What is the solution for y^{''}+5y^'+2=0 ?

The solution for y^{''}+5y^'+2=0 is y=c_{1}+c_{2}e^{-5t}-(2t)/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+5y^'+2=0

y^{^{''}} + 5 y^{^{'}} + 2 = 0

y = c_{1} + c_{2} e^{- 5 t} - \frac{2 t}{5}

Solution steps

y^{''} (t) + 5 y^{'} (t) + 2 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 5 t} - \frac{2 t}{5}

y = c_{1} + c_{2} e^{- 5 t} - \frac{2 t}{5}

y^{^{''}} + 10 y = 10 e^{11 x}

y^{^{''}} + 36 y = 33 cos (5 t), y (0) = 0, y^{^{'}} (0) = 0

D (q) = 2300 - 42 q

y^{^{''}} + 2 y^{^{'}} = x

y^{^{''}} - 2 y^{^{'}} + y = 4 x^{2} - 4 + x - 1 e x

What is the solution for y^{''}+5y^'+2=0 ?
The solution for y^{''}+5y^'+2=0 is y=c_{1}+c_{2}e^{-5t}-(2t)/5