What is the solution for y^'=-(2t(y^2+1))/y ,y(0)=1 ?

The solution for y^'=-(2t(y^2+1))/y ,y(0)=1 is y=sqrt(2e^{-2t^2)-1}

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y^'=-(2t(y^2+1))/y ,y(0)=1

y^{^{'}} = - \frac{2 t ( y ^{2} + 1 )}{y}, y (0) = 1

y = 2 e^{- 2 t^{2}} - 1

Solution steps

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

Solve separable ODE:

y = 2 e^{- 2 t^{2}} - 1

y = 2 e^{- 2 t^{2}} - 1

t y + y^{2} + (t^{2} + t y) \frac{d y}{d t} = 0

y^{^{'}} + y = 2, y (- 2) = 1

\frac{df}{d x} = - 4 x f (x) (x)^{2}

y^{^{'}} = 1 + \frac{( y ) ^{2}}{4}, y (0) = 0.5

yy (x + 1) + y^{^{'}} = 0

What is the solution for y^'=-(2t(y^2+1))/y ,y(0)=1 ?
The solution for y^'=-(2t(y^2+1))/y ,y(0)=1 is y=sqrt(2e^{-2t^2)-1}