What is the solution for (x^2+1)y^'+8x(y-1)=0 ?

The solution for (x^2+1)y^'+8x(y-1)=0 is y=(c_{1})/(x^8+4x^6+6x^4+4x^2+1)+1

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(x^2+1)y^'+8x(y-1)=0

(x^{2} + 1) y^{^{'}} + 8 x (y - 1) = 0

y = \frac{c _{1}}{x ^{8} + 4 x ^{6} + 6 x ^{4} + 4 x ^{2} + 1} + 1

Solution steps

(x^{2} + 1) y^{'} (x) + 8 x (y - 1) = 0

Solve separable ODE:

y = \frac{c _{1}}{x ^{8} + 4 x ^{6} + 6 x ^{4} + 4 x ^{2} + 1} + 1

y = \frac{c _{1}}{x ^{8} + 4 x ^{6} + 6 x ^{4} + 4 x ^{2} + 1} + 1

x^{4} - 8 x^{3}

\frac{\partial}{\partial x} (\frac{3}{3 x + 2 y})

\frac{d y}{d x} = \frac{x y + x ^{2} + y ^{2}}{x ^{2}}

\frac{d}{d x} (3 x^{3} - 4)

8 e^{2 x}

What is the solution for (x^2+1)y^'+8x(y-1)=0 ?
The solution for (x^2+1)y^'+8x(y-1)=0 is y=(c_{1})/(x^8+4x^6+6x^4+4x^2+1)+1