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

The solution for x((dy)/(dx))+4y=x^3-x is y=(x^3)/7-x/5+(c_{1})/(x^4)

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

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

y = \frac{x ^{3}}{7} - \frac{x}{5} + \frac{c _{1}}{x ^{4}}

Solution steps

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

Solve linear ODE:

y = \frac{x ^{3}}{7} - \frac{x}{5} + \frac{c _{1}}{x ^{4}}

y = \frac{x ^{3}}{7} - \frac{x}{5} + \frac{c _{1}}{x ^{4}}

f (x) = 2 x^{2} - 7 x + 5, at x = 1

y = 5 x^{3} - 2 x^{4}, x = 0, x = 2

\frac{d}{d x} (6 x^{2} \cdot cos (2 x))

\frac{d}{d t} (e^{t s i n (2 t)})

\frac{d}{d x} (9 e^{- x})

What is the solution for x((dy)/(dx))+4y=x^3-x ?
The solution for x((dy)/(dx))+4y=x^3-x is y=(x^3)/7-x/5+(c_{1})/(x^4)