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

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

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

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

y^{- 1} = \frac{1}{t ^{2} + 1} + 1

Solution steps

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

Solve Bernoulli ODE:

y^{- 1} = \frac{1}{t ^{2} + 1} + 1

y^{- 1} = \frac{1}{t ^{2} + 1} + 1

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