What is the solution for ((x^2))/(y^2-5)y^'= 1/(2y) ?

The solution for ((x^2))/(y^2-5)y^'= 1/(2y) is y=sqrt(e^{-1/x+c_{1)}+5},y=-sqrt(e^{-1/x+c_{1)}+5}

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((x^2))/(y^2-5)y^'= 1/(2y)

\frac{( x ^{2} )}{y ^{2} - 5} y^{^{'}} = \frac{1}{2 y}

y = e^{- \frac{1}{x} + c_{1}} + 5, y = - e^{- \frac{1}{x} + c_{1}} + 5

Solution steps

\frac{( x ^{2} )}{y ^{2} - 5} y^{'} (x) = \frac{1}{2 y}

Solve separable ODE:

y = e^{- \frac{1}{x} + c_{1}} + 5, y = - e^{- \frac{1}{x} + c_{1}} + 5

y = e^{- \frac{1}{x} + c_{1}} + 5, y = - e^{- \frac{1}{x} + c_{1}} + 5

\int_{6}^{7} t^{2} d t

x \to 2 lim (f (x))

f (x) = \frac{0.3 ( 9 ^{x} )}{x ^{3}}

y^{^{''}} - 4 y^{^{'}} + 4 y = (12 x^{2} - 6 x) e^{2} x

n = 1 \sum \infty \frac{8 ^{n}}{n !}

What is the solution for ((x^2))/(y^2-5)y^'= 1/(2y) ?
The solution for ((x^2))/(y^2-5)y^'= 1/(2y) is y=sqrt(e^{-1/x+c_{1)}+5},y=-sqrt(e^{-1/x+c_{1)}+5}