What is the solution for x^{''}+12x^'+27x=0,x(0)=1,x^'(0)=0 ?

The solution for x^{''}+12x^'+27x=0,x(0)=1,x^'(0)=0 is x=(1+1/2)e^{-3t}-1/2 e^{-9t}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{''}+12x^'+27x=0,x(0)=1,x^'(0)=0

x^{^{''}} + 12 x^{^{'}} + 27 x = 0, x (0) = 1, x^{^{'}} (0) = 0

x = (1 + \frac{1}{2}) e^{- 3 t} - \frac{1}{2} e^{- 9 t}

Solution steps

x^{''} (t) + 12 x^{'} (t) + 27 x = 0

Solve linear ODE:

x = (1 + \frac{1}{2}) e^{- 3 t} - \frac{1}{2} e^{- 9 t}

x = (1 + \frac{1}{2}) e^{- 3 t} - \frac{1}{2} e^{- 9 t}

(4 D^{4} - 45 D^{2} - 70 D - 24) y = 0

y^{^{''}} - 6 y^{^{'}} + 113 y = 0

10 \frac{d ^{2} y}{d x ^{2}} + 20 \frac{d y}{d x} + 10 y = 0

y^{^{''}} + 4 y^{^{'}} + 3 y = 0, y (0) = 6, y^{^{'}} (0) = 2

x^{^{''}} + 5 x^{^{'}} = 0

What is the solution for x^{''}+12x^'+27x=0,x(0)=1,x^'(0)=0 ?
The solution for x^{''}+12x^'+27x=0,x(0)=1,x^'(0)=0 is x=(1+1/2)e^{-3t}-1/2 e^{-9t}