What is the solution for (x^2+1)y^'+6x(y-1)=0 ?

The solution for (x^2+1)y^'+6x(y-1)=0 is y=(c_{1})/(x^6+3x^4+3x^2+1)+1

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(x^2+1)y^'+6x(y-1)=0

(x^{2} + 1) y^{^{'}} + 6 x (y - 1) = 0

y = \frac{c _{1}}{x ^{6} + 3 x ^{4} + 3 x ^{2} + 1} + 1

Solution steps

(x^{2} + 1) y^{'} (x) + 6 x (y - 1) = 0

Solve separable ODE:

y = \frac{c _{1}}{x ^{6} + 3 x ^{4} + 3 x ^{2} + 1} + 1

y = \frac{c _{1}}{x ^{6} + 3 x ^{4} + 3 x ^{2} + 1} + 1

\frac{d}{d x} (\frac{5}{x} + x - 4)

\frac{d}{d x} ((\frac{5 x + 1}{2 x}) d x)

\frac{\partial}{\partial t} (2 r sin (t))

\frac{d}{d x} (x^{3} - 9 x^{2} + 10 x)

\int 1 + cos (y) d y

What is the solution for (x^2+1)y^'+6x(y-1)=0 ?
The solution for (x^2+1)y^'+6x(y-1)=0 is y=(c_{1})/(x^6+3x^4+3x^2+1)+1