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

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

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dx=e^{3x}dy

d x = e^{3 x} d y

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

Solution steps

d x = e^{3 x} d y

Let

y

be the dependent variable. Divide by

d x

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

Substitute

\frac{d y}{d x}

with

y^{'} (x)

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

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}

\int s 8^{s} d s

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

y = 3 - 6 x

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

y^{^{'}} + x y = 1

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