en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

(\partial)/(\partial y)(2x^3y)

\frac{\partial}{\partial y} (2 x^{3} y)

derivative of x^2y+4y

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

area x=0,y=0,y=sqrt((-x^2+8)/9)area

x = 0, y = 0, y = \frac{- x ^{2} + 8}{9}

limit as θ approaches 0 of θ/(sin(θ))

θ \to 0 lim (\frac{θ}{sin ( θ )})

(dy)/(dx)+y/x =x^2y^2

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

Frequently Asked Questions (FAQ)

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