What is the solution for (dy}{dt}=4t(sin(y))^2,y(-1)=\frac{5pi)/4 ?

The solution for (dy}{dt}=4t(sin(y))^2,y(-1)=\frac{5pi)/4 is y=arccot(-2t^2+3)+pin

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(dy}{dt}=4t(sin(y))^2,y(-1)=\frac{5pi)/4

\frac{d y}{d t} = 4 t (sin (y))^{2}, y (- 1) = \frac{5 π}{4}

y = arccot (- 2 t^{2} + 3) + πn

Solution steps

\frac{d y}{d t} = 4 t (sin (y))^{2}

Solve separable ODE:

y = arccot (- 2 t^{2} + 3) + πn

y = arccot (- 2 t^{2} + 3) + πn

y = \frac{x ^{2} ( 2 + x )}{( 1 + x ) ^{2}} y^{^{'}}

\frac{d x}{d t} = 12 - \frac{9}{25} x

[3 x cos (3 x) + sin (3 x) - 3] d x + (2 y + 5) d y = 0

y^{^{'}} = \frac{1 - y ^{2}}{1 - x ^{2}}

x y^{2} y^{^{'}} = x + x^{3}, y (0) = 1

What is the solution for (dy}{dt}=4t(sin(y))^2,y(-1)=\frac{5pi)/4 ?
The solution for (dy}{dt}=4t(sin(y))^2,y(-1)=\frac{5pi)/4 is y=arccot(-2t^2+3)+pin