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

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

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

\frac{d y}{d x} = \frac{1}{y + x}

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

Solution steps

\frac{d y}{d x} = \frac{1}{y + x}

Solve exact ODE:

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

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

f (x) = x^{2} + 7, at x = 3

\frac{\partial}{\partial y} (- 3 x e^{- x y})

\int e^{- 2 x} sin (e^{- x}) d x

\frac{d}{d t} (1 - t^{3})

\frac{\partial}{\partial y} (x arctan (yz))

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