What is the solution for y^'=(4x)/(1+2y),y(9)=0 ?

The solution for y^'=(4x)/(1+2y),y(9)=0 is y=(-1+sqrt(8x^2-647))/2

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y^'=(4x)/(1+2y),y(9)=0

y^{^{'}} = \frac{4 x}{1 + 2 y}, y (9) = 0

y = \frac{- 1 + 8 x ^{2} - 647}{2}

Solution steps

y^{'} (x) = \frac{4 x}{1 + 2 y}

Solve separable ODE:

y = \frac{- 1 + 8 x ^{2} - 647}{2}

y = \frac{- 1 + 8 x ^{2} - 647}{2}

\frac{d x}{x} = \frac{d y}{y}

x y^{^{'}} - (x + 1) y = 0

x y^{2} cos (x) d x - 3 x^{2} y cos (x) d y = 0, y (1) = 1

\frac{d P}{d t} = 2 Pt, P (1) = 2

\frac{d V}{d P} = - \frac{V}{P}

What is the solution for y^'=(4x)/(1+2y),y(9)=0 ?
The solution for y^'=(4x)/(1+2y),y(9)=0 is y=(-1+sqrt(8x^2-647))/2