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