What is the solution for dx-e^{3x}dy=0 ?

The solution for dx-e^{3x}dy=0 is y=-1/3 e^{-3x}+c_{1}

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dx-e^{3x}dy=0

d x - e^{3 x} d y = 0

y = - \frac{1}{3} e^{- 3 x} + c_{1}

Solution steps

d x - e^{3 x} d y = 0

Let

y

be the dependent variable. Divide by

d x

1 - e^{3 x} \frac{d y}{d x} = 0

Substitute

\frac{d y}{d x}

with

y^{'} (x)

1 - e^{3 x} y^{^{'}} (x) = 0

Isolate

y^{'} (x) : y^{'} (x) = \frac{1}{e ^{3 x}}

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

then

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

y = \int \frac{1}{e ^{3 x}} d x

\int \frac{1}{e ^{3 x}} d x = - \frac{1}{3} e^{- 3 x} + c_{1}

y = - \frac{1}{3} e^{- 3 x} + c_{1}

\frac{\partial}{\partial x} (x^{6 y})

\frac{d}{d x} (x^{2} 6 x - 1)

\int_{b}^{d} \frac{1}{r} d r

f (x) = 7 x^{3} + 5 x, at x = 7

\frac{d}{d x} (3 e^{x} (5 x^{5} - 2))

What is the solution for dx-e^{3x}dy=0 ?
The solution for dx-e^{3x}dy=0 is y=-1/3 e^{-3x}+c_{1}