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

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

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

Solution steps

\frac{( 2 x + y + 1 ) 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 z} (\frac{a z ^{2}}{a ^{2} + z ^{2}})

\int x^{3} x^{2} - 7 d x

\int \frac{1}{x ^{3} 9 x ^{2} - 1} d x

2 y^{^{''}} + 3 y^{^{'}} - 4 y = 5 e^{3 x}

n = 1 \sum \infty \frac{( 3 x - 2 ) ^{n}}{n}

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