What is the solution for 4y+9y^{''}=-12y^' ?

The solution for 4y+9y^{''}=-12y^' is y=c_{1}e^{-(2t)/3}+c_{2}e^{-(2t)/3}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4y+9y^{''}=-12y^'

4 y + 9 y^{^{''}} = - 12 y^{^{'}}

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

Solution steps

4 y + 9 y^{''} (t) = - 12 y^{'} (t)

Solve linear ODE:

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

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

9 y^{^{''}} + 12 y^{^{'}} + 4 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 3, y^{^{'}} (0) = 0

\frac{d ^{2}}{d x ^{2}} ((u (x)) e^{2 x}) = 0

y^{^{''}} - 2 y^{^{'}} - 3 y = 0, y^{^{'}} = 13, y (0) = 3

2 y^{^{''}} - 4 y^{^{'}} + 4 y = 0

What is the solution for 4y+9y^{''}=-12y^' ?
The solution for 4y+9y^{''}=-12y^' is y=c_{1}e^{-(2t)/3}+c_{2}e^{-(2t)/3}t