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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dy)/(dx)= 1/(sqrt(4-x^2))

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

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

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}

ln (x (1 - x)^{8})

\frac{d ^{2}}{d x ^{2}} (e^{7 x^{2}})

\frac{d ^{2}}{d r ( T ) d T} (r (T) (T)^{2} \frac{d ^{2}}{d r ( T ) d T} (T))

\frac{d}{d x} (cos (3))

f (t) = - 4 sin (t)

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