What is the solution for (y)dx-(x+sqrt(y^2-x^2))dy=0 ?

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

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(y)dx-(x+sqrt(y^2-x^2))dy=0

(y) d x - (x + y^{2} - x^{2}) d y = 0

y = c_{1}

Solution steps

(y) d x - (x + y^{2} - x^{2}) d y = 0

Let

y

be the dependent variable. Divide by

d x

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

Substitute

\frac{d y}{d x}

with

y^{'} (x)

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

Isolate

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

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}

\int \frac{54}{1 - cos ( 9 x )} d x

(2 x)^{^{'}}

(ln (cos (x)))^{^{'}}

x \to 3 lim (\frac{x ^{2} - 5}{x - 3})

x \to 0 lim (\frac{4 tan ( x )}{x})

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