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}}

5 x^{3} + 2 x + 4, at x = 1

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

f (x) = x^{2} sin (4 x^{2})

\frac{\partial}{\partial x} (2 z)

(\frac{4 - 80 y}{y}) (y - 48)

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)