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 -----------(3)  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 ------------(4)  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