What is the solution for dx+e^3xdy=0 ?

The solution for dx+e^3xdy=0 is y=-1/(e^3)ln(x)+c_{1}

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dx+e^3xdy=0

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

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

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

f (x) = x^{2}, (- 1, - 3)

x^{2} - 4 x, x

x \to 3 lim (5 (x - 3) ln (x - 3))

f (x) = \frac{1}{x} + 3, at x = - 1

x \to 5 lim (- \frac{9}{x + 5})

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