What is the solution for y^'=((y^2+1))/(2y) ?

The solution for y^'=((y^2+1))/(2y) is y=sqrt(e^{t+c_{1)}-1},y=-sqrt(e^{t+c_{1)}-1}

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y^'=((y^2+1))/(2y)

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

y = e^{t + c_{1}} - 1, y = - e^{t + c_{1}} - 1

Solution steps

y^{'} (t) = \frac{( y ^{2} + 1 )}{2 y}

Solve separable ODE:

y = e^{t + c_{1}} - 1, y = - e^{t + c_{1}} - 1

y = e^{t + c_{1}} - 1, y = - e^{t + c_{1}} - 1

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n = 1 \sum \infty \frac{3}{n + 1}

f (x) = sin (8 x + 4)

\int \frac{48 x ^{2}}{( x - 21 ) ( x + 7 ) ^{2}} d x

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What is the solution for y^'=((y^2+1))/(2y) ?
The solution for y^'=((y^2+1))/(2y) is y=sqrt(e^{t+c_{1)}-1},y=-sqrt(e^{t+c_{1)}-1}