What is the solution for xy^2dx-x^2y^2dy=0 ?

The solution for xy^2dx-x^2y^2dy=0 is y=ln(x)+c_{1}

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xy^2dx-x^2y^2dy=0

x y^{2} d x - x^{2} y^{2} d y = 0

y = ln (x) + c_{1}

Solution steps

x y^{2} d x - x^{2} y^{2} d y = 0

Let

y

be the dependent variable. Divide by

d x

x y^{2} - x^{2} y^{2} \frac{d y}{d x} = 0

Substitute

\frac{d y}{d x}

with

y^{'} (x)

x y^{2} - x^{2} y^{2} y^{^{'}} (x) = 0

Isolate

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

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}

\frac{d}{d x} (e^{\frac{( - x ^{2} )}{2}})

\int \frac{1}{16 - t ^{2}} d t

f (x) = ln (1 + x)

\frac{d}{d x} (ln (\frac{2 x}{sin ( x )}))

y = \frac{3}{sin ( 2 x )}

What is the solution for xy^2dx-x^2y^2dy=0 ?
The solution for xy^2dx-x^2y^2dy=0 is y=ln(x)+c_{1}