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

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

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

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

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

Solution steps

\frac{x 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}}

(e^{3 x} cos (2 x))^{^{'}}

y^{^{''}} - 4 y^{^{'}} + 4 y = 16 - 4 t - 6 t e^{2 t}

x \to 5 lim (- 4)

\int x 4^{x^{2}} d x

\frac{d}{d t} (t e^{t})

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