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

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

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x(x^3+1)y^'+(2x^3-1)y=(x^3-2)/x

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

y = \frac{x ^{2}}{x ^{3} + 1} + \frac{1}{x ( x ^{3} + 1 )} + \frac{c _{1} x}{x ^{3} + 1}

Solution steps

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

Solve linear ODE:

y = \frac{x ^{2}}{x ^{3} + 1} + \frac{1}{x ( x ^{3} + 1 )} + \frac{c _{1} x}{x ^{3} + 1}

y = \frac{x ^{2}}{x ^{3} + 1} + \frac{1}{x ( x ^{3} + 1 )} + \frac{c _{1} x}{x ^{3} + 1}

\int x^{2} - 49 d x

y^{^{''}} + 2 y^{^{'}} + 2 = 0

\frac{\partial}{\partial x} (sin (x y^{2}) e^{2 x - 3 y})

\int 5 sin (6 π) x d x

\int 6 e^{- 0.2 x} d x

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