What is the solution for t=(sqrt(1-y^2))/(y^') ?

The solution for t=(sqrt(1-y^2))/(y^') is y=sin(ln(t)+c_{1})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

t=(sqrt(1-y^2))/(y^')

t = \frac{1 - y ^{2}}{y ^{^{'}}}

y = sin (ln (t) + c_{1})

Solution steps

t = \frac{1 - y ^{2}}{y ^{'} ( t )}

Solve separable ODE:

y = sin (ln (t) + c_{1})

y = sin (ln (t) + c_{1})

y = 8

f (x) = \frac{1}{x ^{4}}, at x = - 3

n = 1 \sum \infty (\frac{5}{6})^{n} n

\int 2 \frac{sin ( 2 x )}{sin ( x )} d x

\int (4 x - 3) d x

What is the solution for t=(sqrt(1-y^2))/(y^') ?
The solution for t=(sqrt(1-y^2))/(y^') is y=sin(ln(t)+c_{1})