What is the solution for y^'=(9cos(9x))/(3+2y),y(0)=-1 ?

The solution for y^'=(9cos(9x))/(3+2y),y(0)=-1 is y=(-3+sqrt(4sin(9x)+1))/2

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y^'=(9cos(9x))/(3+2y),y(0)=-1

y^{^{'}} = \frac{9 cos ( 9 x )}{3 + 2 y}, y (0) = - 1

y = \frac{- 3 + 4 sin ( 9 x ) + 1}{2}

Solution steps

y^{'} (x) = \frac{9 cos ( 9 x )}{3 + 2 y}

Solve separable ODE:

y = \frac{- 3 + 4 sin ( 9 x ) + 1}{2}

y = \frac{- 3 + 4 sin ( 9 x ) + 1}{2}

\frac{d y}{d x} y = ln (2 x + 5)

y^{^{'}} = 6 e^{2 x - y}

\frac{d x}{d t} = \frac{5}{x e ^{t + 8 x}}

\frac{d v}{d p} = \frac{- v}{p}

\frac{d y}{d x} - 0.2 x y = 0, y (1) = 1

What is the solution for y^'=(9cos(9x))/(3+2y),y(0)=-1 ?
The solution for y^'=(9cos(9x))/(3+2y),y(0)=-1 is y=(-3+sqrt(4sin(9x)+1))/2