What is the solution for ((1+x^2)dy)/(dx)-(1+y^2)=0 ?

The solution for ((1+x^2)dy)/(dx)-(1+y^2)=0 is y=tan(arctan(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)dy)/(dx)-(1+y^2)=0

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

y = tan (arctan (x) + c_{1})

Solution steps

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

Solve separable ODE:

y = tan (arctan (x) + c_{1})

y = tan (arctan (x) + c_{1})

\int_{0}^{N} e^{- s t} e^{- 5 t} d t

\frac{d}{d x} (\frac{- sin ( x )}{sec ( x ) tan ( x )})

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

\frac{\partial}{\partial x} (4 x^{3} y^{\frac{1}{2}})

\frac{d ^{2}}{d x ^{2}} ((x^{2} + 5)^{6})

What is the solution for ((1+x^2)dy)/(dx)-(1+y^2)=0 ?
The solution for ((1+x^2)dy)/(dx)-(1+y^2)=0 is y=tan(arctan(x)+c_{1})