What is the solution for (dy)/(dx)= 2/(sqrt(16-x^2)) ?

The solution for (dy)/(dx)= 2/(sqrt(16-x^2)) is y=2arcsin(1/4 x)+c_{1}

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(dy)/(dx)= 2/(sqrt(16-x^2))

\frac{d y}{d x} = \frac{2}{16 - x ^{2}}

y = 2 arcsin (\frac{1}{4} x) + c_{1}

Solution steps

\frac{d y}{d x} = \frac{2}{16 - x ^{2}}

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = \frac{2}{16 - x ^{2}}

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

then

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

y = \int \frac{2}{16 - x ^{2}} d x

\int \frac{2}{16 - x ^{2}} d x = 2 arcsin (\frac{1}{4} x) + c_{1}

y = 2 arcsin (\frac{1}{4} x) + c_{1}

\int (e^{1 - x}) d x

\frac{d}{d x} (\frac{1}{10} x - \frac{1}{8})

n = 0 \sum \infty \frac{1}{n ^{- 1}}

\int (x^{- \frac{1}{3}}) d x

\int sin (b x) d x

What is the solution for (dy)/(dx)= 2/(sqrt(16-x^2)) ?
The solution for (dy)/(dx)= 2/(sqrt(16-x^2)) is y=2arcsin(1/4 x)+c_{1}