What is the solution for 0=2x+1-2yy^',y(-2)=-1 ?

The solution for 0=2x+1-2yy^',y(-2)=-1 is y=-sqrt(x^2+x-1)

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0=2x+1-2yy^',y(-2)=-1

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d}{d x} (sin (2 \cdot x))

\frac{d}{d x} (5 + cos (x^{2}))

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