(x, y, t, r)
return [Eq(x(t), sol1), Eq(y(t), sol2)]
r"""
    Differentiating the first equation and substituting the value of '\\frac{d}{d t} y{\\left (t \\right )}'
    from second equation will give a second-order linear equation
    'g \\frac{d^{2}}{d t^{2}}  x{\\left (t \\right )} - \\left(f g + g p + \\frac{d}{d t} g{\\left (t \\right )}' \
    '\\right) \\frac{d}{d t} x{\\left (t \\right )} + \\left(f g p + f \\frac{d}{d t} g{\\left (t \\right )} - ' \
    'g^{2} h - g \\frac{d}{d t} f{\\left (t \\right )}\\right) x{\\left (t \\right )} = 0'

    This above equation can be easily integrated if following conditions are satisfied.
    1. 'f g p + f \\frac{d}{d t} g{\\left (t \\right )} - g^{2} h - g \\frac{d}{d t} f{\\left (t \\right )} = 0'
    2. 'f g p + f \\frac{d}{d t} g{\\left (t \\right )} - g^{2} h - g \\frac{d}{d t} f{\\left (t \\right )} = a g'
    and 'f g + g p + \\frac{d}{d t} g{\\left (t \\right )} = b g'
    If first condition is satisfied then it is solved by current dsolve solver and in second case it becomes
    a constant cofficient differential equation which is also solved by current solver.

    Otherwise if the above condition fails then,
    a particular solution is assumed 'x{\\left (t \\right )} = \\operatorname{x_{0}}{\\left (t \\right )}'.
    Then the general solution is expressed as
    'x{\\left (t \\right )} = C_{1} \\operatorname{x_{0}}{\\left (t \\right )} + C_{2} ' \
    '\\operatorname{x_{0}}{\\left (t \\right )} \\int \\frac{F{\\left (t \\right )} P{\\left (t \\right ' \
    ')}}{\\operatorname{x_{0}}^{2}{\\left (t \\right )}} g{\\left (t \\right )}\\, dt' and
    'y{\\left (t \\right )} = C_{1} \\operatorname{y_{0}}{\\left (t \\right )} + C_{2} \\left(C_{2} ' \
    '\\operatorname{y_{0}}{\\left (t \\right )} \\int \\frac{F{\\left (t \\right )} P{\\left (t \\right )}}' \
    '{\\operatorname{x_{0}}^{2}{\\left (t \\right )}} g{\\left (t \\right )}\\, dt + \\frac{F{\\left ' \
    '(t \\right )} P{\\left (t \\right )}}{\\operatorname{x_{0}}{\\left (t \\right )}}\\right)'
    where C1 and C2 are arbitary constants and 'F = e^{\\int f{\\left (t \\right )}\\, dt}' and
    'P = e^{\\int p{\\left (t \\right )}\\, dt}'
r"""
    Differentiating the first equation and substituting the value of '\\frac{d}{d t} y{\\left (t \\right )}'
    from second equation will give a second-order linear equation
    'g \\frac{d^{2}}{d t^{2}}  x{\\left (t \\right )} - \\left(f g + g p + \\frac{d}{d t} g{\\left (t \\right )}' \
    '\\right) \\frac{d}{d t} x{\\left (t \\right )} + \\left(f g p + f \\frac{d}{d t} g{\\left (t \\right )} - ' \
    'g^{2} h - g \\frac{d}{d t} f{\\left (t \\right )}\\right) x{\\left (t \\right )} = 0'
r Differentiating the first equation and substituting the value of \\ frac { d } { d t } y { \\ left ( t \\ right ) } from second equation will give a second - order linear equation g \\ frac { d^ { 2 }} { d t^ { 2 }} x { \\ left ( t \\ right ) } - \\ left ( f g + g p + \\ frac { d } { d t } g { \\ left ( t \\ right ) } \ \\ right ) \\ frac { d } { d t } x { \\ left ( t \\ right ) } + \\ left ( f g p + f \\ frac { d } { d t } g { \\ left ( t \\ right ) } - \ g^ { 2 } h - g \\ frac { d } { d t } f { \\ left ( t \\ right ) } \\ right ) x { \\ left ( t \\ right ) } = 0
def _linear_2eq_order1_type7(x, y, t, r):
    r"""
    Differentiating the first equation and substituting the value of '\\frac{d}{d t} y{\\left (t \\right )}'
    from second equation will give a second-order linear equation
    'g \\frac{d^{2}}{d t^{2}}  x{\\left (t \\right )} - \\left(f g + g p + \\frac{d}{d t} g{\\left (t \\right )}' \
    '\\right) \\frac{d}{d t} x{\\left (t \\right )} + \\left(f g p + f \\frac{d}{d t} g{\\left (t \\right )} - ' \
    'g^{2} h - g \\frac{d}{d t} f{\\left (t \\right )}\\right) x{\\left (t \\right )} = 0'

