What is the solution for (dy}{dx}=\frac{sqrt(x^2-9))/x ?

The solution for (dy}{dx}=\frac{sqrt(x^2-9))/x is y=-3arcsec(1/3 x)+sqrt(x^2-9)+c_{1}

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(dy}{dx}=\frac{sqrt(x^2-9))/x

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

y = - 3 arcsec (\frac{1}{3} x) + x^{2} - 9 + c_{1}

Solution steps

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

Substitute

\frac{d y}{d x}

with

y^{'} (x)

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

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

then

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

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

\int \frac{x ^{2} - 9}{x} d x = - 3 arcsec (\frac{1}{3} x) + x^{2} - 9 + c_{1}

y = - 3 arcsec (\frac{1}{3} x) + x^{2} - 9 + c_{1}

\int_{0}^{3} x e^{- 4 x} d x

y = 5 e^{x} + 7 ln (x)

\frac{\partial}{\partial x} (1 + x^{3} + x^{4})

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

y^{^{''}} - y^{^{'}} - 6 y = 2 e^{t} - t^{2} - 2 t + 3

What is the solution for (dy}{dx}=\frac{sqrt(x^2-9))/x ?
The solution for (dy}{dx}=\frac{sqrt(x^2-9))/x is y=-3arcsec(1/3 x)+sqrt(x^2-9)+c_{1}