What is the solution for y^'=-y^2t-y^2t^2,y(0)=3 ?

The solution for y^'=-y^2t-y^2t^2,y(0)=3 is y= 6/(2t^3+3t^2+2)

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y^'=-y^2t-y^2t^2,y(0)=3

y^{^{'}} = - y^{2} t - y^{2} t^{2}, y (0) = 3

y = \frac{6}{2 t ^{3} + 3 t ^{2} + 2}

Solution steps

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

Solve separable ODE:

y = \frac{6}{2 t ^{3} + 3 t ^{2} + 2}

y = \frac{6}{2 t ^{3} + 3 t ^{2} + 2}

\int sin (6 x) sin (5 x) d x

x \to \frac{1}{2} - lim (\frac{1}{2 x - 1})

\frac{d}{d x} (ln (\frac{x ^{2}}{x + 1}))

f (x) = (3 a + 1)^{2}

x \to + 6 lim (36 - x^{2})

What is the solution for y^'=-y^2t-y^2t^2,y(0)=3 ?
The solution for y^'=-y^2t-y^2t^2,y(0)=3 is y= 6/(2t^3+3t^2+2)