What is the solution for (dy}{dx}=\frac{x^3-2y)/x ?

The solution for (dy}{dx}=\frac{x^3-2y)/x is y=(x^3)/5+(c_{1})/(x^2)

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(dy}{dx}=\frac{x^3-2y)/x

\frac{d y}{d x} = \frac{x ^{3} - 2 y}{x}

y = \frac{x ^{3}}{5} + \frac{c _{1}}{x ^{2}}

Solution steps

\frac{d y}{d x} = \frac{x ^{3} - 2 y}{x}

Solve linear ODE:

y = \frac{x ^{3}}{5} + \frac{c _{1}}{x ^{2}}

y = \frac{x ^{3}}{5} + \frac{c _{1}}{x ^{2}}

\int \frac{( 3 x + 2 )}{19 - 5 x + x ^{2}} d x

x^{3} - 4 x^{2}

y^{^{'}} = - a y^{2} + b

\frac{d}{d x} (tan (1 + e^{2 x}))

- 2 lo g_{e} (x) + x^{2} - 4 + e^{x} + x^{e}

What is the solution for (dy}{dx}=\frac{x^3-2y)/x ?
The solution for (dy}{dx}=\frac{x^3-2y)/x is y=(x^3)/5+(c_{1})/(x^2)