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

The solution for y^'=(1-2x)/y ,y(1)=-2 is y=-sqrt(2x-2x^2+4)

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y^'=(1-2x)/y ,y(1)=-2

y^{^{'}} = \frac{1 - 2 x}{y}, y (1) = - 2

y = - 2 x - 2 x^{2} + 4

Solution steps

y^{'} (x) = \frac{1 - 2 x}{y}

Solve separable ODE:

y = - 2 x - 2 x^{2} + 4

y = - 2 x - 2 x^{2} + 4

\frac{d y}{d x} = 3 x^{2} y, y (0) = 2

tan (x) sec (y) d x - tan (y) sec (x) d y = 0

\frac{d v}{d t} = 2 - 0.1 v

y^{^{'}} = tan (y) tan (x)

What is the solution for y^'=(1-2x)/y ,y(1)=-2 ?
The solution for y^'=(1-2x)/y ,y(1)=-2 is y=-sqrt(2x-2x^2+4)