What is the solution for y*y^'=-x ?

The solution for y*y^'=-x is y=sqrt(-x^2+c_{1)},y=-sqrt(-x^2+c_{1)}

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y*y^'=-x

y \cdot y^{^{'}} = - x

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

Solution steps

y \cdot y^{'} (x) = - x

Solve separable ODE:

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

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

(1 + x) y^{^{'}} = e^{- y} - 1

(- 3 (2 x - u) + 6 x + 2) d x - (u - 1) (2 d x - d u) = 0

(e^{x} + 1)^{2} e^{- y} d x + (e^{x} + 1)^{3} e^{- x} d y = 0

\frac{d y}{d t} + 2 y = 4

y^{^{'}} = y^{2} e^{- x}

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