What is the solution for (3x^2y+2y)dx+(x^3+2x)dy=0 ?

The solution for (3x^2y+2y)dx+(x^3+2x)dy=0 is y=(c_{1})/(x^3+2x)

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(3x^2y+2y)dx+(x^3+2x)dy=0

(3 x^{2} y + 2 y) d x + (x^{3} + 2 x) d y = 0

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

Solution steps

(3 x^{2} y + 2 y) d x + (x^{3} + 2 x) d y = 0

Solve separable ODE:

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

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

y^{^{'}} = \frac{x - 1}{y ^{2}}

\frac{d y}{d x} = \frac{( 3 e ^{x} )}{( 3 y + 1 )}

\frac{d y}{d x} = \frac{1}{y + x ^{2} y}

\frac{d y}{d x} = 6 - \frac{y}{20}

y^{^{'}} + 3 y = 1, 3 sin (2 t) y (0) = 6

What is the solution for (3x^2y+2y)dx+(x^3+2x)dy=0 ?
The solution for (3x^2y+2y)dx+(x^3+2x)dy=0 is y=(c_{1})/(x^3+2x)