What is the solution for y^'=-x/(4y) ?

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

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y^'=-x/(4y)

y^{^{'}} = - \frac{x}{4 y}

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

Solution steps

y^{'} (x) = - \frac{x}{4 y}

Solve separable ODE:

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

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

y = (sin (3 x)) ln (x)

\frac{d}{d x} (48 x^{\frac{1}{2}})

\int \frac{8}{x ^{2} ( x ^{2} + 25 )} d x

f (x) = \frac{2}{ln ( x )}

x^{2} y y^{^{'}} = ey

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