What is the solution for (1+x^2)y^'+4xy=(1+x^2)^{-2} ?

The solution for (1+x^2)y^'+4xy=(1+x^2)^{-2} is y=(arctan(x))/(1+2x^2+x^4)+(c_{1})/(1+2x^2+x^4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(1+x^2)y^'+4xy=(1+x^2)^{-2}

(1 + x^{2}) y^{^{'}} + 4 x y = (1 + x^{2})^{- 2}

y = \frac{arctan ( x )}{1 + 2 x ^{2} + x ^{4}} + \frac{c _{1}}{1 + 2 x ^{2} + x ^{4}}

Solution steps

(1 + x^{2}) y^{'} (x) + 4 x y = (1 + x^{2})^{- 2}

Solve linear ODE:

y = \frac{arctan ( x )}{1 + 2 x ^{2} + x ^{4}} + \frac{c _{1}}{1 + 2 x ^{2} + x ^{4}}

y = \frac{arctan ( x )}{1 + 2 x ^{2} + x ^{4}} + \frac{c _{1}}{1 + 2 x ^{2} + x ^{4}}

\frac{d}{d x} (e^{2 x^{2} - 1})

\int_{0}^{8} \frac{1}{64 + x ^{2}} d x

\frac{d}{d x} (\frac{x sin ( 2 x )}{cos ( 3 x )})

8 t \frac{d y}{d t} + y = t^{5}

What is the solution for (1+x^2)y^'+4xy=(1+x^2)^{-2} ?
The solution for (1+x^2)y^'+4xy=(1+x^2)^{-2} is y=(arctan(x))/(1+2x^2+x^4)+(c_{1})/(1+2x^2+x^4)