What is the solution for 1/8 x^{''}+x^'+2x=0,x(0)=-1,x^'(0)=14 ?

The solution for 1/8 x^{''}+x^'+2x=0,x(0)=-1,x^'(0)=14 is x=-e^{-4t}+10e^{-4t}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1/8 x^{''}+x^'+2x=0,x(0)=-1,x^'(0)=14

\frac{1}{8} x^{^{''}} + x^{^{'}} + 2 x = 0, x (0) = - 1, x^{^{'}} (0) = 14

x = - e^{- 4 t} + 10 e^{- 4 t} t

Solution steps

\frac{1}{8} x^{''} (t) + x^{'} (t) + 2 x = 0

Solve linear ODE:

x = - e^{- 4 t} + 10 e^{- 4 t} t

x = - e^{- 4 t} + 10 e^{- 4 t} t

4 y^{^{''}} + 24 y^{^{'}} + 37 y = 0

x^{^{''}} - 6 x^{^{'}} + 9 x = 0

y^{^{''}} + 144 y = 0

\frac{5}{8} x^{^{''}} + 40 x = 0

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = 15.451

What is the solution for 1/8 x^{''}+x^'+2x=0,x(0)=-1,x^'(0)=14 ?
The solution for 1/8 x^{''}+x^'+2x=0,x(0)=-1,x^'(0)=14 is x=-e^{-4t}+10e^{-4t}t