What is the solution for y^'=(y^2+y^2cos(x))^2 ?

The solution for y^'=(y^2+y^2cos(x))^2 is y=-\sqrt[3]{4/(3(6x+sin(2x)+8sin(x)+c_{1))}}

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y^'=(y^2+y^2cos(x))^2

y^{^{'}} = (y^{2} + y^{2} cos (x))^{2}

Solution steps

y^{'} (x) = (y^{2} + y^{2} cos (x))^{2}

Solve separable ODE:

\frac{d x}{d t} = 4 (x^{2} + 1), x (\frac{π}{4}) = 1

(4 x - 8) d x - (6 y - 10) d y = 0

\frac{d x}{d t} = 80 - \frac{x}{100}

y^{^{'}} = x y^{2} - x, y (1) = 2

(y + 1) d x + (4 x + 2)^{2} d y = 0

What is the solution for y^'=(y^2+y^2cos(x))^2 ?
The solution for y^'=(y^2+y^2cos(x))^2 is y=-\sqrt[3]{4/(3(6x+sin(2x)+8sin(x)+c_{1))}}