What is the solution for derivative of 2y^2=3-x ?

The solution for derivative of 2y^2=3-x is y=(sqrt(6x+c_{1)-x^2})/2 ,y=-(sqrt(6x+c_{1)-x^2})/2

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derivative of 2y^2=3-x

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

y = \frac{6 x + c _{1} - x ^{2}}{2}, y = - \frac{6 x + c _{1} - x ^{2}}{2}

Solution steps

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

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

2 y^{2} = \int 3 - x d x

\int 3 - x d x = 3 x - \frac{x ^{2}}{2} + c_{1}

2 y^{2} = 3 x - \frac{x ^{2}}{2} + c_{1}

Isolate

y : y = \frac{6 x + 2 c _{1} - x ^{2}}{2}, y = - \frac{6 x + 2 c _{1} - x ^{2}}{2}

y = \frac{6 x + 2 c _{1} - x ^{2}}{2}, y = - \frac{6 x + 2 c _{1} - x ^{2}}{2}

Simplify

y = \frac{6 x + c _{1} - x ^{2}}{2}, y = - \frac{6 x + c _{1} - x ^{2}}{2}

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

y^{^{'}} = e^{y} x^{6}

\frac{d y}{d x} = 64 y x^{15}

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

\frac{d y}{d t} = 1 + y^{2}

What is the solution for derivative of 2y^2=3-x ?
The solution for derivative of 2y^2=3-x is y=(sqrt(6x+c_{1)-x^2})/2 ,y=-(sqrt(6x+c_{1)-x^2})/2