What is the solution for e^ydx+xe^ydy=0 ?

The solution for e^ydx+xe^ydy=0 is y=-ln(x)+c_{1}

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e^ydx+xe^ydy=0

e^{y} d x + x e^{y} d y = 0

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

Solution steps

e^{y} d x + x e^{y} d y = 0

Let

y

be the dependent variable. Divide by

d x

e^{y} + x e^{y} \frac{d y}{d x} = 0

Substitute

\frac{d y}{d x}

with

y^{'} (x)

e^{y} + x e^{y} y^{^{'}} (x) = 0

Isolate

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

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

then

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

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

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

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

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

(x^{2} (x^{2} + 9 x))^{6}

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

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

\int w w^{2} - ln (3) d w

What is the solution for e^ydx+xe^ydy=0 ?
The solution for e^ydx+xe^ydy=0 is y=-ln(x)+c_{1}