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

The solution for y^'=(4x)/(1+4y),y(2)=0 is y=(-1+sqrt(16x^2-63))/4

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

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

y = \frac{- 1 + 16 x ^{2} - 63}{4}

Solution steps

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

Solve separable ODE:

y = \frac{- 1 + 16 x ^{2} - 63}{4}

y = \frac{- 1 + 16 x ^{2} - 63}{4}

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

y^{^{'}} = 0.1 \cdot y \cdot (1 - \frac{y}{80})

5 \frac{d v}{d t} = - 2 v^{2}

y^{^{'}} = 3 x 2 y

\frac{d}{d x} (f (x)^{- 1}) f (x) = \frac{1}{5} x + 5, x = 6

What is the solution for y^'=(4x)/(1+4y),y(2)=0 ?
The solution for y^'=(4x)/(1+4y),y(2)=0 is y=(-1+sqrt(16x^2-63))/4