en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

xy^'+y+x=0

Solution

x y^{^{'}} + y + x = 0

Solution

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

Solution steps

x y^{'} (x) + y + x = 0

Solve linear ODE:

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

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

Popular Examples

area (x-3)x(x+3),3-x,0,2 area

(x - 3) x (x + 3), 3 - x, 0, 2

slope of y=-(x+1/2)^2+7/2 slope

y = - (x + \frac{1}{2})^{2} + \frac{7}{2}

integral of 3cos((npix)/3)

\int 3 cos (\frac{nπ x}{3}) d x

y^'=(x^2)/(y(2+x^3))

y^{^{'}} = \frac{x ^{2}}{y ( 2 + x ^{3} )}

(\partial)/(\partial x)(3e^xcos(y))

\frac{\partial}{\partial x} (3 e^{x} cos (y))

Frequently Asked Questions (FAQ)

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