What is the solution for (dy)/(dx)= 1/(sqrt(4-x^2)),y(0)=pi ?

The solution for (dy)/(dx)= 1/(sqrt(4-x^2)),y(0)=pi is y=arcsin(x/2)+pi

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(dy)/(dx)= 1/(sqrt(4-x^2)),y(0)=pi

\frac{d y}{d x} = \frac{1}{4 - x ^{2}}, y (0) = π

y = arcsin (\frac{x}{2}) + π

Solution steps

\frac{d y}{d x} = \frac{1}{4 - x ^{2}}

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = \frac{1}{4 - x ^{2}}

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

then

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

y = \int \frac{1}{4 - x ^{2}} d x

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

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

Apply initial conditions:

y = arcsin (\frac{x}{2}) + π

y = arcsin (\frac{x}{2}) + π

\int 5 tan^{3} (x) sec (x) d x

y^{^{'}} = (1 + t) \cdot y, y (0) = 1

\int \frac{1}{sec ( x )} d x

\frac{d}{d x} ((x^{2} + 1) (x^{3} + x^{2} + 1))

\frac{\partial}{\partial x} (e^{6 x})

What is the solution for (dy)/(dx)= 1/(sqrt(4-x^2)),y(0)=pi ?
The solution for (dy)/(dx)= 1/(sqrt(4-x^2)),y(0)=pi is y=arcsin(x/2)+pi