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Popular >

1/x dx+dy=0

Solution

\frac{1}{x} d x + d y = 0

Solution

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

Solution steps

\frac{1}{x} d x + d y = 0

Let

y

be the dependent variable. Divide by

d x

:

\frac{1}{x} + \frac{d y}{d x} = 0

Substitute

\frac{d y}{d x}

with

y^{'} (x)

\frac{1}{x} + y^{^{'}} (x) = 0

Isolate

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

If

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

then

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

y = \int - \frac{1}{x} d x

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

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

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x d x = y d y

Frequently Asked Questions (FAQ)

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