What is the solution for sqrt(-2y-y^2)dx+(3+2x-x^2)dy=0 ?

The solution for sqrt(-2y-y^2)dx+(3+2x-x^2)dy=0 is y=sin(c_{1}+(-ln(x+1)+ln(x-3))/4)-1

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sqrt(-2y-y^2)dx+(3+2x-x^2)dy=0

- 2 y - y^{2} d x + (3 + 2 x - x^{2}) d y = 0

y = sin (c_{1} + \frac{- ln ( x + 1 ) + ln ( x - 3 )}{4}) - 1

Solution steps

- 2 y - y^{2} d x + (3 + 2 x - x^{2}) d y = 0

Solve separable ODE:

y = sin (c_{1} + \frac{- ln ( x + 1 ) + ln ( x - 3 )}{4}) - 1

y = sin (c_{1} + \frac{- ln ( x + 1 ) + ln ( x - 3 )}{4}) - 1

y^{^{''}} - y^{^{'}} - 2 y = cos (x) - sin (2 x)

\frac{d}{d x} (- 2 a sin (2 x) + 2 b cos (2 x))

y = 2 + 8 e^{6 x}

\frac{d}{d x} (\frac{( e ^{x} - 2 )}{( 1 - e ^{- x} ) ^{2}})

\int \frac{1}{( x - 1 ) ^{2}} d x

What is the solution for sqrt(-2y-y^2)dx+(3+2x-x^2)dy=0 ?
The solution for sqrt(-2y-y^2)dx+(3+2x-x^2)dy=0 is y=sin(c_{1}+(-ln(x+1)+ln(x-3))/4)-1