What is the solution for xdy-(2x+1)e^{-y}dx=0 ?

The solution for xdy-(2x+1)e^{-y}dx=0 is y=ln(2x+ln(x)+c_{1})

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xdy-(2x+1)e^{-y}dx=0

x d y - (2 x + 1) e^{- y} d x = 0

y = ln (2 x + ln (x) + c_{1})

Solution steps

x d y - (2 x + 1) e^{- y} d x = 0

Solve separable ODE:

y = ln (2 x + ln (x) + c_{1})

y = ln (2 x + ln (x) + c_{1})

\int \frac{1}{( 3 x - 1 )} d x

\frac{\partial}{\partial x} (2 cos (3 x))

\frac{d y}{d x} + y^{5} x + 4 y = 0

\int_{0}^{2} x^{2} e^{- x} d x

\int - 2^{x} d x

What is the solution for xdy-(2x+1)e^{-y}dx=0 ?
The solution for xdy-(2x+1)e^{-y}dx=0 is y=ln(2x+ln(x)+c_{1})