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

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

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

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

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

Solution steps

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

Solve exact ODE:

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

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

\frac{( s - 1 )}{s \cdot ( s ^{2} + s + 1 )}

\int \frac{x ^{5}}{( 1 + x ^{3} ) ^{\frac{5}{2}}} d x

\int cos (x - n x) d x

\int e^{- x} x^{3} d x

y^{^{''}} + 4 y^{^{'}} + 9 y = 0, y (0) = 1, y^{^{'}} (0) = 13

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