What is the solution for dx+e^{-5x}dy=0 ?

The solution for dx+e^{-5x}dy=0 is y=-1/5 e^{5x}+c_{1}

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dx+e^{-5x}dy=0

d x + e^{- 5 x} d y = 0

y = - \frac{1}{5} e^{5 x} + c_{1}

Solution steps

d x + e^{- 5 x} d y = 0

Let

y

be the dependent variable. Divide by

d x

1 + e^{- 5 x} \frac{d y}{d x} = 0

Substitute

\frac{d y}{d x}

with

y^{'} (x)

1 + e^{- 5 x} y^{^{'}} (x) = 0

Isolate

y^{'} (x) : y^{'} (x) = - e^{5 x}

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

then

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

y = \int - e^{5 x} d x

\int - e^{5 x} d x = - \frac{1}{5} e^{5 x} + c_{1}

y = - \frac{1}{5} e^{5 x} + c_{1}

f (x) = 2 ln (6 + x)

\frac{d}{d x} (\frac{3}{lo g _{8} ( x )})

x \cdot y^{^{'}} - y = \frac{5}{4} \cdot x \cdot ln (x)

\int \frac{x}{1 + 2 x ^{2}} d x

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

What is the solution for dx+e^{-5x}dy=0 ?
The solution for dx+e^{-5x}dy=0 is y=-1/5 e^{5x}+c_{1}