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Popular >

(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}

Popular Examples

x y^{^{'}} - 3 y + 12 = 0

derivative of a\sqrt[n]{x}+b\sqrt[3]{x}

tangent of y= 5/(4x+1),(1,1)tangent of

y = \frac{5}{4 x + 1}, (1, 1)

taylor 1/(x-1),23 taylor

\frac{1}{x - 1}, 23

integral of sqrt(9/(5-6x))

\int \frac{9}{5 - 6 x} d x

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}