    This above equation can be easily integrated if following conditions are satisfied.
    1. 'f g p + f \\frac{d}{d t} g{\\left (t \\right )} - g^{2} h - g \\frac{d}{d t} f{\\left (t \\right )} = 0'
    2. 'f g p + f \\frac{d}{d t} g{\\left (t \\right )} - g^{2} h - g \\frac{d}{d t} f{\\left (t \\right )} = a g'
    and 'f g + g p + \\frac{d}{d t} g{\\left (t \\right )} = b g'
    If first condition is satisfied then it is solved by current dsolve solver and in second case it becomes
    a constant cofficient differential equation which is also solved by current solver.

    Otherwise if the above condition fails then,
    a particular solution is assumed 'x{\\left (t \\right )} = \\operatorname{x_{0}}{\\left (t \\right )}'.
    Then the general solution is expressed as
    'x{\\left (t \\right )} = C_{1} \\operatorname{x_{0}}{\\left (t \\right )} + C_{2} ' \
    '\\operatorname{x_{0}}{\\left (t \\right )} \\int \\frac{F{\\left (t \\right )} P{\\left (t \\right ' \
    ')}}{\\operatorname{x_{0}}^{2}{\\left (t \\right )}} g{\\left (t \\right )}\\, dt' and
    'y{\\left (t \\right )} = C_{1} \\operatorname{y_{0}}{\\left (t \\right )} + C_{2} \\left(C_{2} ' \
    '\\operatorname{y_{0}}{\\left (t \\right )} \\int \\frac{F{\\left (t \\right )} P{\\left (t \\right )}}' \
    '{\\operatorname{x_{0}}^{2}{\\left (t \\right )}} g{\\left (t \\right )}\\, dt + \\frac{F{\\left ' \
    '(t \\right )} P{\\left (t \\right )}}{\\operatorname{x_{0}}{\\left (t \\right )}}\\right)'
    where C1 and C2 are arbitary constants and 'F = e^{\\int f{\\left (t \\right )}\\, dt}' and
    'P = e^{\\int p{\\left (t \\right )}\\, dt}'

