What is the solution for (dy)/(dx)+(-x)/(x+1)=0 ?

The solution for (dy)/(dx)+(-x)/(x+1)=0 is y=x+1-ln(x+1)+c_{1}

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(dy)/(dx)+(-x)/(x+1)=0

\frac{d y}{d x} + \frac{- x}{x + 1} = 0

y = x + 1 - ln (x + 1) + c_{1}

Solution steps

\frac{d y}{d x} + \frac{- x}{x + 1} = 0

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) + \frac{- x}{x + 1} = 0

Isolate

y^{'} (x) : y^{'} (x) = \frac{x}{x + 1}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

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

\int \frac{x}{x + 1} d x = x + 1 - ln (x + 1) + c_{1}

y = x + 1 - ln (x + 1) + c_{1}

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

y = x^{2}, y = 8 + x

\int 100 t^{2} + 36 t^{4} d t

9 \frac{d}{d x} (\frac{d y}{d x}) + 10 y = 0

x \to - 3 - lim (\frac{x}{x + 3})

What is the solution for (dy)/(dx)+(-x)/(x+1)=0 ?
The solution for (dy)/(dx)+(-x)/(x+1)=0 is y=x+1-ln(x+1)+c_{1}