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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2x-x^{-2}y)dx+x^{-1}dy=0

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

y = - x^{3} + c_{1} x

Solution steps

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

Solve linear ODE:

y = - x^{3} + c_{1} x

y = - x^{3} + c_{1} x

\int \frac{2 x ^{3} + x ^{2} + 4}{( x ^{2} + 4 ) ^{2}} d x

\frac{\partial}{\partial x} (x e^{y} - ln (x))

- \frac{1}{x ^{2}}

\frac{d}{d x} (\frac{4 x ^{2} - 3}{x - 1})

\frac{d}{d t} ((sin (t) + cos (t))^{2})

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