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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2xy+y)dx+(x^2-x)dy=0

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

y = \frac{c _{1} x}{x ^{3} - 3 x ^{2} + 3 x - 1}

Solution steps

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

Solve separable ODE:

y = \frac{c _{1} x}{x ^{3} - 3 x ^{2} + 3 x - 1}

y = \frac{c _{1} x}{x ^{3} - 3 x ^{2} + 3 x - 1}

\frac{d y}{d x} - \frac{y}{x} = x^{2}

x \to 0 + lim (12 cos (12 x))

x y^{^{'}} + 2 y = sin (x), y (π) = \frac{1}{π}

\int \frac{1}{r ^{2} - x ^{2}} d x

\frac{d y}{d x}, x y = 6

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