en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+1/2 y=8,y(0)=2

Solution

y^{^{'}} + \frac{1}{2} y = 8, y (0) = 2

Solution

y = - 14 e^{- \frac{t}{2}} + 16

Solution steps

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

Solve separable ODE:

y = - 14 e^{- \frac{t}{2}} + 16

y = - 14 e^{- \frac{t}{2}} + 16

Popular Examples

(\partial)/(\partial x)(((xv^2))/((x^2+v^2)))

\frac{\partial}{\partial x} (\frac{( x v ^{2} )}{( x ^{2} + v ^{2} )})

derivative of ((x+2)/x)(7-2x^3)derivative of

(\frac{x + 2}{x}) (7 - 2 x^{3})

integral of 5/((5x+4)ln(5x+4))

\int \frac{5}{( 5 x + 4 ) ln ( 5 x + 4 )} d x

integral of 6x^2-2x+1

\int 6 x^{2} - 2 x + 1 d x

sum from n=1 to infinity of n/(n^3+3)

n = 1 \sum \infty \frac{n}{n ^{3} + 3}

Frequently Asked Questions (FAQ)

What is the solution for y^'+1/2 y=8,y(0)=2 ?
The solution for y^'+1/2 y=8,y(0)=2 is y=-14e^{-t/2}+16