What is the solution for (x^2+2)^2y=(5x(y^2+1)^2)/((x^2+2)y^')-(x^4+4x^2+4)y ?

The solution for (x^2+2)^2y=(5x(y^2+1)^2)/((x^2+2)y^')-(x^4+4x^2+4)y is

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(x^2+2)^2y=(5x(y^2+1)^2)/((x^2+2)y^')-(x^4+4x^2+4)y

(x^{2} + 2)^{2} y = \frac{5 x ( y ^{2} + 1 ) ^{2}}{( x ^{2} + 2 ) y ^{^{'}}} - (x^{4} + 4 x^{2} + 4) y

y = \frac{- 4 x ^{4} - 8 c _{1} x ^{4} - 16 x ^{2} - 32 c _{1} x ^{2} - 32 c _{1} - 11}{- 5 + 8 c _{1} x ^{4} + 32 c _{1} x ^{2} + 32 c _{1}}, y = - \frac{- 4 x ^{4} - 8 c _{1} x ^{4} - 16 x ^{2} - 32 c _{1} x ^{2} - 32 c _{1} - 11}{- 5 + 8 c _{1} x ^{4} + 32 c _{1} x ^{2} + 32 c _{1}}

Solution steps

(x^{2} + 2)^{2} y = \frac{5 x ( y ^{2} + 1 ) ^{2}}{( x ^{2} + 2 ) y ^{'} ( x )} - (x^{4} + 4 x^{2} + 4) y

Solve separable ODE:

y = \frac{- 4 x ^{4} - 8 c _{1} x ^{4} - 16 x ^{2} - 32 c _{1} x ^{2} - 32 c _{1} - 11}{- 5 + 8 c _{1} x ^{4} + 32 c _{1} x ^{2} + 32 c _{1}}, y = - \frac{- 4 x ^{4} - 8 c _{1} x ^{4} - 16 x ^{2} - 32 c _{1} x ^{2} - 32 c _{1} - 11}{- 5 + 8 c _{1} x ^{4} + 32 c _{1} x ^{2} + 32 c _{1}}

y = \frac{- 4 x ^{4} - 8 c _{1} x ^{4} - 16 x ^{2} - 32 c _{1} x ^{2} - 32 c _{1} - 11}{- 5 + 8 c _{1} x ^{4} + 32 c _{1} x ^{2} + 32 c _{1}}, y = - \frac{- 4 x ^{4} - 8 c _{1} x ^{4} - 16 x ^{2} - 32 c _{1} x ^{2} - 32 c _{1} - 11}{- 5 + 8 c _{1} x ^{4} + 32 c _{1} x ^{2} + 32 c _{1}}

\frac{d y}{d x} = \frac{1}{x ^{4} + 4 x ^{2}}

x \to 2 lim (sin (\frac{1}{x - 2}))

x \to \infty lim (\frac{4 x ^{2} + 5 x}{10 + x ^{4}})

3 y^{^{''}} - 12 y = 0

x \to 0 + lim (\frac{1}{x ^{3} - 2 x ^{2}})

What is the solution for (x^2+2)^2y=(5x(y^2+1)^2)/((x^2+2)y^')-(x^4+4x^2+4)y ?
The solution for (x^2+2)^2y=(5x(y^2+1)^2)/((x^2+2)y^')-(x^4+4x^2+4)y is