What is the solution for (x+1)dx-ydy=0 ?

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

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(x+1)dx-ydy=0

(x + 1) d x - y d y = 0

y = x^{2} + 2 x - c_{1}, y = - x^{2} + 2 x - c_{1}

Solution steps

(x + 1) d x - y d y = 0

Solve separable ODE:

y = x^{2} + 2 x - c_{1}, y = - x^{2} + 2 x - c_{1}

y = x^{2} + 2 x - c_{1}, y = - x^{2} + 2 x - c_{1}

\frac{d y}{d t} + t y = y, y (1) = 3

y^{^{'}} = e^{- y} \frac{x ^{5}}{7}

\frac{d y}{d x} = \frac{2 x}{( y + 1 )}

\frac{d y}{d x} = x y^{2} sin (x^{2})

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