What is the solution for y^'=(8cos(8x))/(3+2y),y(0)=-1 ?

The solution for y^'=(8cos(8x))/(3+2y),y(0)=-1 is y=(-3+sqrt(4sin(8x)+1))/2

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y^'=(8cos(8x))/(3+2y),y(0)=-1

y^{^{'}} = \frac{8 cos ( 8 x )}{3 + 2 y}, y (0) = - 1

y = \frac{- 3 + 4 sin ( 8 x ) + 1}{2}

Solution steps

y^{'} (x) = \frac{8 cos ( 8 x )}{3 + 2 y}

Solve separable ODE:

y = \frac{- 3 + 4 sin ( 8 x ) + 1}{2}

y = \frac{- 3 + 4 sin ( 8 x ) + 1}{2}

t^{9} \frac{d y}{d t} + 2 t^{8} y = 0

\frac{d y}{d x} = - 1 \cdot \frac{1}{( x + 9 ) ^{2}} \cdot (y + 1)

\frac{d y}{d x} = 4 + y^{2}

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

\frac{d y}{d x} - 2 y = y, y = 1, x = - 1

What is the solution for y^'=(8cos(8x))/(3+2y),y(0)=-1 ?
The solution for y^'=(8cos(8x))/(3+2y),y(0)=-1 is y=(-3+sqrt(4sin(8x)+1))/2