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

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

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

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

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

Solution steps

(2 x + y + 1) y^{'} (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}

(x^{2} - 1)^{^{'}}

\frac{d}{d x} f (x) = - 9

\frac{d}{d x} (\frac{x}{3 x ^{2} + 1})

(cos^{4} (x))^{^{'}}

x \to 0 lim (\frac{sin ( - 4 x )}{x})

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