en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^'+x=-2

Solution

x^{^{'}} + x = - 2

Solution

x = e^{- t - c_{1}} - 2

Solution steps

x^{'} (t) + x = - 2

Solve separable ODE:

x = e^{- t - c_{1}} - 2

x = e^{- t - c_{1}} - 2

Graph

Popular Examples

(dy)/(dx)=y(2x+1),y(0)=2

\frac{d y}{d x} = y (2 x + 1), y (0) = 2

(dy)/(dx)=(y+1)x^3

\frac{d y}{d x} = (y + 1) x^{3}

(dy)/(dx)=4(y^2+1),x(pi/4)=1

\frac{d y}{d x} = 4 (y^{2} + 1), x (\frac{π}{4}) = 1

2x(dy)/(dx)=(2x+1)y

2 x \frac{d y}{d x} = (2 x + 1) y

(dx)/(dt)=-(sqrt(x))/(50pi)

\frac{d x}{d t} = - \frac{x}{50 π}

Frequently Asked Questions (FAQ)

What is the solution for x^'+x=-2 ?
The solution for x^'+x=-2 is x=e^{-t-c_{1}}-2