What is the solution for y^'=(1+x)/y ?

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

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y^'=(1+x)/y

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

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

Solution steps

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

Solve separable ODE:

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

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

sin (2 x) d x + cos (9 y) d y = 0, y (\frac{π}{2}) = \frac{π}{9}

\frac{d y}{d x} = (9 + y^{2}) (3 + x^{2})

y^{^{'}} = 3 y^{2} - 3 y

x^{3} y^{^{'}} - (8 x^{2} - 5) y = 0, y (1) = 2

\frac{d y}{d x} = cy - 1

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