What is the solution for y^'=(3x+2y)/(3x+2y+2) ?

The solution for y^'=(3x+2y)/(3x+2y+2) is y=(2\W(1/(4e^{(-25x-c_{1))/4}})-3)/(5)-(3x)/2

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y^'=(3x+2y)/(3x+2y+2)

y^{^{'}} = \frac{3 x + 2 y}{3 x + 2 y + 2}

y = \frac{2 W ( \frac{1}{4 e ^{\frac{- 25 x - c _{1}}{4}}} ) - 3}{5} - \frac{3 x}{2}

Solution steps

y^{'} (x) = \frac{3 x + 2 y}{3 x + 2 y + 2}

Solve using substitution:

y = \frac{2 W ( \frac{1}{4 e ^{\frac{- 25 x - c _{1}}{4}}} ) - 3}{5} - \frac{3 x}{2}

y = \frac{2 W ( \frac{1}{4 e ^{\frac{- 25 x - c _{1}}{4}}} ) - 3}{5} - \frac{3 x}{2}

y d y = \frac{2 x}{( 1 + x ^{2} )} d x, y (0) = 2

\frac{\partial}{\partial y} (3 x^{3 y})

\int (5 x - 4)^{2} d x

f (x) = (3 x + 5)^{3}

f (x) = \frac{5}{x + 2}

What is the solution for y^'=(3x+2y)/(3x+2y+2) ?
The solution for y^'=(3x+2y)/(3x+2y+2) is y=(2\W(1/(4e^{(-25x-c_{1))/4}})-3)/(5)-(3x)/2