What is the solution for y^'= y/(x+xe^x) ?

The solution for y^'= y/(x+xe^x) is y=e^{integral of 1/(x+e^xx)}

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y^'= y/(x+xe^x)

y^{^{'}} = \frac{y}{x + x e ^{x}}

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

Solution steps

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

Solve separable ODE:

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

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

(t^{2} + 1) d y + 3 t y d t = 6 t d t, y (0) = 5

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

v^{^{'}} = \frac{( v + \frac{4}{v} )}{t}

\frac{d y}{d x} 6 x y + 1 = 0

(x^{2} + 6)^{\frac{1}{2}} d x + \frac{1}{x csc ( 8 y )} d y = 0

What is the solution for y^'= y/(x+xe^x) ?
The solution for y^'= y/(x+xe^x) is y=e^{integral of 1/(x+e^xx)}