What is the solution for sqrt(1-9x^2)y^'=x ?

The solution for sqrt(1-9x^2)y^'=x is y=-1/9 sqrt(1-9x^2)+c_{1}

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sqrt(1-9x^2)y^'=x

1 - 9 x^{2} y^{^{'}} = x

y = - \frac{1}{9} 1 - 9 x^{2} + c_{1}

Solution steps

1 - 9 x^{2} y^{'} (x) = x

Isolate

y^{'} (x) : y^{'} (x) = \frac{x}{1 - 9 x ^{2}}

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

then

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

y = \int \frac{x}{1 - 9 x ^{2}} d x

\int \frac{x}{1 - 9 x ^{2}} d x = - \frac{1}{9} 1 - 9 x^{2} + c_{1}

y = - \frac{1}{9} 1 - 9 x^{2} + c_{1}

\frac{d}{d x} (\frac{∣ x ∣}{2 - x ^{2}})

\frac{d y}{d t} + 5 y = 5

3 (1 + t^{2}) (\frac{d y}{d t}) = 2 t y (y^{3} - 1)

y^{^{''}} + 81 y = - 81 sec^{2} (9 x)

\int \frac{7 ( 4 - x ^{2} )}{( x - 4 ) ( x - 2 ) ( x + 3 )} d x

What is the solution for sqrt(1-9x^2)y^'=x ?
The solution for sqrt(1-9x^2)y^'=x is y=-1/9 sqrt(1-9x^2)+c_{1}