What is the solution for (9xy^2-4y)+(9x^2y-4x)y^'=0 ?

The solution for (9xy^2-4y)+(9x^2y-4x)y^'=0 is y=(c_{1})/x

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(9xy^2-4y)+(9x^2y-4x)y^'=0

(9 x y^{2} - 4 y) + (9 x^{2} y - 4 x) y^{^{'}} = 0

y = \frac{c _{1}}{x}

Solution steps

(9 x y^{2} - 4 y) + (9 x^{2} y - 4 x) y^{'} (x) = 0

Solve linear ODE:

y = \frac{c _{1}}{x}

y = \frac{c _{1}}{x}

\int (csc (3 x) - cot (3 x))^{2} d x

y^{^{''}} - 2 y^{^{'}} = sin (x), y (0) = 1, y^{^{'}} (0) = 2

\int t^{2} + 4 d t

\frac{d}{d x} (sin (x) + \frac{5}{6} cot (x))

\int_{- 8}^{0} (4 - 9 x)^{- 0.5} d x

What is the solution for (9xy^2-4y)+(9x^2y-4x)y^'=0 ?
The solution for (9xy^2-4y)+(9x^2y-4x)y^'=0 is y=(c_{1})/x