What is the solution for y^{''}+4y^'+3y=15 ?

The solution for y^{''}+4y^'+3y=15 is y=c_{1}e^{-t}+c_{2}e^{-3t}+5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'+3y=15

y^{^{''}} + 4 y^{^{'}} + 3 y = 15

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + 5

Solution steps

y^{''} (t) + 4 y^{'} (t) + 3 y = 15

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + 5

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + 5

(D^{2} + 1) y = sec^{3} (x)

y^{^{''}} - 4 y^{^{'}} + 3 y = x, y (0) = 6, y^{^{'}} (0) = 10

y^{^{''}} + 6 y^{^{'}} + 25 y = x^{2}

y^{^{''}} + 2 y^{^{'}} - 8 y = - 650 sin (3 t)

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

What is the solution for y^{''}+4y^'+3y=15 ?
The solution for y^{''}+4y^'+3y=15 is y=c_{1}e^{-t}+c_{2}e^{-3t}+5