en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y^'+8y=1

Solution

2 y^{^{'}} + 8 y = 1

Solution

y = - \frac{e ^{- 4 t - c_{1}}}{8} + \frac{1}{8}

Solution steps

2 y^{'} (t) + 8 y = 1

Solve separable ODE:

y = - \frac{e ^{- 4 t - c_{1}}}{8} + \frac{1}{8}

y = - \frac{e ^{- 4 t - c_{1}}}{8} + \frac{1}{8}

Popular Examples

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

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

(dx)/(dt)=(sin(t))/(sin(x))

\frac{d x}{d t} = \frac{sin ( t )}{sin ( x )}

(dy)/(dx)=0.1y(1-0.5y),y(0)=5000

\frac{d y}{d x} = 0.1 y (1 - 0.5 y), y (0) = 5000

(dy)/(dt)+0.4ty=4t

\frac{d y}{d t} + 0.4 t y = 4 t

y^'=(1+4x)sqrt(y)

y^{^{'}} = (1 + 4 x) y

Frequently Asked Questions (FAQ)

What is the solution for 2y^'+8y=1 ?
The solution for 2y^'+8y=1 is y=-(e^{-4t-c_{1}})/8+1/8