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

The solution for (x^2+1)dx+1/y dy=0 is y=e^{-(x^3)/3-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)dx+1/y dy=0

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

y = e^{- \frac{x ^{3}}{3} - x + c_{1}}

Solution steps

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

Solve separable ODE:

y = e^{- \frac{x ^{3}}{3} - x + c_{1}}

y = e^{- \frac{x ^{3}}{3} - x + c_{1}}

10 x^{2} y + 2 x^{3} y^{^{'}} - 16 y = 0

1 = y^{^{'}} e^{y} x

\frac{d y}{d x} = \frac{y x + 3 x - y - 3}{x y - 2 x + 4 y - 8}

\frac{d y}{d x} = \frac{9 ( 1 + x )}{x y}

x^{^{'}} = \frac{2 x}{t + 1}

What is the solution for (x^2+1)dx+1/y dy=0 ?
The solution for (x^2+1)dx+1/y dy=0 is y=e^{-(x^3)/3-x+c_{1}}