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

The solution for y^'=y(2+y)t,y(0)=1 is y=(2e^{t^2})/(3-e^{t^2)}

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

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

y = \frac{2 e ^{t^{2}}}{3 - e ^{t^{2}}}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{2 e ^{t^{2}}}{3 - e ^{t^{2}}}

y = \frac{2 e ^{t^{2}}}{3 - e ^{t^{2}}}

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\int \frac{x ^{2} - 25}{x ^{4}} d x

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\int 2 x d x

What is the solution for y^'=y(2+y)t,y(0)=1 ?
The solution for y^'=y(2+y)t,y(0)=1 is y=(2e^{t^2})/(3-e^{t^2)}