en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+8y=1,y(0)=1

Solution

y^{^{'}} + 8 y = 1, y (0) = 1

Solution

y = \frac{1}{8} + \frac{7}{8 e ^{8 t}}

Solution steps

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

Solve linear ODE:

y = \frac{1}{8} + \frac{7}{8 e ^{8 t}}

y = \frac{1}{8} + \frac{7}{8 e ^{8 t}}

Popular Examples

laplacetransform f(t)=e^{2t+1}laplace transform

f (t) = e^{2 t + 1}

derivative of (-3x+8/(2sqrt(4-x)))

\frac{d}{d x} (\frac{- 3 x + 8}{2 4 - x})

derivative of (x-5/(4x^2e^x))

\frac{d}{d x} (\frac{x - 5}{4 x ^{2} e ^{x}})

limit as x approaches 0 of x^4cos(1/x)

x \to 0 lim (x^{4} cos (\frac{1}{x}))

(e^{- 2 t})^{^{'}}

Frequently Asked Questions (FAQ)

What is the solution for y^'+8y=1,y(0)=1 ?
The solution for y^'+8y=1,y(0)=1 is y= 1/8+7/(8e^{8t)}