    """
    C1, C2, C3, C4 = symbols('C1:5')
    e1 = r['a']*r['b']*r['c'] - r['b']**2*r['c'] + r['a']*diff(r['b'],t) - diff(r['a'],t)*r['b']
    e2 = r['a']*r['c']*r['d'] - r['b']*r['c']**2 + diff(r['c'],t)*r['d'] - r['c']*diff(r['d'],t)
    m1 = r['a']*r['b'] + r['b']*r['d'] + diff(r['b'],t)
    m2 = r['a']*r['c'] + r['c']*r['d'] + diff(r['c'],t)
    if e1 == 0:
        sol1 = dsolve(r['b']*diff(x(t),t,t) - m1*diff(x(t),t)).rhs
        sol2 = dsolve(diff(y(t),t) - r['c']*sol1 - r['d']*y(t)).rhs
    elif e2 == 0:
        sol2 = dsolve(r['c']*diff(y(t),t,t) - m2*diff(y(t),t)).rhs
        sol1 = dsolve(diff(x(t),t) - r['a']*x(t) - r['b']*sol2).rhs
    elif not (e1/r['b']).has(t) and not (m1/r['b']).has(t):
        sol1 = dsolve(diff(x(t),t,t) - (m1/r['b'])*diff(x(t),t) - (e1/r['b'])*x(t)).rhs
        sol2 = dsolve(diff(y(t),t) - r['c']*sol1 - r['d']*y(t)).rhs
    elif not (e2/r['c']).has(t) and not (m2/r['c']).has(t):
        sol2 = dsolve(diff(y(t),t,t) - (m2/r['c'])*diff(y(t),t) - (e2/r['c'])*y(t)).rhs
        sol1 = dsolve(diff(x(t),t) - r['a']*x(t) - r['b']*sol2).rhs
    else:
        x0, y0 = symbols('x0, y0')              #x0 and y0 being particular solutions
        F = exp(C.Integral(r['a'],t))
        P = exp(C.Integral(r['d'],t))
        sol1 = C1*x0 + C2*x0*C.Integral(r['b']*F*P/x0**2, t)
        sol2 = C1*y0 + C2(F*P/x0 + y0*C.Integral(r['b']*F*P/x0**2, t))
    return [Eq(x(t), sol1), Eq(y(t), sol2)]
def _linear_2eq_order1_type7 ( x , y , t , r ) : C1 , C2 , C3 , C4 = symbols ( 'C1:5' ) e1 = r [ 'a' ] * r [ 'b' ] * r [ 'c' ] - r [ 'b' ] ** 2 * r [ 'c' ] + r [ 'a' ] * diff ( r [ 'b' ] , t ) - diff ( r [ 'a' ] , t ) * r [ 'b' ] e2 = r [ 'a' ] * r [ 'c' ] * r [ 'd' ] - r [ 'b' ] * r [ 'c' ] ** 2 + diff ( r [ 'c' ] , t ) * r [ 'd' ] - r [ 'c' ] * diff ( r [ 'd' ] , t ) m1 = r [ 'a' ] * r [ 'b' ] + r [ 'b' ] * r [ 'd' ] + diff ( r [ 'b' ] , t ) m2 = r [ 'a' ] * r [ 'c' ] + r [ 'c' ] * r [ 'd' ] + diff ( r [ 'c' ] , t ) if e1 == 0 : sol1 = dsolve ( r [ 'b' ] * diff ( x ( t ) , t , t ) - m1 * diff ( x ( t ) , t ) ) . rhs sol2 = dsolve ( diff ( y ( t ) , t ) - r [ 'c' ] * sol1 - r [ 'd' ] * y ( t ) ) . rhs elif e2 == 0 : sol2 = dsolve ( r [ 'c' ] * diff ( y ( t ) , t , t ) - m2 * diff ( y ( t ) , t ) ) . rhs sol1 = dsolve ( diff ( x ( t ) , t ) - r [ 'a' ] * x ( t ) - r [ 'b' ] * sol2 ) . rhs elif not ( e1 / r [ 'b' ] ) . has ( t ) and not ( m1 / r [ 'b' ] ) . has ( t ) : sol1 = dsolve ( diff ( x ( t ) , t , t ) - ( m1 / r [ 'b' ] ) * diff ( x ( t ) , t ) - ( e1 / r [ 'b' ] ) * x ( t ) ) . rhs sol2 = dsolve ( diff ( y ( t ) , t ) - r [ 'c' ] * sol1 - r [ 'd' ] * y ( t ) ) . rhs elif not ( e2 / r [ 'c' ] ) . has ( t ) and not ( m2 / r [ 'c' ] ) . has ( t ) : sol2 = dsolve ( diff ( y ( t ) , t , t ) - ( m2 / r [ 'c' ] ) * diff ( y ( t ) , t ) - ( e2 / r [ 'c' ] ) * y ( t ) ) . rhs sol1 = dsolve ( diff ( x ( t ) , t ) - r [ 'a' ] * x ( t ) - r [ 'b' ] * sol2 ) . rhs else : x0 , y0 = symbols ( 'x0, y0' ) #x0 and y0 being particular solutions F = exp ( C . Integral ( r [ 'a' ] , t ) ) P = exp ( C . Integral ( r [ 'd' ] , t ) ) sol1 = C1 * x0 + C2 * x0 * C . Integral ( r [ 'b' ] * F * P / x0 ** 2 , t ) sol2 = C1 * y0 + C2 ( F * P / x0 + y0 * C . Integral ( r [ 'b' ] * F * P / x0 ** 2 , t ) ) return [ Eq ( x ( t ) , sol1 ) , Eq ( y ( t ) , sol2 ) ]