What is the solution for 4x(dy)/(dx)=sqrt(x^2-9) ?

The solution for 4x(dy)/(dx)=sqrt(x^2-9) is y=-3/4 arcsec(1/3 x)+1/4 sqrt(x^2-9)+c_{1}

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4x(dy)/(dx)=sqrt(x^2-9)

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

y = - \frac{3}{4} arcsec (\frac{1}{3} x) + \frac{1}{4} x^{2} - 9 + c_{1}

Solution steps

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

Substitute

\frac{d y}{d x}

with

y^{'} (x)

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

Isolate

y^{'} (x) : y^{'} (x) = \frac{x ^{2} - 9}{4 x}

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

then

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

y = \int \frac{x ^{2} - 9}{4 x} d x

\int \frac{x ^{2} - 9}{4 x} d x = - \frac{3}{4} arcsec (\frac{1}{3} x) + \frac{1}{4} x^{2} - 9 + c_{1}

y = - \frac{3}{4} arcsec (\frac{1}{3} x) + \frac{1}{4} x^{2} - 9 + c_{1}

\int_{- 1}^{1} ln (ln (x + 3)) d x

\frac{d x}{d y} = - \frac{( 4 y ^{2} + 6 x y )}{3 y ^{2} + 2 x}

\int_{0}^{2} (6 x^{2} - 4 x + 5) d x

y = e^{\frac{1}{2} x}

x \to a lim (\frac{f ( x )}{p ( x )})

What is the solution for 4x(dy)/(dx)=sqrt(x^2-9) ?
The solution for 4x(dy)/(dx)=sqrt(x^2-9) is y=-3/4 arcsec(1/3 x)+1/4 sqrt(x^2-9)+c_{1}