What is the solution for x derivative of [ln(y)]=1 ?

The solution for x derivative of [ln(y)]=1 is y=c_{1}x

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x derivative of [ln(y)]=1

x \frac{d}{d x} [ln (y)] = 1

y = c_{1} x

Solution steps

x \frac{d}{d x} (ln (y)) = 1

Isolate

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

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

then

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

ln (y) = \int \frac{1}{x} d x

\int \frac{1}{x} d x = ln (x) + c_{1}

ln (y) = ln (x) + c_{1}

Isolate

y : y = e^{c_{1}} x

y = e^{c_{1}} x

Simplify

y = c_{1} x

y = x, y = 4 x, y = - x + 2

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\frac{d}{d x} (24 x - 4 x)

What is the solution for x derivative of [ln(y)]=1 ?
The solution for x derivative of [ln(y)]=1 is y=c_{1}x