What is the solution for y^'=8tcos^2(y),y(1)=(3pi)/4 ?

The solution for y^'=8tcos^2(y),y(1)=(3pi)/4 is y=arctan(4t^2-5)+pin

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y^'=8tcos^2(y),y(1)=(3pi)/4

y^{^{'}} = 8 t cos^{2} (y), y (1) = \frac{3 π}{4}

y = arctan (4 t^{2} - 5) + πn

Solution steps

y^{'} (t) = 8 t cos^{2} (y)

Solve separable ODE:

y = arctan (4 t^{2} - 5) + πn

y = arctan (4 t^{2} - 5) + πn

\frac{d}{d x} (\frac{x ^{8}}{y ^{5}}) = 0

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

y = \frac{3 ( 1 + 3 x ^{2} ) ^{2}}{x} \frac{d y}{d x}

\frac{d y}{d t} y = 3 + csc (t^{6})

6 x - 4 y x^{2} + 1 \frac{d y}{d x} = 0, y (0) = - 5

What is the solution for y^'=8tcos^2(y),y(1)=(3pi)/4 ?
The solution for y^'=8tcos^2(y),y(1)=(3pi)/4 is y=arctan(4t^2-5)+pin