What is the solution for dx+e^{2x}dy=0 ?

The solution for dx+e^{2x}dy=0 is y= 1/2 e^{-2x}+c_{1}

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dx+e^{2x}dy=0

d x + e^{2 x} d y = 0

y = \frac{1}{2} e^{- 2 x} + c_{1}

Solution steps

d x + e^{2 x} d y = 0

Let

y

be the dependent variable. Divide by

d x

1 + e^{2 x} \frac{d y}{d x} = 0

Substitute

\frac{d y}{d x}

with

y^{'} (x)

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

Isolate

y^{'} (x) : y^{'} (x) = - \frac{1}{e ^{2 x}}

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

then

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

y = \int - \frac{1}{e ^{2 x}} d x

\int - \frac{1}{e ^{2 x}} d x = \frac{1}{2} e^{- 2 x} + c_{1}

y = \frac{1}{2} e^{- 2 x} + c_{1}

\frac{A}{y ^{14}} + B e^{y}

x \to 2 - lim (- 3)

\frac{d}{d x} (x sin (tan (x)))

y = 3 x^{2} + 2 x^{2} + 1, (- 1, 0)

x \to 2 lim (x^{2} + 2 x + 1)

What is the solution for dx+e^{2x}dy=0 ?
The solution for dx+e^{2x}dy=0 is y= 1/2 e^{-2x}+c_{1}