en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+2(x+1)y^2=0

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

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

\frac{d y}{d x} - \frac{y}{x + 1} = 0

a y^{^{'}} + y = c

(2y+2)dy-(4x^3+6x)dx=0

(2 y + 2) d y - (4 x^{3} + 6 x) d x = 0

(dx)/(dy)=sin(x)(cos(2y)-cos^2(y))

\frac{d x}{d y} = sin (x) (cos (2 y) - cos^{2} (y))

d/(dθ)(f(θ))f(θ)=sin(θ)+csc(θ)

\frac{d}{d θ} (f (θ)) f (θ) = sin (θ) + csc (θ)

Frequently Asked Questions (FAQ)

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