What is the solution for (1-x^2)y^'-y^2=0 ?

The solution for (1-x^2)y^'-y^2=0 is y=-2/(ln(1+x)-ln(-1+x)+c_{1)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(1-x^2)y^'-y^2=0

(1 - x^{2}) y^{^{'}} - y^{2} = 0

y = - \frac{2}{ln ( 1 + x ) - ln ( - 1 + x ) + c _{1}}

Solution steps

(1 - x^{2}) y^{'} (x) - y^{2} = 0

Solve separable ODE:

y = - \frac{2}{ln ( 1 + x ) - ln ( - 1 + x ) + c _{1}}

y = - \frac{2}{ln ( 1 + x ) - ln ( - 1 + x ) + c _{1}}

\frac{d x}{d t} = 125 - \frac{x}{100}, x (0) = 4000

\frac{x d x}{1 + x ^{2}} + \frac{y d y}{1 + y ^{2}} = 0

\frac{d x}{d t} = t \cdot x^{\frac{1}{2}}

\frac{d y}{d x} = \frac{cos ^{2} ( y )}{4 x - 3}

\frac{d p}{d t} = 5 p (20 - ln (p))

What is the solution for (1-x^2)y^'-y^2=0 ?
The solution for (1-x^2)y^'-y^2=0 is y=-2/(ln(1+x)-ln(-1+x)+c_{1)}