What is the solution for yy^'-xy^2+x=0 ?

The solution for yy^'-xy^2+x=0 is y=sqrt(e^{x^2+c_{1)}+1},y=-sqrt(e^{x^2+c_{1)}+1}

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yy^'-xy^2+x=0

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

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

Solution steps

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

Solve separable ODE:

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

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

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

e^{- y} sin (x) - \frac{d y}{d x} cos^{2} (x) = 0

\frac{d y}{d x} = \frac{17 x y}{( ln ( y ) ) ^{6}}

y^{^{'}} = 2 y + 1, y (0) = 1

y^{^{'}} + 3 y = 7, y (0) = 1

What is the solution for yy^'-xy^2+x=0 ?
The solution for yy^'-xy^2+x=0 is y=sqrt(e^{x^2+c_{1)}+1},y=-sqrt(e^{x^2+c_{1)}+1}