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}}

(1.3) (4.2)

y \frac{d y}{d t} = 2 t + 7

y = x^{\frac{2}{x}}

1 t

\frac{\partial}{\partial x} (x^{^{'}} A 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)