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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

(7 x^{2} + 2) (6 x + x)

y^{^{'}} = - 0.000631594 \cdot y

x \to 1 + lim (\frac{4 x - 1}{x - 1})

f (x) = 2 cos (x)

\int \frac{x ^{2}}{( 9 + x ^{2} ) ^{2}} d x

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