) ) sol2 = exp ( F ) * ( C1 * exp ( G ) - C2 * exp ( - G ) ) return [ Eq ( x ( t ) , sol1 ) , Eq ( y ( t ) , sol2 ) ]