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

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

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

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

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

Solution steps

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

Substitute

\frac{d y}{d x}

with

y^{'} (x)

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

Isolate

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

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

then

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

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

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

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

f (x) = (5 x^{2} - 7 x) e^{x}

f (x) = ln^{3} (x)

\frac{d}{d x} (2 - \frac{2}{3} x)

\int \frac{1}{5} x^{5} d x

h \to 0 lim (\frac{1 - cos ( 3 h )}{h})

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