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

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

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xdy-(2xe^{(-y)/x}+y)dx=0

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

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

Solution steps

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

Solve using substitution:

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

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

\frac{d}{d θ} (8 cos (θ) + 2 sin (θ))

\frac{d y}{d x} - x e^{y} = 2 e^{y}

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + y = e^{- x} \cdot ln (x)

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

h \to 0 lim (\frac{3}{h})

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