What is the solution for y^'=-(2t(1+y^2))/y ?

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

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

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

y = e^{- 2 t^{2} + c_{1}} - 1, y = - e^{- 2 t^{2} + c_{1}} - 1

Solution steps

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

Solve separable ODE:

y = e^{- 2 t^{2} + c_{1}} - 1, y = - e^{- 2 t^{2} + c_{1}} - 1

y = e^{- 2 t^{2} + c_{1}} - 1, y = - e^{- 2 t^{2} + c_{1}} - 1

\int 6 x^{2} - 2 x + 8 d x

\frac{2}{x}

x \to 2 lim (x - 3)

\frac{d}{d v} (2 uv)

f (x) = (1 + 2 x)^{2}, 9, at x = 1

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