What is the solution for (1+x^2)(dy)/(dx)-(4x^3y)/(1-x^2)=1 ?

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

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(1+x^2)(dy)/(dx)-(4x^3y)/(1-x^2)=1

(1 + x^{2}) \frac{d y}{d x} - \frac{4 x ^{3} y}{1 - x ^{2}} = 1

y = - \frac{x ^{3}}{3 ( 1 - x ^{4} )} + \frac{x}{1 - x ^{4}} + \frac{c _{1}}{1 - x ^{4}}

Solution steps

(1 + x^{2}) \frac{d y}{d x} - \frac{4 x ^{3} y}{1 - x ^{2}} = 1

Solve linear ODE:

y = - \frac{x ^{3}}{3 ( 1 - x ^{4} )} + \frac{x}{1 - x ^{4}} + \frac{c _{1}}{1 - x ^{4}}

y = - \frac{x ^{3}}{3 ( 1 - x ^{4} )} + \frac{x}{1 - x ^{4}} + \frac{c _{1}}{1 - x ^{4}}

\int 3 x^{2} + 8 x - 9 d x

\frac{\partial}{\partial x} (ln (t + 2 c x))

f (x) = a x^{b} + d e^{- c x}

\int (4^{x} - sec^{2} (x)) d x

\int \frac{d}{1} (e^{- x} y) d x

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