What is the solution for (dy)/(dx)= 1/((2x+y+1)) ?

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

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(dy)/(dx)= 1/((2x+y+1))

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

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

Solution steps

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

Solve exact ODE:

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

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

\frac{d y}{d t} = \frac{y}{t}

y = \frac{x ^{2} - 4 x - 4}{x ^{2} - 4 x - 5}, y = 0

6 t (\frac{d y}{d t}) + y = t^{3}

\frac{d}{d x} (4 x^{3} + x + 5 sin (x))

\int 2 + 3 x^{2} - 8 x^{3} d x

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