What is the solution for (x^4+1)dy+x(1+4y^2)dx=0 ?

The solution for (x^4+1)dy+x(1+4y^2)dx=0 is y=(tan(-arctan(x^2)+c_{1}))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x^4+1)dy+x(1+4y^2)dx=0

(x^{4} + 1) d y + x (1 + 4 y^{2}) d x = 0

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

Solution steps

(x^{4} + 1) d y + x (1 + 4 y^{2}) d x = 0

Solve separable ODE:

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

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

n = 0 \sum \infty \frac{n ^{2}}{5 ^{n}}

\int e^{- 2 s} d s

1

\frac{d}{d x} (- \frac{18}{x ^{4}})

x \to 13 lim (4)

What is the solution for (x^4+1)dy+x(1+4y^2)dx=0 ?
The solution for (x^4+1)dy+x(1+4y^2)dx=0 is y=(tan(-arctan(x^2)+c_{1}))/2