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Popular Calculus >

(y)dx+(y-x)dy=0

Solution

(y) d x + (y - x) d y = 0

Solution

y = c_{1}

Solution steps

(y) d x + (y - x) d y = 0

Let

y

be the dependent variable. Divide by

d x

:

\frac{d y}{d x} + (y - x) \frac{d y}{d x} = 0

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) + (y - x) y^{^{'}} (x) = 0

Isolate

y^{'} (x) : y^{'} (x) = 0

If

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

then

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

y = \int 0 d x

\int 0 d x = c_{1}

y = c_{1}

Graph

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Frequently Asked Questions (FAQ)

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