Решить дифференциальное уравнение y'(x+y⁵)=y

Y(x^3-y^5) dx - x(x^3+y^5) dy = 0  (x^3y - y^6) dx + (-x^4 -y^5 x) dy = 0 (1)  m(x,y) dx + n(x,y) dy = 0  m(x,y) = x^3y -y^6  n(x,y) = -x^4- xy^5 = 0  let my be the partial derivative of m with respect to y  let nx be the partial derivative of n with respect to x  my = x^3 -6y^5  nx = -4x^3-y^5  since my ≠ nx , the equation is not 'exact'.  compute (my-nx) / n = (x^3-6y^5 +4x^3+y^5) / -(x^4+xy^5) = (5x^3-5y^5) / -(x^4+xy^5) = a function of x and y  compute (nx-my) / m = (-4x^3-y^5-x^3+6y^5) / (x^3y-y^6) = (-5x^3+y^5) / y(x^3-y^5) = -5(x^3-y^5) /y(x^3-y^5) = -5/y  -5/y is a function of y only  therefore, the integrating factor is e^∫-5/y dy = e^-5ln y = e^ln y^(-5) = y^(-5) = 1/y^5  multiply equation (1) by the integrating factor 1/y^5  (x^3 /y^4 - y) dx + ( -x^4 / y^5 - x) dy = 0  (x^3 y^(-4) - y) dx + ( -x^4 y^(-5) - x) dy = 0 (2)  p(x,y) dx + q(x,y) dy = 0  p(x,y) = x^3y^(-4) - y  q(x,y) = -x^4 y^(-5) - x  let py be the partial derivative of p with respect to y  let qx be the partial derivative of q with respect to x  py = (-4) y^(-5) x^3 - 1  qx = -4x^3 y^(-5) - 1 = (-4) y^(-5) x^3 - 1  py = qx , so equation (2) is 'exact.'  solve the exact equation (2)  integrate p(x,y) with respect to x  ∫ ( x^3y^(-4) - y ) dx  = (1/4) y^(-4) x^4 - yx )  integrate q(x,y) with respect to y  ∫ ( -x^4 y^(-5) - x ) dy = -x^4 y^(-5+1) /(-5+1) - yx = (1/4) x^4 y^(-4) - yx )  merge these two expressions (3) & (4), write down each term exactly once, even if a  particular term appears in both results. here the two expressions contain the terms  (1/4) x^4 / y^4 , -yx (both appear twice but written only once)  the solution is  (1/4) x^4 / y^4 - xy = c

(284+326)> (500+302) (800+230)< (156+285) вот и все