What is the solution for y^'=((x+y+1))/((2x+2y+3)) ?

The solution for y^'=((x+y+1))/((2x+2y+3)) is y=-x-4/3+(\W(2/(e^{-9x-c_{1)}}))/6

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y^'=((x+y+1))/((2x+2y+3))

y^{^{'}} = \frac{( x + y + 1 )}{( 2 x + 2 y + 3 )}

y = - x - \frac{4}{3} + \frac{W ( \frac{2}{e ^{- 9 x - c_{1}}} )}{6}

Solution steps

y^{'} (x) = \frac{( x + y + 1 )}{( 2 x + 2 y + 3 )}

Solve using substitution:

y = - x - \frac{4}{3} + \frac{W ( \frac{2}{e ^{- 9 x - c_{1}}} )}{6}

y = - x - \frac{4}{3} + \frac{W ( \frac{2}{e ^{- 9 x - c_{1}}} )}{6}

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What is the solution for y^'=((x+y+1))/((2x+2y+3)) ?
The solution for y^'=((x+y+1))/((2x+2y+3)) is y=-x-4/3+(\W(2/(e^{-9x-c_{1)}}))/6