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

The solution for e^{t^2}y^'+(t+ty^2)=0,y(0)=1 is y=tan(1/2 e^{-t^2}+pi/4-1/2)

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e^{t^2}y^'+(t+ty^2)=0,y(0)=1

e^{t^{2}} y^{^{'}} + (t + t y^{2}) = 0, y (0) = 1

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

Solution steps

e^{t^{2}} y^{'} (t) + (t + t y^{2}) = 0

Solve separable ODE:

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

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

\frac{64 x}{x ^{2} + 64}

\frac{\partial}{\partial y} (2 x^{3} - x^{2} y)

\int (t^{2} - 1) (4 + 3 t) d t

f (x) = x^{4}, at x = 1

y = \frac{6}{x}, (25, \frac{6}{5})

What is the solution for e^{t^2}y^'+(t+ty^2)=0,y(0)=1 ?
The solution for e^{t^2}y^'+(t+ty^2)=0,y(0)=1 is y=tan(1/2 e^{-t^2}+pi/4-1/2)