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

The solution for y^'=-pi/2 t(1+y^2) is y=tan(-(pit^2)/4+c_{1})

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

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

y = tan (- \frac{π t ^{2}}{4} + c_{1})

Solution steps

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

Solve separable ODE:

y = tan (- \frac{π t ^{2}}{4} + c_{1})

y = tan (- \frac{π t ^{2}}{4} + c_{1})

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What is the solution for y^'=-pi/2 t(1+y^2) ?
The solution for y^'=-pi/2 t(1+y^2) is y=tan(-(pit^2)/4+c_{1})