What is the solution for 2x+3y+(6x+9y-3)y^'=0 ?

The solution for 2x+3y+(6x+9y-3)y^'=0 is y=-(2x)/3+(\W(1/(3e^{(-3x-c_{1))/3}}))/(3)+2/3

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2x+3y+(6x+9y-3)y^'=0

2 x + 3 y + (6 x + 9 y - 3) y^{^{'}} = 0

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

Solution steps

2 x + 3 y + (6 x + 9 y - 3) y^{'} (x) = 0

Solve using substitution:

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

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

f (x) = cos (x^{2}) - 2 x^{2} sin (x^{2})

\frac{\partial}{\partial y} ((x + y) (x - y))

\frac{d}{d x} (tan^{40} (sec (cos (x))))

\frac{d}{d x} (x^{e} + \frac{1}{x ^{10}})

- \frac{x}{8 y}

What is the solution for 2x+3y+(6x+9y-3)y^'=0 ?
The solution for 2x+3y+(6x+9y-3)y^'=0 is y=-(2x)/3+(\W(1/(3e^{(-3x-c_{1))/3}}))/(3)+2/3