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

The solution for (x^2+1)y^'+y^2+1=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 >

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

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

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

Solution steps

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

Solve separable ODE:

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

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

y^{^{'}} + \frac{3}{t} y = 1 - 2 t, y (3) = 0

\frac{d}{d x} (sin (2 x + y))

\int_{0}^{2 π} sin (x) cos (x) d x

y^{^{''}} - 3 y^{^{'}} = 4 t

\int 2 e^{7} d x

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