en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+8y=4

Solution

y^{^{'}} + 8 y = 4

Solution

y = - \frac{e ^{- 8 t - c_{1}}}{2} + \frac{1}{2}

Solution steps

y^{'} (t) + 8 y = 4

Solve separable ODE:

y = - \frac{e ^{- 8 t - c_{1}}}{2} + \frac{1}{2}

y = - \frac{e ^{- 8 t - c_{1}}}{2} + \frac{1}{2}

Graph

Popular Examples

taylor e^{4x},0 taylor

e^{4 x}, 0

integral of 2x^3-5x^2+3x+1

\int 2 x^{3} - 5 x^{2} + 3 x + 1 d x

(\partial)/(\partial x)(ln(e^x+xy^3))

\frac{\partial}{\partial x} (ln (e^{x} + x y^{3}))

integral from-infinity to 0 of 8^r

\int_{- \infty}^{0} 8^{r} d r

derivative of 3sqrt(x^2+y^2)

\frac{d}{d x} (3 x^{2} + y^{2})

Frequently Asked Questions (FAQ)

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