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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

(dx)/(dy)=(1+x^2)/(1-y^2)

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

x = tan (\frac{ln ( y + 1 ) - ln ( y - 1 )}{2} + c_{1})

Solution steps

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

Solve separable ODE:

x = tan (\frac{ln ( y + 1 ) - ln ( y - 1 )}{2} + c_{1})

x = tan (\frac{ln ( y + 1 ) - ln ( y - 1 )}{2} + c_{1})

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