What is the solution for y*y^'-2x=sin(x),y(0)=6 ?

The solution for y*y^'-2x=sin(x),y(0)=6 is y=sqrt(-2cos(x)+2x^2+38)

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y*y^'-2x=sin(x),y(0)=6

y \cdot y^{^{'}} - 2 x = sin (x), y (0) = 6

y = - 2 cos (x) + 2 x^{2} + 38

Solution steps

y \cdot y^{'} (x) - 2 x = sin (x)

Solve separable ODE:

y = - 2 cos (x) + 2 x^{2} + 38

y = - 2 cos (x) + 2 x^{2} + 38

\frac{d y}{d x} = \frac{3 y}{2 y ^{2} + 1}, y = 1, x = 3

(t^{2} - 16 t + 15) \frac{d y}{d t} = y

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

y^{^{'}} = x y^{3} (1 + x^{2})^{\frac{1}{2}}

\frac{d y}{d x} = \frac{y ln ( y )}{x}

What is the solution for y*y^'-2x=sin(x),y(0)=6 ?
The solution for y*y^'-2x=sin(x),y(0)=6 is y=sqrt(-2cos(x)+2x^2+38)