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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

\frac{28 s}{( s ^{2} + 11 )}

\frac{d y}{d x} + 4 tan (4 x) (y) = 3 cos (4 x)

\int_{- 2}^{3} (x^{2} + 3 x - 13) d x

x \to 0 lim (arctan (ln (x)))

(cos (x) sin (x))^{^{'}}

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