What is the solution for sqrt(1-6x^2)y^'=x ?

The solution for sqrt(1-6x^2)y^'=x is y=-1/6 sqrt(1-6x^2)+c_{1}

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sqrt(1-6x^2)y^'=x

1 - 6 x^{2} y^{^{'}} = x

y = - \frac{1}{6} 1 - 6 x^{2} + c_{1}

Solution steps

1 - 6 x^{2} y^{'} (x) = x

Isolate

y^{'} (x) : y^{'} (x) = \frac{x}{1 - 6 x ^{2}}

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

then

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

y = \int \frac{x}{1 - 6 x ^{2}} d x

\int \frac{x}{1 - 6 x ^{2}} d x = - \frac{1}{6} 1 - 6 x^{2} + c_{1}

y = - \frac{1}{6} 1 - 6 x^{2} + c_{1}

x = - 2, y = x + 2, y = x^{2}

\int 8 y d y

m = 2 \sum \infty \frac{5}{6 ^{m}}

y^{^{'}} = 8 x^{6} y - y, y (1) = - 4

\frac{3 x}{x}

What is the solution for sqrt(1-6x^2)y^'=x ?
The solution for sqrt(1-6x^2)y^'=x is y=-1/6 sqrt(1-6x^2)+c_{1}