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

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

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

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

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

Solution steps

\frac{x 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}}

\int 8 x e^{2 x} d x

\frac{1}{m + k e ^{m}}

\int \frac{3 x ^{3} + 9 x ^{2} + 2 x + 6}{x ^{4} + x ^{2}} d x

x \cdot sec (x)

\int \frac{1}{x 16 x ^{2} - 9} d x

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