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

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

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

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

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

Solution steps

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

n \to \infty lim (2)

\int 6 sin^{3} (x co) s^{5} x d x

\frac{d}{d x} (\frac{( x ^{2} )}{( 1 + x )})

2 x (x + 3)^{2}

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