What is the solution for yy^'=x+1 ?

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

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yy^'=x+1

y y^{^{'}} = x + 1

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

Solution steps

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

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}

y^{^{'}} \cdot e^{x + y} - x = 0

y^{^{'}} = y + \frac{y}{x ln ( x )}, y (e) = 1

y^{^{'}} = e^{2 x} + y \cdot e^{2 x}

(x - 1)^{2} y^{^{'}} + 8 (x - 1) y^{^{'}} + 12 y = 0

\frac{d y}{d x} = \frac{4 x}{y ^{2}}

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