What is the solution for 1/(2y)=y^'(x^2)/(y^2-8),y(1)=sqrt(9) ?

The solution for 1/(2y)=y^'(x^2)/(y^2-8),y(1)=sqrt(9) is y=sqrt(e^{-1/x+1)+8}

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1/(2y)=y^'(x^2)/(y^2-8),y(1)=sqrt(9)

\frac{1}{2 y} = y^{^{'}} \frac{x ^{2}}{y ^{2} - 8}, y (1) = 9

y = e^{- \frac{1}{x} + 1} + 8

Solution steps

\frac{1}{2 y} = y^{'} (x) \frac{x ^{2}}{y ^{2} - 8}

Solve separable ODE:

y = e^{- \frac{1}{x} + 1} + 8

y = e^{- \frac{1}{x} + 1} + 8

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

- y e^{- x} + x e^{- x} y^{^{'}} = 0

y^{^{'}} = \frac{x}{sin ( y )}

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

\frac{d v}{d t} = 5 - 2 \cdot v

What is the solution for 1/(2y)=y^'(x^2)/(y^2-8),y(1)=sqrt(9) ?
The solution for 1/(2y)=y^'(x^2)/(y^2-8),y(1)=sqrt(9) is y=sqrt(e^{-1/x+1)+8}