(x, y, t, r)
return [Eq(x(t), sol1), Eq(y(t), sol2)]
r"""
    Equations 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 )} = a \\left(a h{\\left (t \\right )} + ' \
    'f{\\left (t \\right )}\\right) x{\\left (t \\right )} + a \\left(g{\\left (t \\right )} - ' \
    'h{\\left (t \\right )}\\right) y{\\left (t \\right )}'

    This is solved by first multiplying the first equation by ``-a`` and adding it to the second
    equation to obtain '- a \\frac{d}{d t} x{\\left (t \\right )} + \\frac{d}{d t} y{\\left (t \\right )} ' \
    '= - a \\left(- a x{\\left (t \\right )} + y{\\left (t \\right )}\\right) h{\\left (t \\right )}'
    Setting ``W`` as '- a x{\\left (t \\right )} + y{\\left (t \\right )}' and integrating the equation
    we arrive at '- a x{\\left (t \\right )} + y{\\left (t \\right )} = C_{1} e^{- a \\int h{\\left (t \\right )}\\, dt}'
    and on substituing the value of y in first equation give rise to first order ODEs. After solving for
    ``x``, we can obtain ``y`` by substituting the value of ``x`` in second equation.
r"""
    Equations 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 )} = a \\left(a h{\\left (t \\right )} + ' \
    'f{\\left (t \\right )}\\right) x{\\left (t \\right )} + a \\left(g{\\left (t \\right )} - ' \
    'h{\\left (t \\right )}\\right) y{\\left (t \\right )}'
r Equations 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 ) } = a \\ left ( a h { \\ left ( t \\ right ) } + \ f { \\ left ( t \\ right ) } \\ right ) x { \\ left ( t \\ right ) } + a \\ left ( g { \\ left ( t \\ right ) } - \ h { \\ left ( t \\ right ) } \\ right ) y { \\ left ( t \\ right ) }
def _linear_2eq_order1_type6(x, y, t, r):
    r"""
    Equations 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 )} = a \\left(a h{\\left (t \\right )} + ' \
    'f{\\left (t \\right )}\\right) x{\\left (t \\right )} + a \\left(g{\\left (t \\right )} - ' \
    'h{\\left (t \\right )}\\right) y{\\left (t \\right )}'

    This is solved by first multiplying the first equation by ``-a`` and adding it to the second
    equation to obtain '- a \\frac{d}{d t} x{\\left (t \\right )} + \\frac{d}{d t} y{\\left (t \\right )} ' \
    '= - a \\left(- a x{\\left (t \\right )} + y{\\left (t \\right )}\\right) h{\\left (t \\right )}'
    Setting ``W`` as '- a x{\\left (t \\right )} + y{\\left (t \\right )}' and integrating the equation
    we arrive at '- a x{\\left (t \\right )} + y{\\left (t \\right )} = C_{1} e^{- a \\int h{\\left (t \\right )}\\, dt}'
    and on substituing the value of y in first equation give rise to first order ODEs. After solving for
    ``x``, we can obtain ``y`` by substituting the value of ``x`` in second equation.

    """
    C1, C2, C3, C4 = symbols('C1:5')
    p = 0
    q = 0
    p1 = cancel(r['c']/cancel(r['c']/r['d']).as_numer_denom()[0])
    p2 = cancel(r['a']/cancel(r['a']/r['b']).as_numer_denom()[0])
    for n, i in enumerate([p1, p2]):
        for j in Mul.make_args(collect_const(i)):
            if not j.has(t):
                q = j
            if q!=0 and n==0:
                if ((r['c']/j - r['a'])/(r['b'] - r['d']/j)) == j:
                    p = 1
                    s = j
                    break
            if q!=0 and n==1:
                if ((r['a']/j - r['c'])/(r['d'] - r['b']/j)) == j:
                    p = 2
                    s = j
                    break
    if p == 1:
        equ = diff(x(t),t) - r['a']*x(t) - r['b']*(s*x(t) + C1*exp(-s*C.Integral(r['b'] - r['d']/s, t)))
        hint1 = classify_ode(equ)[1]
        sol1 = dsolve(equ, hint=hint1+'_Integral').rhs
        sol2 = s*sol1 + C1*exp(-s*C.Integral(r['b'] - r['d']/s, t))
    elif p ==2:
        equ = diff(y(t),t) - r['c']*y(t) - r['d']*s*y(t) + C1*exp(-s*C.Integral(r['d'] - r['b']/s, t))
        hint1 = classify_ode(equ)[1]
        sol2 = dsolve(equ, hint=hint1+'_Integral').rhs
        sol1 = s*sol2 + C1*exp(-s*C.Integral(r['d'] - r['b']/s, t))
    return [Eq(x(t), sol1), Eq(y(t), sol2)]
def _linear_2eq_order1_type6 ( x , y , t , r ) : C1 , C2 , C3 , C4 = symbols ( 'C1:5' ) p = 0 q = 0 p1 = cancel ( r [ 'c' ] / cancel ( r [ 'c' ] / r [ 'd' ] ) . as_numer_denom ( ) [ 0 ] ) p2 = cancel ( r [ 'a' ] / cancel ( r [ 'a' ] / r [ 'b' ] ) . as_numer_denom ( ) [ 0 ] ) for n , i in enumerate ( [ p1 , p2 ] ) : for j in Mul . make_args ( collect_const ( i ) ) : if not j . has ( t ) : q = j if q != 0 and n == 0 : if ( ( r [ 'c' ] / j - r [ 'a' ] ) / ( r [ 'b' ] - r [ 'd' ] / j ) ) == j : p = 1 s = j break if q != 0 and n == 1 : if ( ( r [ 'a' ] / j - r [ 'c' ] ) / ( r [ 'd' ] - r [ 'b' ] / j ) ) == j : p = 2 s = j break if p == 1 : equ = diff ( x ( t ) , t ) - r [ 'a' ] * x ( t ) - r [ 'b' ] * ( s * x ( t ) + C1 * exp ( - s * C . Integral ( r [ 'b' ] - r [ 'd' ] / s , t ) ) ) hint1 = classify_ode ( equ ) [ 1 ] sol1 = dsolve ( equ , hint = hint1 + '_Integral' ) . rhs sol2 = s * sol1 + C1 * exp ( - s * C . Integral ( r [ 'b' ] - r [ 'd' ] / s , t ) ) elif p == 2 : equ = diff ( y ( t ) , t ) - r [ 'c' ] * y ( t ) - r [ 'd' ] * s * y ( t ) + C1 * exp ( - s * C . Integral ( r [ 'd' ] - r [ 'b' ] / s , t ) ) hint1 = classify_ode ( equ ) [ 1 ] sol2 = dsolve ( equ , hint = hint1 + '_Integral' ) . rhs sol1 = s * sol2 + C1 * exp ( - s * C . Integral ( r [ 'd' ] - r [ 'b' ] / s , t ) ) return [ Eq ( x ( t ) , sol1 ) , Eq ( y ( t ) , sol2 ) ]