What is the solution for y^'sqrt(1-4x^2)=x ?

The solution for y^'sqrt(1-4x^2)=x is y=-1/4 sqrt(1-4x^2)+c_{1}

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y^'sqrt(1-4x^2)=x

y^{^{'}} 1 - 4 x^{2} = x

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

Solution steps

y^{'} (x) 1 - 4 x^{2} = x

Isolate

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

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int \frac{x}{1 - 4 x ^{2}} d x

\int \frac{x}{1 - 4 x ^{2}} d x = - \frac{1}{4} 1 - 4 x^{2} + c_{1}

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

y^{4} y^{^{'}} = - 3 x^{2} y^{5} + x^{2}

f (t) = (8 t - 7)^{4} + (8 t + 7)^{4}

\frac{\partial}{\partial x} (\frac{( 7 x + 7 y )}{z})

x \to 2 lim (\frac{5}{x ^{4} - 16})

t \to 0 lim ((\frac{3}{t 1 + t}) - \frac{3}{t})

What is the solution for y^'sqrt(1-4x^2)=x ?
The solution for y^'sqrt(1-4x^2)=x is y=-1/4 sqrt(1-4x^2)+c_{1}