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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d}{d x} (\frac{x - 8}{x + 8})

x \to 4 lim ((x - 4)^{5} - 5)

x \to 3 lim (- 1)

\frac{d}{d x} (\frac{3}{4} x^{\frac{4}{3}} - \frac{3}{8} x^{\frac{2}{3}} + 5)

f (x) = 2 - x

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