What is the solution for sqrt(1-3x^2)y^'=x ?

The solution for sqrt(1-3x^2)y^'=x is y=-1/3 sqrt(1-3x^2)+c_{1}

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sqrt(1-3x^2)y^'=x

1 - 3 x^{2} y^{^{'}} = x

y = - \frac{1}{3} 1 - 3 x^{2} + c_{1}

Solution steps

1 - 3 x^{2} y^{'} (x) = x

Isolate

y^{'} (x) : y^{'} (x) = \frac{x}{1 - 3 x ^{2}}

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

then

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

y = \int \frac{x}{1 - 3 x ^{2}} d x

\int \frac{x}{1 - 3 x ^{2}} d x = - \frac{1}{3} 1 - 3 x^{2} + c_{1}

y = - \frac{1}{3} 1 - 3 x^{2} + c_{1}

\frac{\partial}{\partial y} (e^{x^{2} - y^{2} + 4 y})

x \to 1 lim (\frac{x ^{3} - x}{1 - x})

\frac{d}{d x} ((x^{2} + 2 y^{2} + 3 z^{2}) e^{- (x^{2} + y^{2} + z^{2})})

6 cos (sin (x^{2}))

\frac{d}{d x} (ln (1 + 0))

What is the solution for sqrt(1-3x^2)y^'=x ?
The solution for sqrt(1-3x^2)y^'=x is y=-1/3 sqrt(1-3x^2)+c_{1}