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

The solution for 2y^{'''}+y^{''}=0 is y=c_{1}e^{-t/2}+c_{2}+c_{3}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

7 y^{^{''}} + y^{^{'}} - y = 0

6 y^{^{''}} - 7 y^{^{'}} - 20 y = 0

y^{^{''''}} + 49 y^{^{''}} = 0

y^{^{''''}} - 8 y^{^{'''}} + 23 y^{^{''}} - 28 y^{^{'}} + 12 y = 0

What is the solution for 2y^{'''}+y^{''}=0 ?
The solution for 2y^{'''}+y^{''}=0 is y=c_{1}e^{-t/2}+c_{2}+c_{3}t