What is the solution for x(dy)/(dx)+6y=x^3-x ?

The solution for x(dy)/(dx)+6y=x^3-x is y=(x^3)/9-x/7+(c_{1})/(x^6)

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x(dy)/(dx)+6y=x^3-x

x \frac{d y}{d x} + 6 y = x^{3} - x

y = \frac{x ^{3}}{9} - \frac{x}{7} + \frac{c _{1}}{x ^{6}}

Solution steps

x \frac{d y}{d x} + 6 y = x^{3} - x

Solve linear ODE:

y = \frac{x ^{3}}{9} - \frac{x}{7} + \frac{c _{1}}{x ^{6}}

y = \frac{x ^{3}}{9} - \frac{x}{7} + \frac{c _{1}}{x ^{6}}

y^{^{'}} - cos (t) y = t e^{t^{2} + s i n (t)}

\frac{d}{d x} (20 (1 - e^{- \frac{x}{2}}))

\int - 3 x^{2} + 100 d x

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

4 x y \frac{d y}{d x} = 3

What is the solution for x(dy)/(dx)+6y=x^3-x ?
The solution for x(dy)/(dx)+6y=x^3-x is y=(x^3)/9-x/7+(c_{1})/(x^6)