What is the solution for dx+e^{4x}dy=0 ?

The solution for dx+e^{4x}dy=0 is y= 1/4 e^{-4x}+c_{1}

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dx+e^{4x}dy=0

d x + e^{4 x} d y = 0

y = \frac{1}{4} e^{- 4 x} + c_{1}

Solution steps

d x + e^{4 x} d y = 0

Let

y

be the dependent variable. Divide by

d x

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

Substitute

\frac{d y}{d x}

with

y^{'} (x)

1 + e^{4 x} y^{^{'}} (x) = 0

Isolate

y^{'} (x) : y^{'} (x) = - \frac{1}{e ^{4 x}}

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

then

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

y = \int - \frac{1}{e ^{4 x}} d x

\int - \frac{1}{e ^{4 x}} d x = \frac{1}{4} e^{- 4 x} + c_{1}

y = \frac{1}{4} e^{- 4 x} + c_{1}

\int \frac{8 ( x ^{2} + 1 )}{4 x ^{2} + 9} d x

\frac{1 + 6 s ^{2}}{s ( s ^{2} + 1 )}

\frac{d ^{2} y}{d x ^{2}} = (4 sin (3 x))

\frac{s}{( s ^{2} + 4 ) ( s + 2 )}

\int - 4 x - 1 d x

What is the solution for dx+e^{4x}dy=0 ?
The solution for dx+e^{4x}dy=0 is y= 1/4 e^{-4x}+c_{1}