What is the solution for sqrt(1-8x^2)y^'=x ?

The solution for sqrt(1-8x^2)y^'=x is y=-1/8 sqrt(1-8x^2)+c_{1}

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sqrt(1-8x^2)y^'=x

1 - 8 x^{2} y^{^{'}} = x

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

Solution steps

1 - 8 x^{2} y^{'} (x) = x

Isolate

y^{'} (x) : y^{'} (x) = \frac{x}{1 - 8 x ^{2}}

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

then

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

y = \int \frac{x}{1 - 8 x ^{2}} d x

\int \frac{x}{1 - 8 x ^{2}} d x = - \frac{1}{8} 1 - 8 x^{2} + c_{1}

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

\int sin (x) + c d x

\frac{1}{( s + 1 ) ( ( s + 2 ) ^{2} + 2 )}

\frac{d}{d x} (5 x - sin^{2} (4 x))

\frac{d y}{d x} = 1 + x + y + x y

\frac{d}{d x} ((- 4 x^{2} + 25)^{2})

What is the solution for sqrt(1-8x^2)y^'=x ?
The solution for sqrt(1-8x^2)y^'=x is y=-1/8 sqrt(1-8x^2)+c_{1}