What is the solution for (9x^2+2x)dx+(12y^3-4y)dy=0 ?

The solution for (9x^2+2x)dx+(12y^3-4y)dy=0 is

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(9x^2+2x)dx+(12y^3-4y)dy=0

(9 x^{2} + 2 x) d x + (12 y^{3} - 4 y) d y = 0

y = \frac{1 + - 3 ( 3 x ^{3} + x ^{2} - c _{1} ) + 1}{3}, y = - \frac{1 + - 3 ( 3 x ^{3} + x ^{2} - c _{1} ) + 1}{3}, y = \frac{1 - - 3 ( 3 x ^{3} + x ^{2} - c _{1} ) + 1}{3}, y = - \frac{1 - - 3 ( 3 x ^{3} + x ^{2} - c _{1} ) + 1}{3}

Solution steps

(9 x^{2} + 2 x) d x + (12 y^{3} - 4 y) d y = 0

Solve separable ODE:

y = \frac{1 + - 3 ( 3 x ^{3} + x ^{2} - c _{1} ) + 1}{3}, y = - \frac{1 + - 3 ( 3 x ^{3} + x ^{2} - c _{1} ) + 1}{3}, y = \frac{1 - - 3 ( 3 x ^{3} + x ^{2} - c _{1} ) + 1}{3}, y = - \frac{1 - - 3 ( 3 x ^{3} + x ^{2} - c _{1} ) + 1}{3}

y = \frac{1 + - 3 ( 3 x ^{3} + x ^{2} - c _{1} ) + 1}{3}, y = - \frac{1 + - 3 ( 3 x ^{3} + x ^{2} - c _{1} ) + 1}{3}, y = \frac{1 - - 3 ( 3 x ^{3} + x ^{2} - c _{1} ) + 1}{3}, y = - \frac{1 - - 3 ( 3 x ^{3} + x ^{2} - c _{1} ) + 1}{3}

\frac{d y}{d x} = 2 x^{5} y

\frac{d N}{d t} + N = N \cdot t e^{t + 2}

\frac{d y}{d x} = \frac{( y x + 5 x )}{x ^{2} + 1}

y^{^{'}} - 3 y sin (3 t) = - 3 sin (3 t), y (\frac{π}{6}) = 2

y^{^{'}} = \frac{y ^{2}}{4}

What is the solution for (9x^2+2x)dx+(12y^3-4y)dy=0 ?
The solution for (9x^2+2x)dx+(12y^3-4y)dy=0 is