y^'= 1/(2yt)-y/(2t)

Lösung

y^{^{'}} = \frac{1}{2 y t} - \frac{y}{2 t}

y = \frac{- c _{1} + t}{t}, y = - \frac{- c _{1} + t}{t}

Schritte zur Lösung

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

Solve separable ODE:

y = \frac{- c _{1} + t}{t}, y = - \frac{- c _{1} + t}{t}

y = \frac{- c _{1} + t}{t}, y = - \frac{- c _{1} + t}{t}

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