en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+3y=8,y(0)=0

Solution

y^{^{'}} + 3 y = 8, y (0) = 0

Solution

y = - \frac{8 e ^{- 3 t}}{3} + \frac{8}{3}

Solution steps

y^{'} (t) + 3 y = 8

Solve separable ODE:

y = - \frac{8 e ^{- 3 t}}{3} + \frac{8}{3}

y = - \frac{8 e ^{- 3 t}}{3} + \frac{8}{3}

Popular Examples

derivative of 1/(-3+(2x+1)^2)derivative of

\frac{1}{- 3 + ( 2 x + 1 ) ^{2}}

integral of (5x+8)/(x^2+4x+4)

\int \frac{5 x + 8}{x ^{2} + 4 x + 4} d x

limit as x approaches infinity of (1+1/(2x))^{3x}

x \to \infty lim ((1 + \frac{1}{2 x})^{3 x})

integral from 2 to 6 of t^3ln(4t)

\int_{2}^{6} t^{3} ln (4 t) d t

y^'=5ysin(4x),y(pi/8)=3

y^{^{'}} = 5 y sin (4 x), y (\frac{π}{8}) = 3

Frequently Asked Questions (FAQ)

What is the solution for y^'+3y=8,y(0)=0 ?
The solution for y^'+3y=8,y(0)=0 is y=-(8e^{-3t})/3+8/3